Appendix C SQL Lab with Solution
Download the following script: OrdersAndData.sql.
Part 1 – DDL
Figure C.1. ERD for Orders and Data.- Use the script OrdersAndData.sql that creates the tables and adds the data for the Orders and Data ERD in Figure C.1.
- Create a database called Orders. Modify the script to integrate the PK and referential integrity. Show the CREATE TABLE statements with the modifications including the constraints given in step 3.
- Add the following constraints:
- tblCustomers table: Country – default to Canada
- tblOrderDetails: Quantity – > 0
- tblShippers: CompanyName must be unique.
- tblOrders: ShippedDate must be greater than order date.
Go
Use Orders
Go Use Orders
Go
CREATE TABLE [dbo].[tblCustomers]
[CustomerID] nvarchar(5) NOT NULL,
[CompanyName] nvarchar(40) NOT NULL,
[ContactName] nvarchar(30) NULL,
[ContactTitle] nvarchar(30) NULL,
[Address] nvarchar(60) NULL,
[City] nvarchar(15) NULL,
[Region] nvarchar(15) NULL,
[PostalCode] nvarchar(10) NULL,
[Country] nvarchar(15) NULL
Constraint df_country DEFAULT ‘Canada’,
[Phone] nvarchar(24) NULL,
[Fax] nvarchar(24) NULL,
Primary Key (CustomerID)
); CREATE TABLE [dbo].[tblSupplier] (
[SupplierID] int NOT NULL,
[Name] nvarchar(50) NULL,
[Address] nvarchar(50) NULL,
[City] nvarchar(50) NULL,
[Province] nvarchar(50) NULL,
Primary Key (SupplierID)
); CREATE TABLE [dbo].[tblShippers] (
[ShipperID] int NOT NULL,
[CompanyName] nvarchar(40) NOT NULL,
Primary Key (ShipperID),<
CONSTRAINT uc_CompanyName UNIQUE (CompanyName)
); CREATE TABLE [dbo].[tblProducts] (
[ProductID] int NOT NULL,
[SupplierID] int NULL,
[CategoryID] int NULL,
[ProductName] nvarchar(40) NOT NULL,
[EnglishName] nvarchar(40) NULL,
[QuantityPerUnit] nvarchar(20) NULL,
[UnitPrice] money NULL,
[UnitsInStock] smallint NULL,
[UnitsOnOrder] smallint NULL,
[ReorderLevel] smallint NULL,
[Discontinued] bit NOT NULL,
Primary Key (ProductID),
Foreign Key (SupplierID) References tblSupplier
); CREATE TABLE [dbo].[tblOrders] (
[OrderID] int NOT NULL,
[CustomerID] nvarchar(5) NOT NULL,
[EmployeeID] int NULL,
[ShipName] nvarchar(40) NULL,
[ShipAddress] nvarchar(60) NULL,
[ShipCity] nvarchar(15) NULL,
[ShipRegion] nvarchar(15) NULL,
[ShipPostalCode] nvarchar(10) NULL,
[ShipCountry] nvarchar(15) NULL,
[ShipVia] int NULL,
[OrderDate] smalldatetime NULL,
[RequiredDate] smalldatetime NULL,
[ShippedDate] smalldatetime NULL,
[Freight] money NULL
Primary Key (OrderID),
Foreign Key (CustomerID) References tblCustomers,
Foreign Key (ShipVia) References tblShippers,
Constraint valid_ShipDate CHECK (ShippedDate > OrderDate)
); CREATE TABLE [dbo].[tblOrderDetails] (
[OrderID] int NOT NULL,
[ProductID] int NOT NULL,
[UnitPrice] money NOT NULL,
[Quantity] smallint NOT NULL,
[Discount] real NOT NULL,
Primary Key (OrderID, ProductID),
Foreign Key (OrderID) References tblOrders,
Foreign Key (ProductID) References tblProducts,
Constraint Valid_Qty Check (Quantity > 0)
);
Go
Part 2 – Create the Following SQL Statements
1. Show a list of customers and the orders they generated during 2014. Display customer ID, order ID, order date and date ordered.
Use OrdersGo
SELECT CompanyName, OrderID, RequiredDate as ‘order date’, OrderDate as ‘date ordered’
FROM tblcustomers JOIN tblOrders on tblOrders.CustomerID = tblCustomers.CustomerID
WHERE Year(OrderDate) = 2014
2. Using the ALTER TABLE statement, add a new field (Active) in the tblcustomer. Default it to True.
ALTER TABLE tblCustomersADD Active bit DEFAULT (‘True’)
3. Show all orders purchased before September 1, 2012. Display company name, date ordered and total amount of order (include freight).
SELECT tblOrders.OrderID, OrderDate as ‘Date Ordered’, sum(unitprice*quantity*(1-discount))+ freight as ‘Total Cost’FROM tblOrderDetails join tblOrders on tblOrders.orderID = tblOrderDetails.OrderID
WHERE OrderDate < ‘September 1, 2012’
GROUP BY tblOrders.OrderID, freight, OrderDate
4. Show all orders that have been shipped via Federal Shipping. Display OrderID, ShipName, ShipAddress and CustomerID.
SELECT OrderID, ShipName, ShipAddress, CustomerIDFROM tblOrders join tblShippers on tblOrders.ShipVia = tblShippers.ShipperID
WHERE CompanyName= ‘Federal Shipping’
5. Show all customers who have not made purchases in 2011.
SELECT CompanyNameFROM tblCustomers
WHERE CustomerID not in
( SELECT CustomerID
FROM tblOrders
WHERE Year(OrderDate) = 2011
)
6. Show all products that have never been ordered.
SELECT ProductID from tblProductsExcept
SELECT ProductID from tblOrderDetails
OR
SELECT Products.ProductID,Products.ProductNameFROM Products LEFT JOIN [Order Details]
ON Products.ProductID = [Order Details].ProductID
WHERE [Order Details].OrderID IS NULL
7. Show OrderIDs for customers who reside in London. Use a subquery. Display CustomerID, CustomerName and OrderID.
SELECT Customers.CompanyName,Customers.CustomerID,OrderIDFROM Orders
LEFT JOIN Customers ON Orders.CustomerID = Customers.CustomerID
WHERE Customers.CompanyName IN
(SELECT CompanyName
FROM Customers
WHERE City = ‘London’)
8. Show products supplied by Supplier A and Supplier B. Display product name and supplier name.
SELECT ProductName, NameFROM tblProducts JOIN tblSupplier on tblProducts.SupplierID = tblSupplier.SupplierID
WHERE Name Like ‘Supplier A’ or Name Like ‘Supplier B’
9. Show all products that come in boxes. Display product name and QuantityPerUnit.
SELECT EnglishName, ProductName, QuantityPerUnitFROM tblProducts
WHERE QuantityPerUnit like ‘%box%’
ORDER BY EnglishName
Part 3 – Insert, Update, Delete, Indexes
1. Create an Employee table. The primary key should be EmployeeID (aut\nonumber). Add the following fields: LastName, FirstName, Address, City, Province, Postalcode, Phone, Salary. Show the CREATE TABLE statement and the INSERT statements for the five employees. Join the employee table to the tblOrders. Show the script for creating the table, setting constraints and adding employees.
 Add a field to tblOrders called TotalSales. Show DDL – ALTER TABLE statement.
ALTER TABLE tblOrdersADD Foreign Key (EmployeeID) references tblEmployees (EmployeeID)
3. Using the UPDATE statement, add the total sale for each order based on the order details table.
UPDATE tblOrdersSet TotalSales = (select sum(unitprice*quantity*(1-discount))
FROM tblOrderDetails
WHERE tblOrderDetails.OrderID= tblOrders.OrderID
GROUP BY OrderID


3.8: Communication Barriers
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Learning Objectives
- Understand different ways that the communication process can be sidetracked.
- Understand the problem of poor listening and how to promote active listening.
Barriers to Effective Communication
Communicating can be more of a challenge than you think, when you realize the many things that can stand in the way of effective communication. These include filtering, selective perception, information overload, emotional disconnects, lack of source familiarity or credibility, workplace gossip, semantics, gender differences, differences in meaning between Sender and Receiver, and biased language. Let’s examine each of these barriers.
Filtering
Filtering is the distortion or withholding of information to manage a person’s reactions. Some examples of filtering include a manager who keeps her division’s poor sales figures from her boss, the vice president, fearing that the bad news will make him angry. The old saying, “Don’t shoot the messenger!” illustrates the tendency of Receivers (in this case, the vice president) to vent their negative response to unwanted Messages on the Sender. A gatekeeper (the vice president’s assistant, perhaps) who doesn’t pass along a complete Message is also filtering. The vice president may delete the e-mail announcing the quarter’s sales figures before reading it, blocking the Message before it arrives.
As you can see, filtering prevents members of an organization from getting a complete picture of the way things are. To maximize your chances of sending and receiving effective communications, it’s helpful to deliver a Message in multiple ways and to seek information from multiple sources. In this way, the effect of any one person’s filtering the Message will be diminished.
Since people tend to filter bad news more during upward communication, it is also helpful to remember that those below you in an organization may be wary of sharing bad news. One way to defuse the tendency to filter is to reward employees who clearly convey information upward, regardless of whether the news is good and bad.
Here are some of the criteria that individuals may use when deciding whether to filter a Message or pass it on:
- Past experience: Was the Sender rewarded for passing along news of this kind in the past, or was she criticized?
- Knowledge, perception of the speaker: Has the Receiver’s direct superior made it clear that “no news is good news?”
- Emotional state, involvement with the topic, level of attention: Does the Sender’s fear of failure or criticism prevent him from conveying the Message? Is the topic within his realm of expertise, increasing his confidence in his ability to decode it, or is he out of his comfort zone when it comes to evaluating the Message’s significance? Are personal concerns impacting his ability to judge the Message’s value?
Once again, filtering can lead to miscommunications in business. Each listener translates the Message into his or her own words, creating his or her own version of what was said (Alessandra, 1993).
Selective Perception
Selective perception refers to filtering what we see and hear to suit our own needs. This process is often unconscious. Small things can command our attention when we’re visiting a new place—a new city or a new company. Over time, however, we begin to make assumptions about the way things are on the basis of our past experience. Often, much of this process is unconscious. “We simply are bombarded with too much stimuli every day to pay equal attention to everything so we pick and choose according to our own needs (Pope, 2008).” Selective perception is a time-saver, a necessary tool in a complex culture. But it can also lead to mistakes.
Think back to the earlier example conversation between Bill, who was asked to order more toner cartridges, and his boss. Since Bill found his boss’s to-do list to be unreasonably demanding, he assumed the request could wait. (How else could he do everything else on the list?) The boss, assuming that Bill had heard the urgency in her request, assumed that Bill would place the order before returning to the other tasks on her list.
Both members of this organization were using selective perception to evaluate the communication. Bill’s perception was that the task of ordering could wait. The boss’s perception was that her time frame was clear, though unstated. When two selective perceptions collide, a misunderstanding occurs.
Information Overload
Information overload can be defined as “occurring when the information processing demands on an individual’s time to perform interactions and internal calculations exceed the supply or capacity of time available for such processing (Schick, et. al., 1990).” Messages reach us in countless ways every day. Some are societal—advertisements that we may hear or see in the course of our day. Others are professional—e-mails, and memos, voice mails, and conversations from our colleagues. Others are personal—messages and conversations from our loved ones and friends.
Add these together and it’s easy to see how we may be receiving more information than we can take in. This state of imbalance is known as information overload. Experts note that information overload is “A symptom of the high-tech age, which is too much information for one human being to absorb in an expanding world of people and technology. It comes from all sources including TV, newspapers, and magazines as well as wanted and unwanted regular mail, e-mail and faxes. It has been exacerbated enormously because of the formidable number of results obtained from Web search engines (PC Magazine, 2008).” Other research shows that working in such fragmented fashion has a significant negative effect on efficiency, creativity, and mental acuity (Overholt, 2001).
Going back to our example of Bill. Let’s say he’s in his cubicle on the phone with a supplier. While he’s talking, he hears the chime of e-mail alerting him to an important message from his boss. He’s scanning through it quickly, while still on the phone, when a coworker pokes his head around the cubicle corner to remind Bill that he’s late for a staff meeting. The supplier on the other end of the phone line has just given Bill a choice among the products and delivery dates he requested. Bill realizes he missed hearing the first two options, but he doesn’t have time to ask the supplier to repeat them all or to try reconnecting to place the order at a later time. He chooses the third option—at least he heard that one, he reasons, and it seemed fair. How good was Bill’s decision amid all the information he was processing at the same time?
Emotional disconnects
Emotional disconnects happen when the Sender or the Receiver is upset, whether about the subject at hand or about some unrelated incident that may have happened earlier. An effective communication requires a Sender and a Receiver who are open to speaking and listening to one another, despite possible differences in opinion or personality. One or both parties may have to put their emotions aside to achieve the goal of communicating clearly. A Receiver who is emotionally upset tends to ignore or distort what the Sender is saying. A Sender who is emotionally upset may be unable to present ideas or feelings effectively.
Lack of Source Credibility
Lack of source familiarity or credibility can derail communications, especially when humor is involved.  You and the Receiver lacked the common context that could have made it funny. (Or yes, it could have just been a lousy joke.) Sarcasm and irony are subtle, and potentially hurtful, commodities in business. It’s best to keep these types of communications out of the workplace as their benefits are limited, and their potential dangers are great. Lack of familiarity with the Sender can lead to misinterpreting humor, especially in less-rich information channels like e-mail. For example, an e-mail from Jill that ends with, “Men, like hens, should boil in vats of oil,” could be interpreted as antimale if the Receiver didn’t know that Jill has a penchant for rhyme and likes to entertain coworkers by making up amusing sayings.
Similarly, if the Sender lacks credibility or is untrustworthy, the Message will not get through. Receivers may be suspicious of the Sender’s motivations (“Why am I being told this?”). Likewise, if the Sender has communicated erroneous information in the past, or has created false emergencies, his current Message may be filtered.
Workplace gossip, also known as the grapevine , is a lifeline for many employees seeking information about their company (Kurland & Pelled, 2000). Researchers agree that the grapevine is an inevitable part of organizational life. Research finds that 70% of all organizational communication occurs at the grapevine level (Crampton, 1998).
Employees trust their peers as a source of Messages, but the grapevine’s informal structure can be a barrier to effective communication from the managerial point of view. Its grassroots structure gives it greater credibility in the minds of employees than information delivered through official channels, even when that information is false.
Some downsides of the office grapevine are that gossip offers politically minded insiders a powerful tool for disseminating communication (and self-promoting miscommunications) within an organization. In addition, the grapevine lacks a specific Sender, which can create a sense of distrust among employees—who is at the root of the gossip network? When the news is volatile, suspicions may arise as to the person or persons behind the Message. Managers who understand the grapevine’s power can use it to send and receive Messages of their own. They also decrease the grapevine’s power by sending official Messages quickly and accurately, should big news arise.
Semantics
Semantics is the study of meaning in communication. Words can mean different things to different people, or they might not mean anything to another person. For example, companies often have their own acronyms and buzzwords (called business jargon) that are clear to them but impenetrable to outsiders. For example, at IBM, GBS is focusing on BPTS, using expertise acquired from the PwC purchase (which had to be sold to avoid conflicts of interest in light of SOX) to fend other BPO providers and inroads by the Bangalore tiger. Does this make sense to you? If not, here’s the translation: IBM’s Global Business Services (GBS) division is focusing on offering companies Business Process Transformation Services (BPTS), using the expertise it acquired from purchasing the management consulting and technology services arm of PricewaterhouseCoopers (PwC), which had to sell the division because of the Sarbanes-Oxley Act (SOX, enacted in response to the major accounting scandals like the Enron). The added management expertise puts it above business process outsourcing (BPO) vendors who focus more on automating processes rather than transforming and improving them. Chief among these BPO competitors is Wipro, often called the “Bangalore tiger” because of its geographic origin and aggressive growth.
Given the amount of Messages we send and receive every day, it makes sense that humans try to find shortcuts—a way to communicate things in code. In business, this code is known as jargon . Jargon is the language of specialized terms used by a group or profession. It is common shorthand among experts and if used sensibly can be a quick and efficient way of communicating.  Every profession, trade, and organization has its own specialized terms (Wright, 2008). At first glance, jargon seems like a good thing—a quicker way to send an effective communication, the way text message abbreviations can send common messages in a shorter, yet understandable way. But that’s not always how things happen. Jargon can be an obstacle to effective communication, causing listeners to tune out or fostering ill-feeling between partners in a conversation. When jargon rules the day, the Message can get obscured.
A key question to ask before using jargon is, “Who is the Receiver of my Message?” If you are a specialist speaking to another specialist in your area, jargon may be the best way to send a message while forging a professional bond—similar to the way best friends can communicate in code. For example, an information technology (IT) systems analyst communicating with another IT employee may use jargon as a way of sharing information in a way that reinforces the pair’s shared knowledge. But that same conversation should be held in standard English, free of jargon, when communicating with staff members outside the IT group.
Online Follow-Up
Here is a Web site of 80 buzz words in business:
http://www.amanet.org/movingahead/editorial2002_2003/nov03_80buzzwords.htm
and a discussion of why slang is a problem:
http://sbinfocanada.about.com/od/speakforsuccesscourse/a/speechlesson5.htm .
Gender Differences
Gender differences in communication have been documented by a number of experts, including linguistics professor Deborah Tannen in her best-selling book You Just Don’t Understand: Women and Men in Conversation (Tannen, 1991). Men and women work together every day. But their different styles of communication can sometimes work against them. Generally speaking, women like to ask questions before starting a project, while men tend to “jump right in.” A male manager who’s unaware of how many women communicate their readiness to work may misperceive a ready employee as not ready.
Another difference that has been noticed is that men often speak in sports metaphors, while many women use their home as a starting place for analogies. Women who believe men are “only talking about the game” may be missing out on a chance to participate in a division’s strategy and opportunities for teamwork and “rallying the troops” for success (Krotz, 2008).
“It is important to promote the best possible communication between men and women in the workplace,” notes gender policy adviser Dee Norton, who provided the above example. “As we move between the male and female cultures, we sometimes have to change how we behave (speak the language of the other gender) to gain the best results from the situation. Clearly, successful organizations of the future are going to have leaders and team members who understand, respect and apply the rules of gender culture appropriately (Norton, 2008).”
Being aware of these gender differences can be the first step in learning to work with them, as opposed to around them. For example, keep in mind that men tend to focus more on competition, data, and orders in their communications, while women tend to focus more on cooperation, intuition, and requests. Both styles can be effective in the right situations, but understanding the differences is a first step in avoiding misunderstandings based on them.
Differences in meaning often exist between the Sender and Receiver. “Mean what you say, and say what you mean .” It’s an easy thing to say. But in business, what do those words mean? Different words mean different things to different people. Age, education, and cultural background are all factors that influence how a person interprets words. The less we consider our audience, the greater our chances of miscommunication will be. When communication occurs in the cross-cultural context, extra caution is needed given that different words will be interpreted differently across cultures and different cultures have different norms regarding nonverbal communication. Eliminating jargon is one way of ensuring that our words will convey real-world concepts to others. Speaking to our audience, as opposed to about ourselves, is another. Nonverbal Messages can also have different meanings.
Table \(\PageIndex{1}\) Gestures Around the Globe
|
Figure \(\PageIndex{1}\) |
1. “V” for victory . Use this gesture with caution! While in North America it signs victory or peace, in England and Australia it means something closer to “take this!” |
|
Figure \(\PageIndex{2}\) |
2. The “OK” gesture . While in North America it means things are going well, in France it means a person is thought to be worthless, in Japan it refers to money, and in Brazil, Russia, and Germany it means something really not appropriate for the workplace. |
|
Figure \(\PageIndex{3}\) |
3. The “thumbs up ” means one in Germany, five in Japan, but a good job in North America. This can lead to confusion. |
|
Figure \(\PageIndex{4}\) |
4. “Hook ‘em horns .” This University of Texas rallying call looks like the horns of a bull. However, in Italy it means you are being tricked, while in Brazil and Venezuela it means you are warding off evil. |
|
Figure \(\PageIndex{5}\) |
5. Waving your hand . In much of Europe waving your hand indicates a disagreement. However, in North America it is routinely used as a way to signal greetings or to get someone’s attention. |
Adapted from information in Axtell, R. E. (1998). Gestures: The do’s and taboos of body language around the world . New York: John Wiley.
Managers who speak about “long-term goals and profits” to a staff that has received scant raises may find their core Message (“You’re doing a great job—and that benefits the folks in charge!”) has infuriated the group they hoped to inspire. Instead, managers who recognize the “contributions” of their staff and confirm that this work is contributing to company goals in ways “that will benefit the source of our success—our employees as well as executives,” will find their core Message (“You’re doing a great job—we really value your work”) is received as opposed to being misinterpreted.
Biased language can offend or stereotype others on the basis of their personal or group affiliation. The figure below provides a list of words that have the potential to be offensive in the left-hand column. The right-hand column provides more neutral words that you can use instead (Ashcraft & Mumby, 2003; Swift, 2007).
Effective communication is clear, factual, and goal-oriented. It is also respectful. Referring to a person by one adjective (a brain , a diabetic , an invalid ) reduces that person to that one characteristic. Language that belittles or stereotypes a person poisons the communication process. Language that insults an individual or group based on age, ethnicity, sexual preference, or political beliefs violates public and private standards of decency, ranging from civil rights to corporate regulations.
The effort to create a neutral set of terms to refer to heritage and preferences has resulted in a debate over the nature of “political correctness.” Proponents of political correctness see it as a way to defuse the volatile nature of words that stereotyped groups and individuals in the past. Critics of political correctness see its vocabulary as stilted and needlessly cautious.
Many companies offer new employees written guides on standards of speech and conduct. These guides, augmented by common sense and courtesy, are solid starting points for effective, respectful workplace communication. Tips for appropriate workplace speech include but are not limited to
- Alternating the use of “he” and “she” when referring to people in general.
- Relying on human resources–generated guidelines.
- Remembering that terms that feel respectful or comfortable to us may not be comfortable or respectful to others.
Poor Listening and Active Listening
Former Chrysler CEO Lee Iacocca lamented, “I only wish I could find an institute that teaches people how to listen. After all, a good manager needs to listen at least as much as he needs to talk (Iacocca & Novak, 1984).” Research shows that listening skills are related to promotions (Sypher, et. al., 1989). A Sender may strive to deliver a Message clearly. But the Receiver’s ability to listen effectively is equally vital to effective communication. The average worker spends 55% of her workdays listening. Managers listen up to 70% each day. But listening doesn’t lead to understanding in every case. Listening takes practice, skill, and concentration.
According to University of San Diego professor Phillip Hunsaker, “The consequences of poor listening are lower employee productivity, missed sales, unhappy customers, and billions of dollars of increased cost and lost profits. Poor listening is a factor in low employee morale and increased turnover because employees do not feel their managers listen to their needs, suggestions, or complaints (Alessandra, et. al., 1993).” Clearly, if you hope to have a successful career in management, it behooves you to learn to be a good listener.
Alan Gulick, a Starbucks spokesperson, puts better listening to work in pursuit of better profits. If every Starbucks employee misheard one $10 order each day, he calculates, their errors would cost the company a billion dollars annually. To teach its employees to listen, Starbucks created a code that helps employees taking orders hear the size, flavor, and use of milk or decaf coffee. The person making the drink echoes the order aloud.
How can you improve your listening skills? The Roman philosopher Cicero said, “Silence is one of the great arts of conversation.” How often have we been in conversation with someone else where we are not really listening but itching to convey our portion? This behavior is known as “rehearsing.” It suggests the Receiver has no intention of considering the Sender’s Message and intends to respond to an earlier point instead. Clearly, rehearsing is an impediment to the communication process. Effective communication relies on another kind of listening: active listening.
Active listening can be defined as giving full attention to what other people are saying, taking time to understand the points being made, asking questions as appropriate, and not interrupting at inappropriate times (Onet Center, 2008).Active listening creates a real-time relationship between the Sender and the Receiver by acknowledging the content and receipt of a Message. As we’ve seen in the Starbucks example, repeating and confirming a Message’s content offers a way to confirm that the correct content is flowing between colleagues. The process creates a bond between coworkers while increasing the flow and accuracy of messaging.
Carl Rogers, founder of the “person-centered” approach to psychology, formulated five rules for active listening:
- Listen for message content
- Listen for feelings
- Respond to feelings
- Note all cues
- Paraphrase and restate
The good news is that listening is a skill that can be learned (Brownell, 1990). The first step is to decide that we want to listen. Casting aside distractions, such as by reducing background or internal noise, is critical. The Receiver takes in the Sender’s Message silently, without speaking. Second, throughout the conversation, show the speaker that you’re listening. You can do this nonverbally by nodding your head and keeping your attention focused on the speaker. You can also do it verbally, by saying things like, “Yes,” “That’s interesting,” or other such verbal cues. As you’re listening, pay attention to the Sender’s body language for additional cues about how they’re feeling. Interestingly, silence plays a major role in active listening. During active listening, we are trying to understand what has been said, and in silence, we can consider the implications. We can’t consider information and reply to it at the same time. That’s where the power of silence comes into play. Finally, if anything is not clear to you, ask questions. Confirm that you’ve heard the message accurately, by repeating back a crucial piece like, “Great, I’ll see you at 2 p.m. in my office.” At the end of the conversation, a “thank you” from both parties is an optional but highly effective way of acknowledging each other’s teamwork.
In summary, active listening creates a more dynamic relationship between a Receiver and a Sender. It strengthens personal investment in the information being shared. It also forges healthy working relationships among colleagues by making Speakers and Listeners equally valued members of the communication process.
Key Takeaway
Many barriers to effective communication exist. Examples include filtering, selective perception, information overload, emotional disconnects, lack of source familiarity or credibility, workplace gossip, semantics, gender differences, differences in meaning between Sender and Receiver, and biased language. The Receiver can enhance the probability of effective communication by engaging in active listening, which involves (1) giving one’s full attention to the Sender and (2) checking for understanding by repeating the essence of the Message back to the Sender.
Exercises
- Most people are poor listeners. Do you agree or disagree with this statement? Please support your position.
- Please share an example of how differences in shared meaning have affected you.
- Give an example of selective perception.
- Do you use jargon at or in your classes? If so, do you think it helps or hampers communication? Why or why not?
- In your experience, how is silence used in communication? How does your experience compare with the recommended use of silence in active listening?
References
Alessandra, T. (1993). Communicating at work . New York: Fireside.
Alessandra, T., Garner, H., & Hunsaker, P. L. (1993). Communicating at work . New York: Simon & Schuster.
Ashcraft, K., & Mumby, D. K. (2003). Reworking gender . Thousand Oaks, CA, Sage; Miller, C., &.
Brownell, J. (1990). Perceptions of effective listeners: A management study. Journal of Business Communications , 27 , 401–415.
Crampton, S. M. (1998). The informal communication network: factors influencing grapevine activity. Public Personnel Management . Retrieved July 2, 2008, from www.allbusiness.com/management/735210-1.html.
Iacocca, L., & Novak, W. (1984). Iacocca: An autobiography . New York: Bantam Press.
Krotz, J. L. (n.d.). 6 tips for bridging the communication gap. Retrieved July 2, 2008, from Microsoft Small Business Center Web site, www.microsoft.com/smallbusiness/resources/management/leadership-training/women-vs-men-6-tips-for-bridging-the-communication-gap.aspx.
Kurland, N. B., & Pelled, L. H. (2000). Passing the word: Toward a model of gossip and power in the workplace. Academy of Management Review , 25 , 428–438.
Norton, D. Gender and communication—finding common ground. Retrieved July 2, 2008, from www.uscg.mil/leadership/gender.htm.
O*NET Resource Center, the nation’s primary source of occupational information. Retrieved July 2, 2008, from http://online.onetcenter.org/skills .
Overholt, A. (2001, February). Intel’s got (too much) mail. Fast Company . Retrieved July 2, 2008, from www.fastcompany.com/online/44/intel.html and http://blogs.intel.com/it/2006/10/information_overload.php .
PC Magazine, retrieved July 1, 2008, from PC Magazine encyclopedia Web site, www.pcmag.com/encyclopedia_term/0,2542,t=information+overload&i=44950,00.asp, and reinforced by information in Dawley, D. D., & Anthony, W. P. (2003). User perceptions of e-mail at work. Journal of Business and Technical Communication , 17 , 170–200.
Pope, R. R. Selective perception. Illinois State University. Retrieved December 1, 2008, from lilt.ilstu.edu/rrpope/rrpopepwd/articles/perception3.html.
Schick, A. G., Gordon, L. A., & Haka, S. (1990). Information overload: A temporal approach. Accounting, Organizations, and Society, 15 , 199–220.
Swift, K. (1980). The handbook of nonsexist writing . New York: Lippincott & Crowell; Procter, M. (2007, September 11). Unbiased language . Retrieved July 2, 2008, from www.utoronto.ca/writing/unbias.html.
Sypher, B. D., Bostrom, R. N., & Seibert, J. H. (1989). Listening, communication abilities, and success at work. Journal of Business Communication , 26 , 293–303.
Tannen, D. (1991). You just don’t understand: Women and men in conversation. New York: Ballantine.
Wright, N. Keep it jargon-free . Retrieved July 2, 2008, from http://www.plainlanguage.gov/howto/wordsuggestions/jargonfree.cfm .


23.1: Faraday’s Law
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In the previous chapter, we described how an electric current produces a magnetic field. In this chapter, we describe how an electric current can be produced (or rather, “induced”) by a magnetic field. The most important aspect of electromagnetic induction is that it always involves quantities that change with time. In past chapters, we have only dealt with static electric and magnetic fields, static charges (for electric fields), and static currents (for magnetic fields).
Faraday’s Law connects the flux of a time-varying magnetic field to an induced voltage (rather than a current). For historical reasons, the induced voltage is also called an induced “electromotive force” (emf), even if it is a voltage and not a force. Faraday’s Law is as follows:
\[\Delta V=-\frac{d\Phi_{B}}{dt}\]
where \(\Delta V\) is the induced voltage, and \(\Phi_B\) is the flux of the magnetic field through an open surface, defined in the same way as the flux of the electric field (Section 17.1):
\[\begin{aligned} \Phi_B = \int_S \vec B\cdot d\vec A\end{aligned}\]
If the magnetic field has a constant magnitude over the surface, \(S\) , and always makes the same angle with the surface, then the flux can be written as:
\[\begin{aligned} \Phi_B = \vec B\cdot\vec A\end{aligned}\]
where the magnitude of the vector \(\vec A\) is equal to the area of the surface, and the vector \(\vec A\) is normal to the surface.
The surface, \(S\) , is defined by a closed path. The induced voltage can be thought of as an ideal battery placed in the closed path that defines the surface (right-hand panel of Figure \(\PageIndex{1}\)). The minus sign indicates the direction of the current associated with the induced voltage. It is important to note that an induced voltage only exists if the flux of the magnetic field changes (since the induced voltage is given by the time-derivative of the flux). Remember, induction is all about time-varying fields! This is better illustrated with an example.
Consider a loop of wire that is immersed in a uniform magnetic field, \(\vec B\) , that is perpendicular to the plane of the loop, as illustrated in Figure \(\PageIndex{1}\). As time goes by, the magnetic field increases in strength, as shown in going from the left panel to the right panel. The flux of the magnetic field through the loop increases in magnitude, and a voltage is thus induced across the wire (illustrated by the ideal battery on the loop in the right panel), leading to an induced current, \(I\) .
When calculating the flux of the magnetic field, we have to choose the surface element vector, \(d\vec A\) , to be perpendicular to the surface over which we calculate the flux. There are two choices 1 (upwards or downwards, referring to Figure \(\PageIndex{1}\)); we chose to define \(d\vec A\) to point upwards. Thus, the magnetic flux is positive in both panels, and increases with time. The derivative, \(d\vec B/dt\) , is thus positive, and the right-hand side of Faraday’s equation is negative because of the negative sign in front. Had we chosen to define \(d\vec A\) to point downwards, the right-hand side of Faraday’s Law would be negative.
We can describe the direction of the induced current, \(I\) , in terms of its magnetic dipole moment (Section 21.4), \(\vec\mu_I\) , also shown in Figure \(\PageIndex{1}\). The overall sign on the right-hand side of Faraday’s Law is determined by our (arbitrary) choice of the direction \(d\vec A\) . With this choice, we found that the right-hand side of Faraday’s Law is negative:
\[\begin{aligned} \Delta V = -\frac{d\Phi_B}{dt}=\text{a negative number}\end{aligned}\]
The overall sign of \(\Delta V\) indicates whether the magnetic moment of the induced current is parallel ( \(\Delta V\) positive) or anti-parallel ( \(\Delta V\) negative) to \(d\vec A\) . This allows us to determine the direction of the induced current, and thus the direction of the ideal battery that represents the induced voltage. In general, when possible, it is common to choose the direction of \(d\vec A\) to be parallel to the magnetic field vector, so that the flux is positive (although this does not guarantee that its derivative is positive).
Lenz’ Law
The minus sign in Faraday’s Law is sometimes called “Lenz’s Law”, and ultimately comes from the conservation of energy. In Figure \(\PageIndex{1}\) above, we found that as the magnetic flux increases through the loop, a current is induced. That induced current will also produce a magnetic field (in the direction of its magnetic dipole moment vector, \(\vec \mu_I\) ).
Lenz’s Law states that the “induced current will always be such that the magnetic field that it produces counteracts the changing magnetic field that induced the current”. In Figure \(\PageIndex{1}\), the magnetic field points in the upwards direction, and increases in magnitude with time. The induced current produces a magnetic field that points downwards to counteract the changing magnetic field, and preserve a constant flux through the loop. If this were not the case, the induced current would be in the opposite direction, contributing to the increasing magnetic flux through the loop, inducing more current, producing more flux, inducing more current, etc. Clearly, this would lead to an infinite current and solve the world’s energy crisis. Unfortunately, conservation of energy (expressed here as Lenz’s Law) prevents this from happening.
You can use Lenz’s Law to determine the direction of induced currents. In general:
- If the magnitude of the magnetic flux is increasing in the loop, then the induced current produces a magnetic field that is in the opposite direction from the original magnetic field.
- If the magnitude of the magnetic flux is decreasing in the loop, then the induced current produces a magnetic field that is in the same direction as the original magnetic field.
The negative sign in Faraday’s Law is not arbitrary (as we saw above, it gives the correct direction for the magnetic moment of the induced current, given our arbitrary choice of direction for \(d\vec A\) ). In practice, one can often use Lenz’s Law to determine the direction of the induced current (so that it counteracts the changing flux), and Faraday’s Law to determine the magnitude of the induced voltage.
Exercise \(\PageIndex{1}\)
A loop of wire is immersed in a constant and uniform magnetic field out of the page, perpendicular to the plane of the loop, as shown in Figure \(\PageIndex{2}\). If the radius of the loop increases with time, in which direction will be the current induced in the loop?
- Since the magnetic field is constant, there is no induced current.
- Clockwise.
- Counter-clockwise.
- Answer
-
Exercise \(\PageIndex{2}\)
A loop of wire is immersed in a constant and uniform magnetic field out of the page, perpendicular to the plane of the loop, as shown in Figure \(\PageIndex{3}\). If the loop is pulled out of the region of magnetic field, as shown, in which direction is the induced current in the loop?
- Since the magnetic field is constant, there is no induced current.
- Clockwise.
- Counter-clockwise.
- Answer
Example \(\PageIndex{1}\)
A uniform time-varying magnetic field is given by:
\[\begin{aligned} \vec B(t) = B_0(1+at)\hat z\end{aligned}\]
where \(B_0\) and \(a\) are positive constants. A coil, made of \(N\) circular loops of radius, \(r\) , lies in the \(x-y\) plane. If the coil has a total resistance, \(R\) , what is the magnitude and direction of the current induced in the coil?
Solution
The coil is made of \(N\) loops of wire. Each loop of wire can be treated independently, and each will have its own induced voltage across it. Since each loop is the same, they will all have the same induced voltage, and the total voltage induced across the coil, \(\Delta V\) , will be given by:
\[\begin{aligned} \Delta V = -N \frac{d\Phi_B}{dt}\end{aligned}\]
where \(\Phi_B\) is the flux through any one of the loops. That is, each loop is similar to an ideal battery, and the coil is similar to placing all of these batteries in series, so that the voltages from each battery sum together.
The coil lies the \(x-y\) plane, perpendicular to the increasing magnetic field, similar to the situation depicted in Figure \(\PageIndex{1}\). Since the magnetic field is uniform over the surface of the coil, we do not need an integral to determine the flux. We define the area vector, \(\vec A\) , to be in the positive \(z\) direction (parallel to the magnetic field):
\[\begin{aligned} \vec A = A \hat z = \pi r^2 \hat z\end{aligned}\]
The flux through one circular loop of radius, \(r\) , is given by:
\[\begin{aligned} \Phi_B (t) &= \vec B \cdot \vec A = ( B_0(1+at)\hat z) \cdot (\pi r^2 \hat z) =B_0(1+at) (\pi r^2)\end{aligned}\]
We can apply Faraday’s Law to determine the induced voltage:
\[\begin{aligned} \Delta V &= -N \frac{d\Phi_B}{dt} = -N \frac{d}{dt} B_0(1+at) (\pi r^2)\\[4pt] &=-NB_0a\pi r^2\end{aligned}\]
Since the induced voltage is negative, the magnetic moment of the induced current points in the negative \(z\) direction (opposite to our choice of direction for \(\vec A\) ). This is consistent with Len’z Law, since the magnetic field increases in the positive \(z\) direction, the induced current will produce a magnetic field in the negative \(z\) direction to counteract the changing flux. The magnitude of the induced current is given by Ohm’s Law:
\[\begin{aligned} I = \frac{\Delta V}{R}=\frac{NB_0a\pi r^2}{R}\end{aligned}\]
Discussion
In this example, we determined the induced voltage and current in a coil made of \(N\) identical loops. We argued that one can sum the induced voltages from the \(N\) loops, as these can be thought of as ideal batteries in series. We found that the direction of the induced current as obtained from Faraday’s Law was consistent with the expectation from Lenz’s Law.
Footnotes
1. Recall that this ambiguity is resolved when using Gauss’ Law by always choosing \(d\vec A\) to point “outwards”, which only makes sense when the surface is closed. With an open surface, there is no inside or outside, and we are left with the ambiguity.


1.5: Measurements
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Learning Objectives
By the end of this section, you will be able to:
- Explain the process of measurement
- Identify the three basic parts of a quantity
- Describe the properties and units of length, mass, volume, density, temperature, and time
- Perform basic unit calculations and conversions in the metric and other unit systems
Measurements provide much of the information that informs the hypotheses, theories, and laws describing the behavior of matter and energy in both the macroscopic and microscopic domains of chemistry. Every measurement provides three kinds of information: the size or magnitude of the measurement (a number); a standard of comparison for the measurement (a unit); and an indication of the uncertainty of the measurement. While the number and unit are explicitly represented when a quantity is written, the uncertainty is an aspect of the measurement result that is more implicitly represented and will be discussed later.
The number in the measurement can be represented in different ways, including decimal form and scientific notation. (Scientific notation is also known as exponential notation; a review of this topic can be found in Appendix B.) For example, the maximum takeoff weight of a Boeing 777-200ER airliner is 298,000 kilograms, which can also be written as 2.98 10 5 kg. The mass of the average mosquito is about 0.0000025 kilograms, which can be written as 2.5 10 −6 kg.
Units , such as liters, pounds, and centimeters, are standards of comparison for measurements. A 2-liter bottle of a soft drink contains a volume of beverage that is twice that of the accepted volume of 1 liter. The meat used to prepare a 0.25-pound hamburger weighs one-fourth as much as the accepted weight of 1 pound. Without units, a number can be meaningless, confusing, or possibly life threatening. Suppose a doctor prescribes phenobarbital to control a patient’s seizures and states a dosage of “100” without specifying units. Not only will this be confusing to the medical professional giving the dose, but the consequences can be dire: 100 mg given three times per day can be effective as an anticonvulsant, but a single dose of 100 g is more than 10 times the lethal amount.
The measurement units for seven fundamental properties (“base units”) are listed in Table \(\PageIndex{1}\). The standards for these units are fixed by international agreement, and they are called the International System of Units or SI Units (from the French, Le Système International d’Unités ). SI units have been used by the United States National Institute of Standards and Technology (NIST) since 1964. Units for other properties may be derived from these seven base units.
| Property Measured | Name of Unit | Symbol of Unit |
|---|---|---|
| length | meter | m |
| mass | kilogram | kg |
| time | second | s |
| temperature | kelvin | K |
| electric current | ampere | A |
| amount of substance | mole | mol |
| luminous intensity | candela | cd |
Everyday measurement units are often defined as fractions or multiples of other units. Milk is commonly packaged in containers of 1 gallon (4 quarts), 1 quart (0.25 gallon), and one pint (0.5 quart). This same approach is used with SI units, but these fractions or multiples are always powers of 10. Fractional or multiple SI units are named using a prefix and the name of the base unit. For example, a length of 1000 meters is also called a kilometer because the prefix kilo means “one thousand,” which in scientific notation is 10 3 (1 kilometer = 1000 m = 10 3 m). The prefixes used and the powers to which 10 are raised are listed in Table \(\PageIndex{2}\).
| Prefix | Symbol | Factor | Example |
|---|---|---|---|
| femto | f | 10 −15 | 1 femtosecond (fs) = 1 x 10 −15 s (0.000000000000001 s) |
| pico | p | 10 −12 | 1 picometer (pm) = 1 x 10 −12 m (0.000000000001 m) |
| nano | n | 10 −9 | 4 nanograms (ng) = 4 x 10 −9 g (0.000000004 g) |
| micro | µ | 10 −6 | 1 microliter (μL) = 1 x 10 −6 L (0.000001 L) |
| milli | m | 10 −3 | 2 millimoles (mmol) = 2 x 10 −3 mol (0.002 mol) |
| centi | c | 10 −2 | 7 centimeters (cm) = 7 x 10 −2 m (0.07 m) |
| deci | d | 10 −1 | 1 deciliter (dL) = 1 x 10 −1 L (0.1 L ) |
| kilo | k | 10 3 | 1 kilometer (km) = 1 x 10 3 m (1000 m) |
| mega | M | 10 6 | 3 megahertz (MHz) = 3 x 10 6 Hz (3,000,000 Hz) |
| giga | G | 10 9 | 8 gigayears (Gyr) = 8 x 10 9 yr (8,000,000,000 yr) |
| tera | T | 10 12 | 5 terawatts (TW) = 5 x 10 12 W (5,000,000,000,000 W) |
Link to Learning
Need a refresher or more practice with scientific notation? Visit this site to go over the basics of scientific notation.
SI Base Units
The initial units of the metric system, which eventually evolved into the SI system, were established in France during the French Revolution. The original standards for the meter and the kilogram were adopted there in 1799 and eventually by other countries. This section introduces four of the SI base units commonly used in chemistry. Other SI units will be introduced in subsequent chapters.
Length
The standard unit of length in both the SI and original metric systems is the meter (m) . A meter was originally specified as 1/10,000,000 of the distance from the North Pole to the equator. It is now defined as the distance light in a vacuum travels in 1/299,792,458 of a second. A meter is about 3 inches longer than a yard (Figure \(\PageIndex{1}\)); one meter is about 39.37 inches or 1.094 yards. Longer distances are often reported in kilometers (1 km = 1000 m = 10 3 m), whereas shorter distances can be reported in centimeters (1 cm = 0.01 m = 10 −2 m) or millimeters (1 mm = 0.001 m = 10 −3 m).
Mass
The standard unit of mass in the SI system is the kilogram (kg) . The kilogram was previously defined by the International Union of Pure and Applied Chemistry ( IUPAC ) as the mass of a specific reference object. This object was originally one liter of pure water, and more recently it was a metal cylinder made from a platinum-iridium alloy with a height and diameter of 39 mm (Figure \(\PageIndex{2}\)). In May 2019, this definition was changed to one that is based instead on precisely measured values of several fundamental physical constants. 1 . One kilogram is about 2.2 pounds. The gram (g) is exactly equal to 1/1000 of the mass of the kilogram (10 −3 kg).
Temperature
Temperature is an intensive property. The SI unit of temperature is the kelvin (K) . The IUPAC convention is to use kelvin (all lowercase) for the word, K (uppercase) for the unit symbol, and neither the word “degree” nor the degree symbol (°). The degree Celsius (°C) is also allowed in the SI system, with both the word “degree” and the degree symbol used for Celsius measurements. Celsius degrees are the same magnitude as those of kelvin, but the two scales place their zeros in different places. Water freezes at 273.15 K (0 °C) and boils at 373.15 K (100 °C) by definition, and normal human body temperature is approximately 310 K (37 °C). The conversion between these two units and the Fahrenheit scale will be discussed later in this chapter.
Time
The SI base unit of time is the second (s) . Small and large time intervals can be expressed with the appropriate prefixes; for example, 3 microseconds = 0.000003 s = 3 x 10 −6 and 5 megaseconds = 5,000,000 s = 5 x 10 6 s. Alternatively, hours, days, and years can be used.
Derived SI Units
We can derive many units from the seven SI base units. For example, we can use the base unit of length to define a unit of volume, and the base units of mass and length to define a unit of density.
Volume
Volume is the measure of the amount of space occupied by an object. The standard SI unit of volume is defined by the base unit of length (Figure \(\PageIndex{3}\)). The standard volume is a cubic meter (m 3 ) , a cube with an edge length of exactly one meter. To dispense a cubic meter of water, we could build a cubic box with edge lengths of exactly one meter. This box would hold a cubic meter of water or any other substance.
A more commonly used unit of volume is derived from the decimeter (0.1 m, or 10 cm). A cube with edge lengths of exactly one decimeter contains a volume of one cubic decimeter (dm 3 ). A liter (L) is the more common name for the cubic decimeter. One liter is about 1.06 quarts.
A cubic centimeter (cm 3 ) is the volume of a cube with an edge length of exactly one centimeter. The abbreviation cc (for c ubic c entimeter) is often used by health professionals. A cubic centimeter is equivalent to a milliliter (mL) and is 1/1000 of a liter.
Density
We use the mass and volume of a substance to determine its density. Thus, the units of density are defined by the base units of mass and length.
The density of a substance is the ratio of the mass of a sample of the substance to its volume. The SI unit for density is the kilogram per cubic meter (kg/m 3 ). For many situations, however, this is an inconvenient unit, and we often use grams per cubic centimeter (g/cm 3 ) for the densities of solids and liquids, and grams per liter (g/L) for gases. Although there are exceptions, most liquids and solids have densities that range from about 0.7 g/cm 3 (the density of gasoline) to 19 g/cm 3 (the density of gold). The density of air is about 1.2 g/L. Table \(\PageIndex{3}\) shows the densities of some common substances.
| Solids | Liquids | Gases (at 25 °C and 1 atm) |
|---|---|---|
| ice (at 0 °C) 0.92 g/cm 3 | water 1.0 g/cm 3 | dry air 1.20 g/L |
| oak (wood) 0.60–0.90 g/cm 3 | ethanol 0.79 g/cm 3 | oxygen 1.31 g/L |
| iron 7.9 g/cm 3 | acetone 0.79 g/cm 3 | nitrogen 1.14 g/L |
| copper 9.0 g/cm 3 | glycerin 1.26 g/cm 3 | carbon dioxide 1.80 g/L |
| lead 11.3 g/cm 3 | olive oil 0.92 g/cm 3 | helium 0.16 g/L |
| silver 10.5 g/cm 3 | gasoline 0.70–0.77 g/cm 3 | neon 0.83 g/L |
| gold 19.3 g/cm 3 | mercury 13.6 g/cm 3 | radon 9.1 g/L |
While there are many ways to determine the density of an object, perhaps the most straightforward method involves separately finding the mass and volume of the object, and then dividing the mass of the sample by its volume. In the following example, the mass is found directly by weighing, but the volume is found indirectly through length measurements.
\[\text { density }=\frac{\text { mass }}{\text { volume }}\]
Example \(\PageIndex{1}\): Calculation of Density
Gold—in bricks, bars, and coins—has been a form of currency for centuries. In order to swindle people into paying for a brick of gold without actually investing in a brick of gold, people have considered filling the centers of hollow gold bricks with lead to fool buyers into thinking that the entire brick is gold. It does not work: Lead is a dense substance, but its density is not as great as that of gold, 19.3 g/cm 3 . What is the density of lead if a cube of lead has an edge length of 2.00 cm and a mass of 90.7 g?
Solution
The density of a substance can be calculated by dividing its mass by its volume. The volume of a cube is calculated by cubing the edge length.
\[\text {volume of lead cube }=2.00 \; \text{cm} \times 2.00 \; \text{cm} \times 2.00 \; \text{cm} =8.00 \; \text{cm}^3 \nonumber \]
\[\text { density }=\dfrac{\text{mass }}{\text {volume }}=\dfrac{90.7\; \text{g} }{8.00 \; \text{cm}^3}=11.3\; \text{g} / \text{cm}^3 \nonumber \]
(We will discuss the reason for rounding to the first decimal place in the next section.)
Exercise \(\PageIndex{1}\)
- To three decimal places, what is the volume of a cube (cm 3 ) with an edge length of 0.843 cm?
- If the cube in part (a) is copper and has a mass of 5.34 g, what is the density of copper to two decimal places?
- Answer
-
- 0.599 cm 3
- 8.91 g/cm 3
Link to Learning
To learn more about the relationship between mass, volume, and density, use this interactive simulator to explore the density of different materials.
Example \(\PageIndex{2}\): Using Displacement of Water to Determine Density
This exercise uses a simulation to illustrate an alternative approach to the determination of density that involves measuring the object’s volume via displacement of water. Use the simulator to determine the densities iron and wood.
Solution
Click the “turn fluid into water” button in the simulator to adjust the density of liquid in the beaker to 1.00 g/mL. Remove the red block from the beaker and note the volume of water is 25.5 mL. Select the iron sample by clicking “iron” in the table of materials at the bottom of the screen, place the iron block on the balance pan, and observe its mass is 31.48 g. Transfer the iron block to the beaker and notice that it sinks, displacing a volume of water equal to its own volume and causing the water level to rise to 29.5 mL. The volume of the iron block is therefore:
\[v_{\text {iron }}=29.5 mL -25.5 mL =4.0 mL \nonumber \]
The density of the iron is then calculated to be:
\[\text { density }=\frac{\text { mass }}{\text { volume }}=\frac{31.48\,\text{g} }{4.0\,\text{mL} }=7.9\, \text{g} / \text{mL} \nonumber \]
Remove the iron block from the beaker, change the block material to wood, and then repeat the mass and volume measurements. Unlike iron, the wood block does not sink in the water but instead floats on the water’s surface. To measure its volume, drag it beneath the water’s surface so that it is fully submerged.
\[\text { density }=\frac{\text { mass }}{\text { volume }}=\frac{1.95\, \text{g} }{3.0\, \text{mL }}=0.65 \,\text{g} / \text{mL} \nonumber \]
Note: The sink versus float behavior illustrated in this example demonstrates the property of “buoyancy”.
Exercise \(\PageIndex{2}\)
Following the water displacement approach, use the simulator to measure the density of the foam sample.
- Answer
-
0.230 g/mL


6.1: Objectives
After completing this chapter, you should
Finding the Factors of a Monomial
- be reminded of products of polynomials
- be able to determine a second factor of a polynomial given a first factor
Factoring a Monomial from a Polynomial
- be able to factor a monomial from a polynomial
The Greatest Common Factor
- understand more clearly the factorization process
- be able to determine the greatest common factor of two or more terms
Factoring by Grouping
- know how to factor a polynomial using the grouping method and when to try the grouping method
Factoring Two Special Products
- know the fundamental rules of factoring
- be able to factor the difference of two squares and perfect square trinomials
Factoring Trinomials with Leading Coefficient 1
- be able to factor trinomials with leading coefficient 1
- become familiar with some factoring hints
Factoring Trinomials with Leading Coefficient Other Than 1
- be able to factor trinomials with leading coefficient other than 1


12.2: Linear Equations
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Linear regression for two variables is based on a linear equation with one independent variable. The equation has the form:
\[y = a + b\text{x}\nonumber \]
where \(a\) and \(b\) are constant numbers. The variable \(x\) is the independent variable, and \(y\) is the dependent variable. Typically, you choose a value to substitute for the independent variable and then solve for the dependent variable.
Example \(\PageIndex{1}\)
The following examples are linear equations.
\[y = 3 + 2\text{x}\nonumber \]
\[y = -0.01 + 1.2\text{x}\nonumber \]
Exercise \(\PageIndex{1}\)
Is the following an example of a linear equation?
\[y = -0.125 - 3.5\text{x}\nonumber \]
- Answer
-
yes
The graph of a linear equation of the form \(y = a + b\text{x}\) is a straight line . Any line that is not vertical can be described by this equation.
Example \(\PageIndex{2}\)
Graph the equation \(y = -1 + 2\text{x}\).
Exercise \(\PageIndex{2}\)
Is the following an example of a linear equation? Why or why not?
- Answer
-
No, the graph is not a straight line; therefore, it is not a linear equation.
Example \(\PageIndex{3}\)
Aaron's Word Processing Service (AWPS) does word processing. The rate for services is $32 per hour plus a $31.50 one-time charge. The total cost to a customer depends on the number of hours it takes to complete the job.
Find the equation that expresses the total cost in terms of the number of hours required to complete the job.
Answer
Let \(x =\) the number of hours it takes to get the job done.
Let \(y =\) the total cost to the customer.
The $31.50 is a fixed cost. If it takes \(x\) hours to complete the job, then \((32)(x)\) is the cost of the word processing only. The total cost is: \(y = 31.50 + 32\text{x}\)
Exercise \(\PageIndex{3}\)
Emma’s Extreme Sports hires hang-gliding instructors and pays them a fee of $50 per class as well as $20 per student in the class. The total cost Emma pays depends on the number of students in a class. Find the equation that expresses the total cost in terms of the number of students in a class.
- Answer
-
\(y = 50 + 20\text{x}\)
Slope and Y-Intercept of a Linear Equation
For the linear equation \(y = a + b\text{x}\), \(b =\) slope and \(a = y\)-intercept. From algebra recall that the slope is a number that describes the steepness of a line, and the \(y\)-intercept is the \(y\) coordinate of the point \((0, a)\) where the line crosses the \(y\)-axis.
Example \(\PageIndex{4}\)
Svetlana tutors to make extra money for college. For each tutoring session, she charges a one-time fee of $25 plus $15 per hour of tutoring. A linear equation that expresses the total amount of money Svetlana earns for each session she tutors is \(y = 25 + 15\text{x}\).
What are the independent and dependent variables? What is the \(y\)-intercept and what is the slope? Interpret them using complete sentences.
Answer
The independent variable (\(x\)) is the number of hours Svetlana tutors each session. The dependent variable (\(y\)) is the amount, in dollars, Svetlana earns for each session.
The \(y\)-intercept is 25 (\(a = 25\)). At the start of the tutoring session, Svetlana charges a one-time fee of $25 (this is when \(x = 0\)). The slope is 15 (\(b = 15\)). For each session, Svetlana earns $15 for each hour she tutors.
Exercise \(\PageIndex{4}\)
Ethan repairs household appliances like dishwashers and refrigerators. For each visit, he charges $25 plus $20 per hour of work. A linear equation that expresses the total amount of money Ethan earns per visit is \(y = 25 + 20\text{x}\).
What are the independent and dependent variables? What is the \(y\)-intercept and what is the slope? Interpret them using complete sentences.
- Answer
-
The independent variable (\(x\)) is the number of hours Ethan works each visit. The dependent variable (\(y\)) is the amount, in dollars, Ethan earns for each visit.
The y -intercept is 25 (\(a = 25\)). At the start of a visit, Ethan charges a one-time fee of $25 (this is when \(x = 0\)). The slope is 20 (\(b = 20\)). For each visit, Ethan earns $20 for each hour he works.
Summary
The most basic type of association is a linear association. This type of relationship can be defined algebraically by the equations used, numerically with actual or predicted data values, or graphically from a plotted curve. (Lines are classified as straight curves.) Algebraically, a linear equation typically takes the form \(y = mx + b\), where \(m\) and \(b\) are constants, \(x\) is the independent variable, \(y\) is the dependent variable. In a statistical context, a linear equation is written in the form \(y = a + bx\), where \(a\) and \(b\) are the constants. This form is used to help readers distinguish the statistical context from the algebraic context. In the equation \(y = a + b\text{x}\), the constant b that multiplies the \(x\) variable (\(b\) is called a coefficient) is called the slope . The constant a is called the \(y\)-intercept.
The slope of a line is a value that describes the rate of change between the independent and dependent variables. The slope tells us how the dependent variable (\(y\)) changes for every one unit increase in the independent (\(x\)) variable, on average. The \(y\) -intercept is used to describe the dependent variable when the independent variable equals zero.
Formula Review
\(y = a + b\text{x}\) where a is the \(y\)-intercept and \(b\) is the slope. The variable \(x\) is the independent variable and \(y\) is the dependent variable.


8.4: The Law of Mass Action for Related and Simultaneous Equilibria
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Learning Objectives
- To know the relationship between the equilibrium constant and the rate constants for the forward and reverse reactions.
- To write an equilibrium constant expression for any reaction.
Relationship among Equilibrium Expressions
Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions.
To illustrate this procedure, let’s consider the reaction of \(N_2\) with \(O_2\) to give \(NO_2\). This reaction is an important source of the \(NO_2\) that gives urban smog its typical brown color. The reaction normally occurs in two distinct steps. In the first reaction (1), \(N_2\) reacts with \(O_2\) at the high temperatures inside an internal combustion engine to give \(NO\). The released \(NO\) then reacts with additional \(O_2\) to give \(NO_2\) (2). The equilibrium constant for each reaction at 100°C is also given.
- \(N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)}\;\; K_1=2.0 \times 10^{−25}\)
- \(2NO_{(g)}+O_{2(g)} \rightleftharpoons 2NO_{2(g)}\;\;\;K_2=6.4 \times 10^9\)
Summing reactions (1) and (2) gives the overall reaction of \(N_2\) with \(O_2\):
- \(N_{2(g)}+2O_{2(g)} \rightleftharpoons 2NO_{2(g)} \;\;\;K_3=?\)
The equilibrium constant expressions for the reactions are as follows:
\[K_1=\dfrac{[NO]^2}{[N_2][O_2]}\;\;\; K_2=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\;\;\; K_3=\dfrac{[NO_2]^2}{[N_2][O_2]^2}\]
What is the relationship between \(K_1\), \(K_2\), and \(K_3\), all at 100°C? The expression for \(K_1\) has \([NO]^2\) in the numerator, the expression for \(K_2\) has \([NO]^2\) in the denominator, and \([NO]^2\) does not appear in the expression for \(K_3\). Multiplying \(K_1\) by \(K_2\) and canceling the \([NO]^2\) terms,
\[ K_1K_2=\dfrac{\cancel{[NO]^2}}{[N_2][O_2]} \times \dfrac{[NO_2]^2}{\cancel{[NO]^2}[O_2]}=\dfrac{[NO_2]^2}{[N_2][O_2]^2}=K_3\]
Thus the product of the equilibrium constant expressions for \(K_1\) and \(K_2\) is the same as the equilibrium constant expression for \(K_3\):
\[K_3 = K_1K_2 = (2.0 \times 10^{−25})(6.4 \times 10^9) = 1.3 \times 10^{−15}\]
The equilibrium constant for a reaction that is the sum of two or more reactions is equal to the product of the equilibrium constants for the individual reactions. In contrast, recall that according to Hess’s Law , \(ΔH\) for the sum of two or more reactions is the sum of the ΔH values for the individual reactions.
Note
To determine \(K\) for a reaction that is the sum of two or more reactions, add the reactions but multiply the equilibrium constants.
Example \(\PageIndex{6}\)
The following reactions occur at 1200°C:
- \(CO_{(g)}+3H_{2(g)} \rightleftharpoons CH_{4(g)}+H_2O_{(g)} \;\;\;K_1=9.17 \times 10^{−2}\)
- \(CH_{4(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+4H_{2(g})\;\;\; K_2=3.3 \times 10^4\)
Calculate the equilibrium constant for the following reaction at the same temperature.
- \(CO_{(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)}\;\;\; K_3=?\)
Given : two balanced equilibrium equations, values of \(K\), and an equilibrium equation for the overall reaction
Asked for : equilibrium constant for the overall reaction
Strategy :
Arrange the equations so that their sum produces the overall equation. If an equation had to be reversed, invert the value of \(K\) for that equation. Calculate \(K\) for the overall equation by multiplying the equilibrium constants for the individual equations.
Solution :
The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and 2:
\[CO_{(g)}+ \cancel{3H_{2(g)}} \rightleftharpoons \cancel{CH_{4(g)}} + H_2O_{(g)}\]
\[\cancel{CH_{4(g)}} +2H_2S_{(g)} \rightleftharpoons CS_{2(g)} + \cancel{3H_{2(g)}} + H_{2(g)}\]
\[ CO_{(g)} + 3H_{2(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)}\]
The values for \(K_1\) and \(K_2\) are given, so it is straightforward to calculate \(K_3\):
\[K_3 = K_1K_2 = (9.17 \times 10^{−2})(3.3 \times 10^4) = 3.03 \times 10^3\]
Exercise \(\PageIndex{6}\)
In the first of two steps in the industrial synthesis of sulfuric acid, elemental sulfur reacts with oxygen to produce sulfur dioxide. In the second step, sulfur dioxide reacts with additional oxygen to form sulfur trioxide. The reaction for each step is shown, as is the value of the corresponding equilibrium constant at 25°C. Calculate the equilibrium constant for the overall reaction at this same temperature.
- \(\frac{1}{8}S_{8(s)}+O_{2(g)} \rightleftharpoons SO_{2(g)}\;\;\; K_1=4.4 \times 10^{53}\)
- \(SO_{2(g)}+\frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_2=2.6 \times 10^{12}\)
- \(\frac{1}{8}S_{8(s)}+\frac{3}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_3=?\)
- Answer
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\(K_3 = 1.1 \times 10^{66}\)
Summary
An equilibrium system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions.


9.4: Preview
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When you hear the word “curriculum” what comes to mind?
As we have learned in previous chapters, developing relationships, as well as understanding the developmental stages and individual interests and skills of children is crucial to effective teaching. This is accomplished through interactions and both informal and formal observations with the children in our care. This information will form the cornerstone of what is called “curriculum”, which includes both the planned and unplanned experiences that occur throughout the day. While you will devote an entire course to understanding curriculum in ECE 104 (Introduction to Curriculum in Early Childhood), we will visit some of the basic concepts below.


8.5: Equilibrium
- Define chemical equilibrium and the equilibrium constant, K .
- Recognize chemical equilibrium as a dynamic process.
Consider the following reaction occurring in a closed container (so that no material can go in or out):
H 2 + I 2 → 2HI
This is the reaction between elemental hydrogen and elemental iodine to make hydrogen iodide. The way the equation is written, we are led to believe that the reaction goes to completion, that all the H 2 and the I 2 react to make HI.
However, this is not the case. The reverse chemical reaction is also taking place:
2HI → H 2 + I 2
It acts to undo what the first reaction does. Eventually, the reverse reaction proceeds so quickly that it matches the speed of the forward reaction. When that happens, the reaction has reached chemical equilibrium , the point at which the forward and reverse processes balance each other's progress.
Because two opposing processes are occurring at once, it is conventional to represent an equilibrium using a double arrow, like this:
\[H_{2}+I_{2}\rightleftharpoons 2HI\nonumber \]
The double arrow implies that the reaction is going in both directions. Note that the reaction must still be balanced.
Write the equilibrium equation that exists between calcium carbonate as a reactant and calcium oxide and carbon dioxide as products.
Solution
As this is an equilibrium situation, a double arrow is used. The equilibrium equation is written as follows:
\[CaCO_{3}\rightleftharpoons CaO+CO_{2}\nonumber \]
Write the equilibrium equation between elemental hydrogen and elemental oxygen as reactants and water as the product.
- Answer
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\[2H_{2}+O_{2}+\rightleftharpoons 2H_{2}O\nonumber \]
One thing to note about equilibrium is that the reactions do not stop; both the forward reaction and the reverse reaction continue to occur. They both occur at the same rate, so any overall change by one reaction is canceled by the reverse reaction. We say that chemical equilibrium is dynamic , rather than static.The concentration of reactants and products does not change over time.
Both reactions are occurring simultaneously, the equilibrium can be written backward. For example, representing an equilibrium as
\[H_{2}+I_{2}\rightleftharpoons 2HI\nonumber \]
is the same thing as representing the same equilibrium as
\[2HI\rightleftharpoons H_{2}+I_{2}\nonumber \]
The reaction must be at equilibrium for this to be the case.
The Equilibrium Constant
In the mid 1860s, Norwegian scientists C. M. Guldberg and P. Waage noted a peculiar relationship between the amounts of reactants and products in an equilibrium. No matter how many reactants they started with, a certain ratio of reactants and products was achieved at equilibrium. Today, we call this observation the law of mass action . It relates the amounts of reactants and products at equilibrium for a chemical reaction. For a general chemical reaction occurring in solution,
\[aA+bB\rightleftharpoons cC+dD\nonumber \]
the equilibrium constant , also known as K eq , is defined by the following expression:
\[K_{eq}=\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\nonumber \]
where [A] is the molar concentration of species A at equilibrium, and so forth. The coefficients a , b , c , and d in the chemical equation become exponents in the expression for K eq . The K eq is a characteristic numerical value for a given reaction at a given temperature; that is, each chemical reaction has its own characteristic K eq . The concentration of each reactant and product in a chemical reaction at equilibrium is related ; the concentrations cannot be random values, but they depend on each other. The numerator of the expression for K eq has the concentrations of every product (however many products there are), while the denominator of the expression for K eq has the concentrations of every reactant, leading to the common products over reactants definition for the K eq .
Balance the reaction equation and then write the corresponding equilibrium constant expression:
CO(g) + O 2 (g) ⇌ CO 2 (g)
Solution
The balanced chemical equation is:
2CO(g) + O 2 (g) ⇌ 2CO 2 (g)
The equilibrium constant expression is:
Many chemical reactions involve substances in the solid or liquid phases. For example, a particular chemical reaction is represented as follows:
\[\ce{2NaHCO3(s) <=> Na2CO3(s) + CO2(g) + H2O(l)} \]
This chemical equation includes all three phases of matter. This kind of equilibrium is called a heterogeneous equilibrium because there is more than one phase present.
The rule for heterogeneous equilibria is as follows: Do not include the concentrations of pure solids and pure liquids in K eq expressions. See example 3 and 4 below.
Write the equilibrium constant expression for the following reaction:
C(s) + H 2 O(g) ⇌ CO(g) + H 2 (g)
Solution
The equilibrium constant expression is:
The concentration of the solid C(s) is the number of moles present in a given volume, a value that never changes at a given temperature, so the concentration of solid reactants or products are omitted in the equilibrium constant expression.
Write the equilibrium constant expression for the following reaction:
H 2 SO 4 (aq) + H 2 O(l) ⇌ HSO 4 - (aq) + H 3 O + (aq)
Solution
The equilibrium constant expression is:
The solvent (H 2 O) is left out of the equilibrium constant expression because the concentration of the solvent is very high and does not change significantly during a reaction.


5.4: Effusion and Diffusion of Gases
- Define and explain effusion and diffusion
- State Graham’s law and use it to compute relevant gas properties
If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule
In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in Figure \(\PageIndex{1}\)). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly—regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in Figure \(\PageIndex{1}\). The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs).
We are often interested in the rate of diffusion , the amount of gas passing through some area per unit time:
The diffusion rate depends on several factors: the concentration gradient (the increase or decrease in concentration from one point to another); the amount of surface area available for diffusion; and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation.
A process involving movement of gaseous species similar to diffusion is effusion , the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum (Figure \(\PageIndex{1}\)). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same.
If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (Figure \(\PageIndex{2}\)). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham’s law of effusion : The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles :
This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles:
Application: Use of Diffusion for Uranium Enrichment
Gaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only 0.72% of 235 U, the kind of uranium that is “fissile,” that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is 2–5% 235 U, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham’s law. In a gaseous diffusion enrichment plant, uranium hexafluoride (UF 6 , the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The 235 UF 6 molecules have a higher average speed and diffuse through the barrier a little faster than the heavier 238 UF 6 molecules. The gas that has passed through the barrier is slightly enriched in 235 UF 6 and the residual gas is slightly depleted. The small difference in molecular weights between 235 UF 6 and 238 UF 6 only about 0.4% enrichment, is achieved in one diffuser (Figure \(\PageIndex{4}\)). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained.
Figure \(\PageIndex{4}\): In a diffuser, gaseous UF 6 is pumped through a porous barrier, which partially separates 235 UF 6 from 238 UF 6 The UF 6 must pass through many large diffuser units to achieve sufficient enrichment in 235 U.
The large scale separation of gaseous 235 UF 6 from 238 UF 6 was first done during the World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about 10 –6 cm in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive UF 6 .
Summary
Gaseous atoms and molecules move freely and randomly through space. Diffusion is the process whereby gaseous atoms and molecules are transferred from regions of relatively high concentration to regions of relatively low concentration. Effusion is a similar process in which gaseous species pass from a container to a vacuum through very small orifices. The rates of effusion of gases are inversely proportional to the square roots of their densities or to the square roots of their atoms/molecules’ masses (Graham’s law).
Key Equations
- \(\textrm{rate of diffusion}=\dfrac{\textrm{amount of gas passing through an area}}{\textrm{unit of time}}\)
- \(\dfrac{\textrm{rate of effusion of gas B}}{\textrm{rate of effusion of gas A}}=\dfrac{\sqrt{m_B}}{\sqrt{m_A}}=\dfrac{\sqrt{ℳ_B}}{\sqrt{ℳ_A}}\)
Summary
- diffusion
- movement of an atom or molecule from a region of relatively high concentration to one of relatively low concentration (discussed in this chapter with regard to gaseous species, but applicable to species in any phase)
- effusion
- transfer of gaseous atoms or molecules from a container to a vacuum through very small openings
- Graham’s law of effusion
- rates of diffusion and effusion of gases are inversely proportional to the square roots of their molecular masses
- mean free path
- average distance a molecule travels between collisions
- rate of diffusion
- amount of gas diffusing through a given area over a given time
Contributors and Attributions
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Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>).


10.2: Appendix B - Solutions to Selected Exercises
1: Mathematics and Problem-Solving
1. 18/230 = 0.07826 = about 7.8%
3. €250(0.23) =€ 57.50 in VAT
5. $15000(5.57) = $83,550
7. absolute increase: 1050. Relative: 1050/3250 = 0.323 = 32.3% increase
9. a. 2200 – 2200(0.15) = 2200(0.85) = $1870
b. Yes, their goal was to decrease by at least 15%. They exceeded their goal.
11. Dropping by 6% is the same as keeping 94%. a(0.94) = 300. a = 319.15. Attendance was about 319 before the drop.
13. a. Kaplan’s enrollment was 64.3% larger than Walden’s. 30510
b. Walden’s enrollment was 39.1% smaller than Kaplan’s.
c. Walden’s enrollment was 60.9% of Kaplan’s.
15. If the original price was $100, the basic clearance price would be $100 – $100(0.60) = $40. The additional markdown would bring it to $40 - $40(0.30) = $28. This is 28% of the original price.
17. These are not comparable; “a” is using a base of all Americans and is talking about health insurance from any source, while “b” is using a base of adults and is talking specifically about health insurance provided by employers.
21. These statements are equivalent, if we assume the claim in “a” is a percentage point increase, not a relative change. Certainly these messages are phrased to convey different opinions of the levy. We are told the new rate will be $9.33 per $1000, which is 0.933% tax rate. If the original rate was 0.833% (0.1 percentage point lower), then this would indeed be a 12% relative increase.
23. 20% of 30% is 30%(0.20) = 6%, a 6 percentage point decrease.
25. Probably not, unless the final is worth 50% of the overall class grade. If the final was worth 25% of the overall grade, then a 100% would only raise her average to 77.5%
27. $4/10 pounds = $0.40 per pound (or 10 pounds/$4 = 2.5 pounds per dollar)
29. x = 15
31. 2.5 cups
33. 74 turbines
35. 96 inches
37. $6000
39. 55.6 meters
43. The population density of the US is 84 people per square mile. The density of India is about 933 people per square mile. The density of India is about 11 times greater than that of the U.S.
49. The oil in the spill could produce 93.1 million gallons of gasoline. Each car uses about 600 gallons a year. That would fuel 155,167 cars for a year.
53. An answer around 100-300 gallons would be reasonable
57. 156 million miles
59. The time it takes the light to reach you is so tiny for any reasonable distance that we can safely ignore it. 750 miles/hr is about 0.21 miles/sec. If the sound takes 4 seconds to reach you, the lightning is about 0.84 miles away. In general, the lightning will be 0.21 n miles away, which is often approximated by dividing the number of seconds by 5.
61. About 8.2 minutes
63. Four cubic yards (or 3.7 if they sell partial cubic yards)
3: Mathematics and Finance
1. A = 200 + .05(200) = $210
3. I=200. t = 13/52 (13 weeks out of 52 in a year). \(P_0\) = 9800
200 = 9800(r)(13/52) r = 0.0816 = 8.16% annual rate
5. \(P_{10}=300(1+.05 / 1)^{10(1)}=\$ 488.67\)
7. a. \(P_{20}=2000(1+.03 / 12)^{20(12)}=\$ 3641.51\) in 20 years
b. 3641.51 – 2000 = $1641.51 in interest
9. \(P_{8}=P_{0}(1+.06 / 12)^{8(12)}=6000 \). \(P_{0}=\$ 3717.14\) would be needed
11. a. \(P_{30}=\dfrac{200\left((1+0.03 / 12)^{30(12)}-1\right)}{0.03 / 12}=\$ 116,547.38\)
b. 200(12)(30) = $72,000
c. $116,547.40 - $72,000 = $44,547.38 of interest
13. a. \(P_{30}=800,000=\dfrac{d\left((1+0.06 / 12)^{30(12)}-1\right)}{0.06 / 12}\) d = $796.40 each month
b. $796.40(12)(30) = $286,704
c. $800,000 - $286,704 = $513,296 in interest
15. a. \(P_{0}=\dfrac{30000\left(1-(1+0.08 / 1)^{-25(1)}\right)}{0.08 / 1}=\$ 320,243.29\)
b. 30000(25) = $750,000
c. $750,000 - $320,243.29 = $429,756.71
17. \(P_{0}=500,000=\dfrac{d\left(1-(1+0.06 / 12)^{-20(12)}\right)}{0.06 / 12}\) d = $3582.16 each month
19. a. \(P_{0}=\dfrac{700\left(1-(1+0.05 / 12)^{-30(12)}\right)}{0.05 / 12}=\) a $130,397.13 loan
b. 700(12)(30) = $252,000
c. $252,200 - $130,397.13 = $121,602.87 in interest
21. \(P_{0}=25,000=\dfrac{d\left(1-(1+0.02 / 12)^{-48}\right)}{0.02 / 12}=\$ 542.38\) a month
23. a. Down payment of 10% is $20,000, leaving $180,000 as the loan amount
b. \(P_{0}=180,000=\dfrac{d\left(1-(1+0.05 / 12)^{-30(12)}\right)}{0.05 / 12}\), d = $966.28 a month
c. \(P_{0}=180,000=\dfrac{d\left(1-(1+0.06 / 12)^{-30(12)}\right)}{0.06 / 12}\), d = $1079.19 a month
25. First we find the monthly payments:
\(P_{0}=24,000=\dfrac{d\left(1-(1+0.03 / 12)^{-5(12)}\right)}{0.03 / 12}\), d = $431.25
Remaining balance: \(P_{0}=\dfrac{431.25\left(1-(1+0.03 / 12)^{-2(12)}\right)}{0.03 / 12}=\$ 10,033.45\)
27. \(\begin{aligned}
6000(1+0.04 / 12)^{12 N}&=10000\\
(1.00333)^{12 N}&=1.667\\
\log \left((1.00333)^{12 N}\right)&=\log (1.667)\\
12 N \log (1.00333)&=\log (1.667)\\
N&=\dfrac{\log (1.667)}{12 \log (1.00333)}=\text { about } 12.8 \text { year }
\end{aligned} \)
29. \(\begin{aligned}
3000&=\dfrac{60\left(1-(1+0.14 / 12)^{-12 N}\right)}{0.14 / 12}\\
3000(0.14 / 12)&=60\left(1-(1.0117)^{-12 N}\right)\\
\dfrac{3000(0.14 / 12)}{60}&=0.5833=1-(1.0117)^{-12 N}\\
0.5833-1&=-(1.0117)^{-12 N}\\
-(0.5833-1)&=(1.0117)^{-12 N}\\
\log (0.4167)&=\log \left((1.0117)^{-12 N}\right)\\
\log (0.4167)&=-12 N \log (1.0117)\\
N&=\dfrac{\log (0.4167)}{-12 \log (1.0117)}=\text { about } 6.3 \text { years }
\end{aligned}\)
31. First 5 years: \(P_{5}=\dfrac{50\left((1+0.08 / 12)^{5(12)}-1\right)}{0.08 / 12}\) = $3673.84
Next 25 years: \(3673.84(1+.08 / 12)^{25(12)}=\$ 26,966.65\)
33. Working backwards, \(P_{0}=\dfrac{10000\left(1-(1+0.08 / 4)^{-10(4)}\right)}{0.08 / 4}=\$ 273,554.79\) needed at retirement. To end up with that amount of money, \(273,554.70=\dfrac{d\left((1+0.08 / 4)^{15(4)}-1\right)}{0.08 / 4}\). He’ll need to contribute d = $2398.52 a quarter.
5: Data and Statistics
1. a. Population is the current representatives in the state’s congress
b. 106
c. the 28 representatives surveyed
d. 14 out of 28 = ½ = 50%
e. We might expect 50% of the 106 representatives = 53 representatives
3. This suffers from leading question bias
5. This question would likely suffer from a perceived lack of anonymity
7. This suffers from leading question bias
9. Quantitative
11. Observational study
13. Stratified sample
15. a. Group 1, receiving the vaccine
b. Group 2 is acting as a control group. They are not receiving the treatment (new vaccine).
c. The study is at least blind. We are not provided enough information to determine if it is double-blind.
d. This is a controlled experiment
17. a. Census
b. Observational study
6: Growth and Decay
1. a. \(P_{0}=20\). \(P_{n}=P_{n-1}+5\)
b. \(P_{n}=20+5 n\)
3. a. \(P_{1}=P_{0}+15=40+15=55\). \(P_{2}=55+15=70\)
b. \(P_{n}=40+15 n\)
c. \(P_{10}=40+15(10)=190\) thousand dollars
d. \(40 + 15n = 100\) when \(n = 4\) years.
5. Grew 64 in 8 weeks: 8 per week
a. \(P_{n}=3+8 n\)
b. \(187 = 3 + 8n\). \(n = 23\) weeks
7. a. \(P_0 = 200\) (thousand), \(P_{n}=(1+.09) P_{n-1}\) where \(n\) is years after 2000
b. \(P_{n}=200(1.09)^{n}\)
c. \(P_{16}=200(1.09)^{16}=794.061\) (thousand) = 794,061
d. \(200(1.09)^{n}=400\). \(n=\log (2) / \log (1.09)=8.043\). In 2008.
9. Let \(n=0\) be 1983. Pn = 1700(2.9)n. 2005 is n=22. P22 = 1700(2.9)22 = 25,304,914,552,324 people. Clearly not realistic, but mathematically accurate.
11. If \(n\) is in hours, better to start with the explicit form. \(P_{0}=300\). \(P_{4}=500=300(1+r)^{4}\)
\(500 / 300=(1+r)^{4}\). \(1+r=1.136\). \(r=0.136\)
a. \(P_{0}=300\). \(P_{n}=(1.136) P_{n-1}\)
b. \(P_{n}=300(1.136)^{n}\)
c. \(P_{24}=300(1.136)^{24}=6400\) bacteria
d. \(300(1.136)^{n}=900\). \(n=\log (3) / \log (1.136)=\) about 8.62 hours
13. a. \(P_{0}=100 \quad P_{n}=P_{n-1}+0.70\left(1-P_{n-1} / 2000\right) P_{n-1}\)
b. \(P_{1}=100+0.70(1-100 / 2000)(100)=166.5\)
c. \(P_{2}=166.5+0.70(1-166.5 / 2000)(166.5)=273.3\)
15. To find the growth rate, suppose \(n=0\) was 1968. Then \(P_0\) would be 1.60 and \(P_{8}=2.30=1.60(1+r)^{8}\), \(r = 0.0464\). Since we want \(n=0\) to correspond to 1960, then we don’t know \(P_0\), but \(P_8\) would \(1.60 = P_{0}(1.0464)^{8}\). \(P_{0}=1.113\)
a. \(P_{n}=1.113(1.0464)^{n}\)
b. \(P_{0}=\$ 1.113\), or about $1.11
c. 1996 would be \(n=36\). \(P_{36}=1.113(1.0464)^{36}=\$ 5.697\). Actual is slightly lower.
17. The population in the town was 4000 in 2005, and is growing by 4% per year.
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A logarithm is an exponent. Specifically, it is the exponent to which a base \(b\) is raised to produce a given value. In the expressions given, the base \(b\) has the same value. The exponent, \(y\), in the expression by can also be written as the logarithm, \(\log _{b} x\), and the value of \(x\) is the result of raising \(b\) to the power of \(y\).
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Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation \(b^{y}=x\), and then properties of exponents can be applied to solve for \(x\).
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The natural logarithm is a special case of the logarithm with base b in that the natural log always has base \(e\). Rather than notating the natural logarithm as \(\log _{e}(x)\), the notation used is \(\ln (x)\).
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\(a^c=b \)
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\(x^y=64\)
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\(x^y=64\)
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\(15^b=a\)
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\(13^a=142\)
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\(e^n=w\)
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2
8: Mathematics and Politics
Apportionment
1. a. Math: 6, English: 5, Chemistry: 3, Biology: 1
b. Math: 7, English: 5, Chemistry: 2, Biology: 1
c. Math: 6, English: 5, Chemistry: 3, Biology: 1
d. Math: 6, English: 5, Chemistry: 3, Biology: 1
e. Math: 6, English: 5, Chemistry: 2, Biology: 2
3. a. Morning: 1, Midday: 5, Afternoon: 6, Evening: 8
b. Morning: 1, Midday: 4, Afternoon: 7, Evening: 8
c. Morning: 1, Midday: 5, Afternoon: 6, Evening: 8
d. Morning: 1, Midday: 5, Afternoon: 6, Evening: 8
e. Morning: 2, Midday: 5, Afternoon: 6, Evening: 7
5. a. Alice: 18, Ben: 14, Carlos: 4
b. Alice: 19, Ben: 14, Carlos: 3
c. Alice: 19, Ben: 14, Carlos: 3
d. Alice: 19, Ben: 14, Carlos: 3
e. Alice: 18, Ben: 14, Carlos: 4
7. a. A: 40, B: 24, C: 15, D: 30, E: 10
b. A: 41, B: 24, C: 14, D: 30, E: 10
c. A: 40, B: 24, C: 15, D: 30, E: 10
d. A: 40, B: 24, C: 15, D: 30, E: 10
e. A: 40, B: 24, C: 15, D: 29, E: 11
Describing Data
1. a. Different tables are possible
|
Score |
Frequency |
|---|---|
|
30 |
1 |
|
40 |
0 |
|
50 |
4 |
|
60 |
3 |
|
70 |
6 |
|
80 |
5 |
|
90 |
2 |
|
100 |
3 |
b. This is technically a bar graph, not a histogram:
c.
3. a. 5+3+4+2+1 = 15
b. 5/15 = 0.3333 = 33.33%
5. Bar is at 25%. 25% of 20 = 5 students earned an A
7. a. (7.25+8.25+9.00+8.00+7.25+7.50+8.00+7.00)/8 = $7.781
b. In order, 7.50 and 8.00 are in middle positions. Median = $7.75
c. 0.25*8 = 2. Q1 is average of 2nd and 3rd data values: $7.375
0.75*8 = 6. Q3 is average of 6th and 7th data values: $8.125
5-number summary: $7.00, $7.375, $7.75, $8.125, $9.00
d. 0.637
9. a. (5*0 + 3*1 + 4*2 + 2*3 + 1*5)/15 = 1.4667
b. Median is 8th data value: 1 child
c. 0.25*15 = 3.75. Q1 is 4th data value: 0 children
0.75*15 = 11.25. Q3 is 12th data value: 2 children
5-number summary: 0, 0, 1, 2, 5
d.
11. Kendra makes $90,000. Kelsey makes $40,000. Kendra makes $50,000 more.
Fractals
3.
5.
7.
11. Four copies of the Koch curve are needed to create a curve scaled by 3. \[D=\dfrac{\log (4)}{\log (3)} \approx 1.262 \nonumber \]
13. Eight copies of the shape are needed to make a copy scaled by 3.
Voting Theory
1.
|
Number of voters |
3 |
3 |
1 |
3 |
2 |
|
1st choice |
A |
A |
B |
B |
C |
|
2nd choice |
B |
C |
A |
C |
A |
|
3rd choice |
C |
B |
C |
A |
B |
3. a. 9+19+11+8 = 47
b. 24 for majority; 16 for plurality (though a choice would need a minimum of 17 votes to actually win under the Plurality method)
c. Atlanta, with 19 first-choice votes
d. Atlanta 94, Buffalo 111, Chicago 77. Winner: Buffalo
e. Chicago eliminated, 11 votes go to Buffalo. Winner: Buffalo
f. A vs B: B. A vs C: A. B vs C: B. B gets 2 pts, A 1 pt. Buffalo wins.
5. a. 120+50+40+90+60+100 = 460
b. 231 for majority; 116 for plurality
c. A with 150 first choice votes
d. A 1140, B 1060, C 1160, D 1240. Winner: D
e. B eliminated, votes to C. D eliminated, votes to A. Winner: A
f. A vs B: B. A vs C: A. A vs D: D. B vs C: C. B vs D: D. C vs D: C
A 1pt, B 1pt, C 2pt, D 2pt. Tie between C and D.
Winner would probably be C since C was preferred over D
7. a. 33
b. 17
9. Yes, B
11. B, with 17 approvals
13. Independence of Irrelevant Alternatives Criterion
15. Condorcet Criterion
Weighted Voting
1. a. 9 players
b. 10+9+9+5+4+4+3+2+2 = 48
c. 47
3. a. 9, a majority of votes
b. 17, the total number of votes
c. 12, which is 2/3 of 17, rounded up
5. a. P1 is a dictator (can reach quota by themselves)
b. P1, since dictators also have veto power
c. P2, P3, P4
7. a. none
b. P1
c. none
9. a. 11+7+2 = 20
b. P1 and P2 are critical
11. Winning coalitions, with critical players underlined:
{ P1 , P2 } { P1 ,P2,P3} { P1,P2 ,P4} { P1 ,P2,P3,P4} { P1,P3 } { P1,P3 ,P4}
P1: 6 times, P2: 2 times, P3: 2 times, P4: 0 times. Total: 10 times
Power: P1: 6/10 = 60%, P2: 2/10 = 20%, P3: 2/10 = 20%, P4: 0/10 = 0%
13. a. { P1 } { P1 ,P2} { P1 ,P3} { P1 ,P4} { P1 ,P2,P3} { P1 ,P2,P4} { P1 ,P3,P4} { P1 ,P2,P3,P4}
P1: 100%, P2: 0%, P3: 0%, P4: 0%
b. { P1,P2 } { P1,P3 } { P1,P4 } { P1 ,P2,P3} { P1 ,P2,P4} { P1 ,P3,P4} { P1 ,P2,P3,P4}
P1: 7/10 = 70%, P2: 1/10 = 10%, P3: 1/10 = 10%, P4: 1/10 = 10%
c. { P1,P2 } { P1,P3 } { P1 ,P2,P3} { P1 , P2 ,P4} { P1 , P3 ,P4} { P1 ,P2,P3,P4}
P1: 6/10 = 60%, P2: 2/10 = 20%, P3: 2/10 = 20%, P4: 0/10 = 0%
15. P3 = 5. P3+P2 = 14. P3+P2+P1 = 27, reaching quota. P1 is critical.
17. Sequential coalitions with pivotal player underlined
<P1, P2 ,P3> <P1, P3 ,P2> <P2, P1 ,P3> <P2, P3 ,P1> <P3, P1 ,P2> <P3, P2 ,P1>
P1: 2/6 = 33.3%, P2: 2/6 = 33.3%, P3: 2/6 = 33.3%
19. a. 6, 7
b. 8, given P1 veto power
c. 9, given P1 and P2 veto power
21. If adding a player to a coalition could cause it to reach quota, that player would also be critical in that coalition, which means they are not a dummy. So a dummy cannot be pivotal.
23. We know P2+P3 can’t reach quota, or else P1 wouldn’t have veto power.
P1 can’t reach quota alone.
P1+P2 and P1+P3 must reach quota or else P2/P3 would be dummy.
a. { P1 , P2 } { P1 , P3 } { P1 ,P2,P3}. P1: 3/5, P2: 1/5, P3: 1/5
b. <P1, P2 ,P3> <P1, P3 ,P2> <P2, P1 ,P3> <P2,P3, P1 > <P3, P1 ,P2> <P3,P2, P1 >
P1: 4/6, P2: 1/6, P3: 1/6
25. [4: 2, 1, 1, 1] is one of many possibilities
27. [56: 30, 30, 20, 20, 10]
29. [54: 10, 10, 10, 10, 10, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] is one of many possibilities
Contributors and Attributions
-
Saburo Matsumoto
CC-BY-4.0


4.1: Types of views used in drawings
Types of views used in drawings
The two main types of views (or “projections”) used in drawings are:
- pictorial
- orthographic
Pictorial views
Pictorial views show a 3-D view of how something should look when completed. There are three types of pictorial views:
- perspective
- isometric
- oblique
Perspective view
A perspective view presents a building or an object just as it would look to you. A perspective view has a vanishing point ; that is, lines that move away from you come together in the distance. For example, in Figure 1, we see a road and line of telephone poles. Even though the poles get smaller in their actual measurement, we recognize them as being the same size but more distant.