1.7: Water Treatment Facilities and Operations
Water treatment is one of the most critical steps in the drinking water process. It is critical to remove unwanted particles, inactivate harmful organisms, and treat to a level that complies with the Safe Drinking Water Act (SDWA). Most water treated within drinking water treatment plants comes from surface supply sources. As snow melts and rainfall moves along the ground surface into rivers, streams, and lakes, particles are picked up along the way. Therefore, ALL surface water to be used for domestic drinking water purposes must be treated. Surface water is also more susceptible to contamination because much of this contamination comes from “runoff”. Every time it rains or as snow melts, water is washed down from mountains, hillsides, roads, and other areas, picking up anything along the way. For example, imagine a farm next to a stream that feeds a local lake used as a drinking water storage reservoir. There is a potential for animal waste among other things making its way to the storage reservoir. Just the sediment from a lake’s hillside and the natural organic matter in a lake makes the water not suitable for drinking. Therefore, there are regulations within the Safe Drinking Water Act called Surface Water Treatment Rule (SWTR), which specifically addresses surface water treatment requirements. There have been several iterations of this rule to improve the quality of drinking water. The SWTR is crafted to prevent waterborne diseases caused by microorganisms. The rule requires water systems to filter and disinfect water from surface water sources to reduce the occurrence of unsafe levels of disease causing microbes.
One of the primary constituents the SWTR requires treatment and monitoring for is turbidity. Turbidity is the cloudiness or haziness seen in water samples and prevents the performance of filtration systems. Low turbidity requirements are used to protect against certain microbial contaminants, in particularly Cryptosporidium . Cryptosporidium is a protozoan that can cause gastrointestinal illness in humans, which is similar to many disease causing microorganisms. The lower the turbidity levels, the less likely there will be disease causing microorganisms. The SWTR also requires disinfection to protect against these pathogens. Disinfection chemicals, such as chlorine, are used to inactivate (kill) pathogens during the treatment process through a required residual present in the water throughout the distribution system.
Surface waters also have to be treated for aesthetic qualities. An aesthetic quality doesn’t typically have an effect on public health. These types of qualities are usually related to the appearance and palatability of the water. Does the water taste good, look good, and smell good? Taste, odor, and color are three qualities surface water quality is treated to improve.
Particles in Water
Particles in water can be broken into three (3) general categories: suspended solids, colloidal compounds, and dissolved solids. Microorganisms fall under the colloidal compound category. One of the main goals of a surface water treatment plant is to prevent the outbreak of disease from microorganisms. These disease-causing agents are referred to as pathogens. The three main pathogen categories are distinguished as bacteria, virus, and protozoa. In drinking water quality, these organisms are not specifically analyzed in samples. Specific organisms, viruses, and protozoa can be very costly and require extensive sampling techniques. Some of the more common pathogens, which can be found in water, are:
- Bacteria – Escherichia coli (E. coli) and Vibrio cholera
- Viruses – Enterovirus and Coronavirus
- Protozoa – Cryptosporidium parvum and Giardia lamblia
Instead of analyzing for these organisms specifically, a class of organisms called total coliforms (TC) is used analyzed instead. TC is referred to collectively as indicator organisms. The presence or absence of TC bacteria indicates the presence or absence of pathogens. The analytical test is easy, inexpensive, and very effective. Microorganisms are one class of “particles” removed or inactivated from water. Microorganisms are difficult to remove from water, therefore they are usually destroyed or inactivated through the treatment process.
In contrast to colloidal compounds, suspended solids are more easily removed. Much of the suspended material found in surface water supplies is referred to as turbidity. This turbidity comes from decaying plant material and soil as water runs across land and makes its way to streams and lakes. While suspended solids are not typically associated with disease, they can shield microorganisms from treatment processes and provide an overall poor quality to the water supply.
Dissolved compounds are the most difficult to remove because they are dissolved in the water. Think about sugar or salt mixed in a glass of water. Once it dissolves and mixes in the water a treatment process is needed to remove it from the water. These treatment processes are more complex than the standard drinking water treatment plant. There are a variety of processes including things such as ion exchange, membrane filtration, chemical absorption, and others. If the dissolved compounds do not pose a health threat or if they are below their corresponding Maximum Contaminant Level (MCL), they are not typically treated for in the drinking water treatment process.
Pre-Storage
Prior to entering a drinking water treatment plant, water is typically held in large storage reservoirs. These “reservoirs” (lakes) provide an area for large amounts of water to collect so the treatment plant has a constant source of supply. These reservoirs can also allow for some settling of solids and many double as a recreation space for fishing, boating, and water-skiing, etc.
Entering the Treatment Plant
As discussed in earlier chapters, surface water used for drinking is commonly stored in above ground lakes. These storage reservoirs are sometimes used for recreation such as fishing, boating, and water skiing. This recreational use is an important part of the economy and public use. However, they can add additional water quality problems for the water treatment plant. Water leaving these storage reservoirs needs to be free of large debris such as plants fish, trash, wood, etc. In order to take water from the storage reservoir and leave the larger debris behind there are “intake” structures. A surface water intake structure is a screened structure to let water through and keep larger items from entering the treatment plant. The screen size can vary depending on the use of the storage reservoir and the quality of the raw water. Raw water is the term used to identify water before treatment. The picture below shows an example.
The water entering the intake structure can either be pumped or can flow by gravity to the treatment plant depending on the location of the plant compared to the storage reservoir. Valves can be used to adjust the flow into the plant and the water is sometimes kept in an onsite storage tank leading into the plant.
Inside a Water Treatment Plant
Pre-Disinfection
Once raw water enters a treatment plant there may be a pre-disinfection process. Some plants use a disinfectant to kill pathogens and other microorganisms prior to the water going through the treatment process. This can be done to prevent certain organisms from entering the plant and having an effect on the various treatment processes. Or it can be done to remove precursors, such as total organic carbon, to prevent disinfection by-products from forming. Newer treatment plants will use ozone as the disinfectant primarily because it is an effective oxidizer and it does not leave a residual in the water throughout the treatment process.
Ozone is a trivalent form of oxygen. Simply put it is a compound of three (3) oxygen molecules (O3). If you have ever smelled an electrical spark or a lightning strike, chances are it is ozone that you smelled. It must be used at the point of generation. Unlike chlorine, which can be processed at a chemical plant (for example), a treatment plant using ozone must generate it on site. Ozone has a number of benefits over chlorine.
Ozone Compared to Chlorine
- Stronger disinfectant
- Does not contribute to disinfection by-products
- Kills a wider range of organisms
- Achieves removal of unwanted tastes and odors
- Reactions are more rapid
However, as mentioned previously, ozone does not leave a disinfectant residual in the water. This might be a good thing for a treatment plant, but ozone is not used in distribution systems because a residual is needed.
Conventional Treatment vs. Direct Filtration
There are two main types of drinking water treatment plants. There are conventional treatment plants and direct filtration plants. Each has benefits and drawbacks. All of which will be discussed in this text. It is important to remember that this text is an introductory text for a very general overview of the waterworks industry. Therefore, details of each treatment process discussed may be omitted and are provided in more specialized courses. The basic difference between a conventional water treatment plant and a direct filtration water treatment plant is a sedimentation basin. A conventional drinking water treatment plant has a sedimentation basin and a direct filtration plant does not. So, why would one be used over another? Below is an example of some of the “pros” and “cons” of a sedimentation basin.
Sedimentation Basin Pros
- Allows solids to settle out of the water prior to entering the filtration process
- Reduces the amount and duration of backwashing filters
Sedimentation Basin Cons
- Sedimentation basins are large and require a bigger treatment plant
- The sludge at the bottom of a sedimentation basin needs to be removed from time to time.
We will now look at the other processes typically found in a conventional and direct filtration drinking water treatment plants. The photos provided in this section are courtesy of the Michigan Department of Environmental Quality.
Coagulation
Coagulation is the process of chemical addition such as “Alum” to the water supply in order for small suspended particles to “stick” together forming floc. The chemicals added produce positive charges to neutralize the negative charges on the particles. The particles stick together becoming larger and larger during this process.
Flocculation
As the particles begin to stick together, the water is then sent through a series of tanks with “paddles.” These paddles are designed to slowly mix the water, bringing the particles together to form larger and larger particles called “floc.” The mixing process must be gentle enough to not break apart the floc back into smaller particles.
Sedimentation
In conventional drinking water treatment plants the sedimentation process allows for the forces of gravity to allow the floc to “settle” to the bottom of the basin. Not all the floc will settle. As you might expect, the large particles settle more rapidly than small particles. Also, the slower the water moves through the basin the more particles will settle out. Water in a direct filtration plant moves straight from the flocculation tanks to the next step…filtration.
Filtration
The filtration process is one of the most critical steps. Filters are commonly constructed in concrete boxes and contain sand and gravel. Sometimes other filter media is used, but many times sand is sufficient enough to remove the remaining suspended particles. The purpose of the gravel is to support the sand and prevent it from leaving the filter. Under the gravel is a structure called an “underdrain.” The purpose of the underdrain is to allow clear filtered water out of the filter while supporting the gravel and sand filter media. The process of backwashing will be discussed later.
Post-Disinfection
Prior to the water entering the distribution system, it is usually disinfected with a chlorine based disinfectant. Chlorine gas was and still is commonly used to provide a “free” chlorine residual throughout the distribution system. However, because of the water quality risk of disinfection byproducts, many water treatment plants are using a “total” chlorine residual by mixing chlorine and ammonia together. This disinfection process is referred to as chloramination.
How much disinfectant is required before water can be provided to customers for domestic use?
There are two “standards” when it comes to disinfection. There is a maximum residual disinfectant level (MRDL) and a minimum disinfectant level, which should be provided to the furthest areas within a distribution system. The MRDL is 4.0 mg/L and this level should not be exceeded in the water entering a distribution system. Typical free (chlorine) and total chlorine (chloramination) levels in water leaving a treatment plant can vary, but are commonly between 2.5 and 3.5 mg/L. The minimum residual level, which should be maintained in the furthest areas of a distribution system is 0.2 mg/L. Since microorganisms have the ability to multiply, a “residual” helps prevent regrowth and keeps the water free of pathogens. The previously mentioned dosage values typically provide enough disinfectant to kill/inactivate (disinfectant demand) remaining microorganisms and allow for a constant and minimal residual within the distribution system. The previous sentence used three (3) common terms associated with disinfection; dosage, demand, and residual.
- Dosage – is the amount of a disinfectant added.
- Demand – is the amount of disinfectant “used up” by the disinfection reducing agents in the water (microorganisms, organic matter, etc.)
- Residual – is the amount of disinfectant left in the water supply.
Therefore, the following formula is commonly used.
Dosage = Demand + Residual
The amount of disinfectant added to a water supply (dosage) minus the amount “used up” (demand) equals the amount remaining, which is referred to as the residual.
Storage
In order for treatment plants to provide large quantities of water to distribution systems, a large amount of storage is commonly required and provided at the treatment plant. This “post” storage can and often times be in the millions to tens of millions of gallons. Therefore, treatment plants require large areas of land for the entire treatment process and storage.
Additional Treatment Processes
If the influent quality of water coming into a treatment plant or if the water supply a distribution system is producing, (often times from groundwater wells), is contaminated and not within drinking water standards, the water will require additional treatment. Some of these processes include but are not limited to, ion exchange, membrane filtration, and air stripping. Ion exchange, for example, is the process of removing ions using water with opposite charged ions. In Water 050 and 052, you will take a more in depth look at each process and analyze the benefits and challenges a water treatment operator might encounter.
Careers in Drinking Water Treatment
Because many drinking water treatment plants are wholesale water providers, meaning they sell water to various water agencies, the extent of career opportunities and paths can be quite extensive. They can have careers ranging from field technicians, maintenance workers, treatment plant operators, engineers, water resource professionals, as well as a variety of office and policy related opportunities. For example, Metropolitan Water District (MWD) of California deliveries is one of the largest wholesaler water suppliers in the country. MWD provides approximately 1.7 billion gallons of water per day and is the largest contractor of the State Water Project. In addition to the responsibilities associated with operating a water treatment plant, MWD has an extensive distribution system. They also have the important task of securing sources of supply to ensure there is always enough water for the water retailers who purchase water from them. They have additional responsibilities in the areas of water conservation, legal matters, public relations, human resources, finance, administration…the list goes on and on. They employ almost 2,000 people with jobs ranging from laborers to scientists, to engineers and more. As you can imagine, trying to list and discuss the vast amount of opportunities for an organization like this would be a lengthy process. For an introductory course like this one it would be impractical. Therefore, we will just look at a smaller example of a drinking water treatment plant and focus on career opportunities available for students such as you.
Treatment Plant Operator
Water Treatment Plant Operators (WTPO) typically work various assigned shifts. It is not usually a typical “8 to 5” type of job. After all, treated, safe drinking water needs to be available to customers 24 hours a day, 7 days a week. Therefore, WTPO need to work around the clock to make sure the treatment process is functioning efficiently. Below is a list of some of the more common tasks and responsibilities of a WTPO.
- Ensure the treatment plant is operating
- Perform biological, chemical, and physical laboratory tests on water
- Interpret test results
- Monitor and read gauges, meters, charts, and other treatment performance indicators
- Monitor and inspect treatment process equipment and instrumentation
- Make adjustments and take corrective action on treatment process equipment
- Maintain operational and water quality records
- Prepare reports
- Participate in training programs
In addition, other responsibilities may include cleaning and disinfecting storage tanks, flushing pipelines, and conducting public tours of the treatment plant. Often, plants will have a variety of automated processes and the WTPO will monitor these processes on a computer screen making adjustments through a Supervisory Control and Data Acquisition System (SCADA).
Experience and educational qualifications will also depend on the size of the facility, but many entry level positions will require a minimum of a Grade 3 Treatment Operator certification (T3) or a Grade 2 Treatment Operator certification with the ability to obtain a T3 within a year of employment. Since treatment plants may also have an associated distribution system to get water to the utilities purchasing water from them, a Distribution Operator certification may also be required. The more education and experience you can obtain will always help your chances in landing a job. An associate’s or bachelor’s degree is sometimes listed as a “desirable” qualification.
Water Treatment Plant Maintenance Worker
Just as a distribution system needs maintaining, a water treatment plant also requires maintenance. Pumps and motors can malfunction. Pipes can leak. System process and equipment can breakdown requiring repairs or replacement. Therefore, most treatment plants will have a maintenance crew as part of the staff.
Water Quality and Laboratory Staff
Many water treatment plants will have their own drinking water laboratory to analyze samples throughout the treatment process ensuring proper function and for adjusting chemical doses. The staff might be in charge of collecting samples and running analytical instrumentation in the lab. They might also be responsible for writing reports and keeping track of water quality data for regulatory compliance.


8.1.1: Social Cognitions
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The culture that we live in has a significant impact on the way we think about and perceive our social worlds, so it is not surprising that people in different cultures would think about people and things somewhat differently. Social cognitions are the way we think about others, pay attention to social information, and use the information in our lives (consciously or unconsciously). In this section we will review several types of social cognitions including schemas, attributions, confirmation bias and the fundamental attribution error. We will also revisit analytic perception and holistic perception that we learned earlier in this chapter.


20.9.4: Nonrandom Mating and Environmental Variance
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Learning Objectives
- Explain how environmental variance and nonrandom mating can change gene frequencies in a population
Nonrandom Mating
If individuals nonrandomly mate with other individuals in the population, i.e. they choose their mate, choices can drive evolution within a population. There are many reasons nonrandom mating occurs. One reason is simple mate choice or sexual selection; for example, female peahens may prefer peacocks with bigger, brighter tails. Traits that lead to more matings for an individual lead to more offspring and through natural selection, eventually lead to a higher frequency of that trait in the population. One common form of mate choice, called positive assortative mating, is an individual’s preference to mate with partners that are phenotypically similar to themselves.
Another cause of nonrandom mating is physical location. This is especially true in large populations spread over large geographic distances where not all individuals will have equal access to one another. Some might be miles apart through woods or over rough terrain, while others might live immediately nearby.
Environmental Variance
Genes are not the only players involved in determining population variation. Phenotypes are also influenced by other factors, such as the environment. A beachgoer is likely to have darker skin than a city dweller, for example, due to regular exposure to the sun, an environmental factor. Some major characteristics, such as gender, are determined by the environment for some species.  TSD means that individuals develop into males if their eggs are incubated within a certain temperature range, or females at a different temperature range.
Geographic separation between populations can lead to differences in the phenotypic variation between those populations. Such geographical variation is seen between most populations and can be significant. One type of geographic variation, called a cline, can be seen as populations of a given species vary gradually across an ecological gradient.
Geographic variation in moose : This graph shows geographical variation in moose; body mass increase positively with latitude. Bergmann’s Rule is an ecologic principle which states that as latitude increases the body mass of a particular species increases. The data are taken from a Swedish study investigating the size of moose as latitude increases as shows the positive relationship between the two, supporting Bergmann’s Rule.
Species of warm-blooded animals, for example, tend to have larger bodies in the cooler climates closer to the earth’s poles, allowing them to better conserve heat. This is considered a latitudinal cline. Alternatively, flowering plants tend to bloom at different times depending on where they are along the slope of a mountain, known as an altitudinal cline.
If there is gene flow between the populations, the individuals will likely show gradual differences in phenotype along the cline. Restricted gene flow, on the other hand, can lead to abrupt differences, even speciation.
Contributions and Attributions
- Structural Biochemistry/Organismic and Evolutionary Biology. Provided by : Wikibooks. Located at : en.wikibooks.org/wiki/Structu...ionary_Biology . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Biology. October 22, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...ol11448/latest . License : CC BY: Attribution
- A-level Biology/Central Concepts/Classification, selection and evolution. Provided by : Wikibooks. Located at : en.wikibooks.org/wiki/A-level..._and_evolution . License : CC BY-SA: Attribution-ShareAlike
- Structural Biochemistry/Evolution of Populations. Provided by : Wikibooks. Located at : en.wikibooks.org/wiki/Structu...etic_Variation . License : CC BY-SA: Attribution-ShareAlike
- crossing over. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/crossing%20over . License : CC BY-SA: Attribution-ShareAlike
- genetic variation. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/genetic%20variation . License : CC BY-SA: Attribution-ShareAlike
- phenotypic variation. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/phenotypic%20variation . License : CC BY-SA: Attribution-ShareAlike
- Genetic Diversity. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Genetic_diversity . License : CC BY-SA: Attribution-ShareAlike
- Coquina variation3. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Co...variation3.jpg . License : CC BY-SA: Attribution-ShareAlike
- Cheetah genetic diversity. Provided by : Wikimedia. Located at : en.Wikipedia.org/wiki/File:Ch..._diversity.jpg . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Biology. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...ol11448/latest . License : CC BY: Attribution
- Genetic Drift. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Genetic_drift . License : CC BY-SA: Attribution-ShareAlike
- founder effect. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/founder_effect . License : CC BY-SA: Attribution-ShareAlike
- genetic drift. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/genetic_drift . License : CC BY-SA: Attribution-ShareAlike
- Coquina variation3. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Co...variation3.jpg . License : CC BY-SA: Attribution-ShareAlike
- Cheetah genetic diversity. Provided by : Wikimedia. Located at : en.Wikipedia.org/wiki/File:Ch..._diversity.jpg . License : CC BY-SA: Attribution-ShareAlike
- Founder effect with drift. Provided by : Wikimedia. Located at : en.Wikipedia.org/wiki/File:Fo...with_drift.jpg . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Population Genetics. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...e_19_02_02.png . License : CC BY: Attribution
- Random genetic drift chart. Provided by : Wikimedia. Located at : en.Wikipedia.org/wiki/Genetic...rift_chart.png . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Population Genetics. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...e_19_02_03.jpg . License : CC BY: Attribution
- OpenStax College, Biology. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...ol11448/latest . License : CC BY: Attribution
- OpenStax College, Biology. October 23, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...ol11448/latest . License : CC BY: Attribution
- gene flow. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/gene%20flow . License : CC BY-SA: Attribution-ShareAlike
- mutation. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/mutation . License : CC BY-SA: Attribution-ShareAlike
- Coquina variation3. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Co...variation3.jpg . License : CC BY-SA: Attribution-ShareAlike
- Cheetah genetic diversity. Provided by : Wikimedia. Located at : en.Wikipedia.org/wiki/File:Ch..._diversity.jpg . License : CC BY-SA: Attribution-ShareAlike
- Founder effect with drift. Provided by : Wikimedia. Located at : en.Wikipedia.org/wiki/File:Fo...with_drift.jpg . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Population Genetics. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...e_19_02_02.png . License : CC BY: Attribution
- Random genetic drift chart. Provided by : Wikimedia. Located at : en.Wikipedia.org/wiki/Genetic...rift_chart.png . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Population Genetics. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...e_19_02_03.jpg . License : CC BY: Attribution
- Portulaca grandiflora mutant1. Provided by : Wikimedia. Located at : en.Wikipedia.org/wiki/File:Po...ra_mutant1.jpg . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Population Genetics. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...e_19_02_04.jpg . License : CC BY: Attribution
- OpenStax College, Biology. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...ol11448/latest . License : CC BY: Attribution
- OpenStax College, Biology. October 23, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...ol11448/latest . License : CC BY: Attribution
- cline. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/cline . License : CC BY-SA: Attribution-ShareAlike
- Sexual Selection. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Sexual_selection . License : CC BY-SA: Attribution-ShareAlike
- assortative mating. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/assortative_mating . License : CC BY-SA: Attribution-ShareAlike
- Coquina variation3. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Co...variation3.jpg . License : CC BY-SA: Attribution-ShareAlike
- Cheetah genetic diversity. Provided by : Wikimedia. Located at : en.Wikipedia.org/wiki/File:Ch..._diversity.jpg . License : CC BY-SA: Attribution-ShareAlike
- Founder effect with drift. Provided by : Wikimedia. Located at : en.Wikipedia.org/wiki/File:Fo...with_drift.jpg . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Population Genetics. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...e_19_02_02.png . License : CC BY: Attribution
- Random genetic drift chart. Provided by : Wikimedia. Located at : en.Wikipedia.org/wiki/Genetic...rift_chart.png . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Population Genetics. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...e_19_02_03.jpg . License : CC BY: Attribution
- Portulaca grandiflora mutant1. Provided by : Wikimedia. Located at : en.Wikipedia.org/wiki/File:Po...ra_mutant1.jpg . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Population Genetics. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...e_19_02_04.jpg . License : CC BY: Attribution
- OpenStax College, Population Genetics. October 16, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44584/latest...e_19_02_05.jpg . License : CC BY: Attribution
- American Robin Close-Up. Provided by : Wikimedia. Located at : en.Wikipedia.org/wiki/America...n_Close-Up.JPG . License : CC BY-SA: Attribution-ShareAlike
Key Points
- Nonrandom mating can occur when individuals prefer mates with particular superior physical characteristics or by the preference of individuals to mate with individuals similar to themselves.
- Nonrandom mating can also occur when mates are chosen based on physical accessibility; that is, the availability of some mates over others.
- Phenotypes of individuals can also be influenced by the environment in which they live, such as temperature, terrain, or other factors.
- A cline occurs when populations of a given species vary gradually across an ecological gradient.
Key Terms
- cline : a gradation in a character or phenotype within a species or other group
- sexual selection : a mode of natural selection in which some individuals out-reproduce others of a population because they are better at securing mates
- assortative mating : between males and females of a species, the mutual attraction or selection, for reproductive purposes, of individuals with similar characteristics


1.5: Formal Charges
After completing this section, you should be able to
- calculate the formal charge of an atom in an organic molecule or ion.
- identify and recognize the bonding patterns for atoms of carbon, hydrogen, oxygen, nitrogen and the halogens that have a formal charge of zero.
Make certain that you can define, and use in context, the key term below.
- valence electrons
- bonding and non-bonding electrons
- formal charge
- carbocations
It is more important that students learn to easily identify atoms that have formal charges of zero, than it is to actually calculate the formal charge of every atom in an organic compound. Students will benefit by memorizing the "normal" number of bonds and non-bonding electrons around atoms whose formal charge is equal to zero.
Determining the Formal Charge on an Atom
A formal charge (\(FC\)) compares the number of electrons around a "neutral atom" (an atom not in a molecule) versus the number of electrons around an atom in a molecule. Formal charge is assigned to an atom in a molecule by assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity. To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules:
- Non-bonding electrons are assigned to the atom on which they are located.
- Bonding electrons are divided equally between the two bonded atoms, so one electron from each bond goes to each atom.
The formal charge of each atom in a molecule can be calculated using the following equation:
\[FC = \text{(# of valence electrons in free atom)} − \text{(# of non-bonded electrons)} −\dfrac{1}{2} (\text{# of bonds}) \label{2.3.1}\]
To illustrate this method, let’s calculate the formal charge on the atoms in ammonia (\(\ce{NH3}\)) whose Lewis structure is as follows:
A neutral nitrogen atom has five valence electrons (it is in group 15). From the Lewis structure, the nitrogen atom in ammonia has one lone pair and three bonds with hydrogen atoms. Substituting into Equation \ref{2.3.1}, we obtain
\[\begin{align*} FC (N) &= (\text{5 valence electrons}) − (\text{2 non-bonding electrons}) − \dfrac{1}{2} (\text{3 bonds}) \\[4pt] &= 0 \end{align*}\]
A neutral hydrogen atom has one valence electron. Each hydrogen atom in the molecule has no non-bonding electrons and one bond. Using Equation \ref{2.3.1} to calculate the formal charge on hydrogen, we obtain
\[\begin{align*} FC (H) &= (\text{1 valence electron}) − (\text{0 non-bonding electrons}) − \dfrac{1}{2} (\text{1 bond}) \\[4pt] &= 0 \end{align*}\]
The sum of the formal charges of each atom must be equal to the overall charge of the molecule or ion. In this example, the nitrogen and each hydrogen has a formal charge of zero. When summed the overall charge is zero, which is consistent with the overall neutral charge o f the \(\ce{NH3}\) molecu le.
Typically, the structure with the most formal charges of zero on atoms is the more stable Lewis structure. In cases where there MUST be positive or negative formal charges on various atoms, the most stable structures generally have negative formal charges on the more electronegative atoms and positive formal charges on the less electronegative atoms. The next example further demonstrates how to calculate formal charges for polyatomic ions.
Calculate the formal charges on each atom in the \(\ce{NH4^{+}}\) ion.
Given: chemical species
Asked for: formal charges
Strategy:
Identify the number of valence electrons in each atom in the \(\ce{NH4^{+}}\) ion. Use the Lewis electron structure of \(\ce{NH4^{+}}\) to identify the number of bonding and non-bonding electrons associated with each atom and then use Equation \ref{2.3.1} to calculate the formal charge on each atom.
Solution:
The Lewis electron structure for the \(\ce{NH4^{+}}\) ion is as follows:
The nitrogen atom in ammonium has zero non-bonding electrons and 4 bonds. Using Equation \ref{2.3.1}, the formal charge on the nitrogen atom is therefore
\[\begin{align*} FC (N) &= (\text{5 valence electrons}) − (\text{0 non-bonding electrons}) − \dfrac{1}{2} (\text{4 bonds}) \\[4pt] &= +1 \end{align*}\]
Each hydrogen atom in has one bond and zero non-bonding electrons. The formal charge on each hydrogen atom is therefore
\[\begin{align*} FC (H) &= (\text{1 valence electron}) − (\text{0 non-bonding electrons}) − \dfrac{1}{2} (\text{1 bond}) \\[4pt] &= 0 \end{align*}\]
The formal charges on the atoms in the \(\ce{NH4^{+}}\) ion are thus
Adding together the formal charges on the atoms should give us the total charge on the molecule or ion. In this case, the sum of the formal charges is 0 + 1 + 0 + 0 + 0 = 1+, which is the same as the total charge of the ammonium polyatomic ion.
Write the formal charges on all atoms in \(\ce{BH4^{−}}\).
- Answer
-
.
Determining the Charge of Atoms in Organic Structures
The calculation method reviewed above for determining formal charges on atoms is an essential starting point for a novice organic chemist, and works well when dealing with small structures. But this method becomes unreasonably time-consuming when dealing with larger structures. It would be exceptionally tedious to determine the formal charges on each atom in 2'-deoxycytidine (one of the four nucleoside building blocks that make up DNA) using Equation \ref{2.3.1}. As you get more experience with organic structures, you will be able to quickly look at this type of complicated structure and determine charges on each atom.
2'-deoxycytidine
You need to develop the ability to quickly and efficiently draw large structures and determine formal charges. Fortunately, this only requires some practice with recognizing common bonding patterns.
Organic chemistry only deals with a small part of the periodic table, so much so that it becomes convenient to be able to recognize the bonding forms of these atoms. The figure below contains the most important bonding forms. These will be discussed in detail below. An important idea to note is most atoms in a molecule are neutral. Pay close attention to the neutral forms of the elements below because that is how they will appear most of the time.
| Atom | Positive | Neutral | Negative |
|---|---|---|---|
| C | |||
| N | |||
| O | |||
| Cl (halogens) |
Carbon
Carbon, the most important element for organic chemists. In the structures of methane, methanol, ethane, ethene, and ethyne, there are four bonds to the carbon atom. And each carbon atom has a formal charge of zero. In other words, carbon is tetravalent , meaning that it commonly forms four bonds.
Carbon usually makes four bonds
Carbon is tetravalent in most organic molecules, but there are exceptions. Later in this chapter and throughout this book are examples of organic ions called ‘carbocations’ and carbanions’, in which a carbon atom has a positive or negative formal charge, respectively. Carbocations occur when a carbon has only three bonds and no lone pairs of electrons. Carbocations have only 3 valence electrons and a formal charge of 1+. Carbanions occur when the carbon atom has three bonds plus one lone pair of electrons. Carbanions have 5 valence electrons and a formal charge of 1−.
| carbocation: 3 bonds & no lone pair | carbanion: 3 bonds & one lone pair | carbon radical: 3 bonds & one unpaired electron |
Two other possibilities are carbon radicals and carbenes, both of which have a formal charge of zero. A carbon radical has three bonds and a single, unpaired electron. Carbon radicals have 4 valence electrons and a formal charge of zero. Carbenes are a highly reactive species, in which a carbon atom has two bonds and one lone pair of electrons, giving it a formal charge of zero. Though carbenes are rare, you will encounter them in section 8.10 Addition of Carbenes to Alkenes .
You should certainly use the methods you have learned to check that these formal charges are correct for the examples given above. More importantly, you will need, before you progress much further in your study of organic chemistry, to simply recognize these patterns (and the patterns described below for other atoms) and be able to identify carbons that bear positive and negative formal charges by a quick inspection.
Hydrogen
The common bonding pattern for hydrogen is easy: hydrogen atoms in organic molecules typically have only one bond, no unpaired electrons and a formal charge of zero. The exceptions to this rule are the proton, H + , the hydride ion, H - , and the hydrogen radical, H . . The proton is a hydrogen with no bonds and no lone pairs and a formal charge of 1+. The hydride ion is a is a hydrogen with no bonds, a pair of electrons, and a formal charge of 1−. The hydrogen radical is a hydrogen atom with no bonds, a single unpaired electron and a formal charge of 0. Because this book concentrates on organic chemistry as applied to living things, however, we will not be seeing ‘naked’ protons and hydrides as such, because they are too reactive to be present in that form in aqueous solution. Nonetheless, the idea of a proton will be very important when we discuss acid-base chemistry, and the idea of a hydride ion will become very important much later in the book when we discuss organic oxidation and reduction reactions. As a rule, though, all hydrogen atoms in organic molecules have one bond, and no formal charge.
Neutral Hydrogen: one bond, no lone pair
Oxygen
The common arrangement of oxygen that has a formal charge of zero is when the oxygen atom has 2 bonds and 2 lone pairs. Other arrangements are oxygen with 1 bond and 3 lone pairs, that has a 1− formal charge, and oxygen with 3 bonds and 1 lone pair that has a formal charge of 1+. All three patterns of oxygen fulfill the octet rule.
| neutral oxygen: 2 bonds & 2 lone pairs | negative oxygen: 1 bond & 3 lone pairs | positive oxygen: 3 bonds & one lone pair |
If it has two bonds and two lone pairs, as in water, it will have a formal charge of zero. If it has one bond and three lone pairs, as in hydroxide ion, it will have a formal charge of 1−. If it has three bonds and one lone pair, as in hydronium ion, it will have a formal charge of 1+.
Oxygen can also exist as a radical, such as where an oxygen atom has one bond, two lone pairs, and one unpaired (free radical) electron, giving it a formal charge of zero. For now, however, concentrate on the three main non-radical examples, as these will account for most oxygen containing molecules you will encounter in organic chemistry.
Nitrogen
Nitrogen has two major bonding patterns, both of which fulfill the octet rule:
| neutral nitrogen: 3 bonds & 1 lone pair | positive nitrogen: 4 bonds | negative nitrogen: 2 bonds & 2 lone pairs |
If a nitrogen has three bonds and a lone pair, it has a formal charge of zero. If it has four bonds (and no lone pair), it has a formal charge of 1+. In a fairly uncommon bonding pattern, negatively charged nitrogen has two bonds and two lone pairs.
Phosphorus and Sulfur
Two third row elements are commonly found in biological organic molecules: phosphorus and sulfur. Although both of these elements have other bonding patterns that are relevant in laboratory chemistry, in a biological context sulfur almost always follows the same bonding/formal charge pattern as oxygen, while phosphorus is present in the form of phosphate ion (PO 4 3 − ), where it has five bonds (almost always to oxygen), no lone pairs, and a formal charge of zero. Remember that elements in the third row of the periodic table have d orbitals in their valence shell as well as s and p orbitals, and thus are not bound by the octet rule.
Halogens
The halogens (fluorine, chlorine, bromine, and iodine) are very important in laboratory and medicinal organic chemistry, but less common in naturally occurring organic molecules. Halogens in organic compounds usually are seen with one bond, three lone pairs, and a formal charge of zero. Sometimes, especially in the case of bromine, we will encounter reactive species in which the halogen has two bonds (usually in a three-membered ring), two lone pairs, and a formal charge of 1+.

Once you have gotten the hang of drawing Lewis structures, it is not always necessary to draw lone pairs on heteroatoms, as you can assume that the proper number of electrons are present around each atom to match the indicated formal charge (or lack thereof). Occasionally, though, lone pairs are drawn if doing so helps to make an explanation more clear.
Please identify an atom with a non-neutral charge in the following atom:
- Answer
-
Drawing the Lewis Structure of Ionic Molecular Compounds
The hydroxide ion, OH - , is drawn simply by showing the oxygen atom with its six valence electrons, then adding one more electron to account for the negative charge. By changing the number of valence electrons the bonding characteristic of oxygen are now changed. Now the oxygen has three non-bonding lone pairs, and can only form one bond to a hydrogen.
To draw a Lewis structure of the hydronium ion, H 3 O + , you again start with the oxygen atom with its six valence electrons, then take one away to account for the positive charge to give oxygen five valence electrons. The oxygen has one non-bonding lone pair and three unpaired electrons which can be used to form bonds to three hydrogen atoms.
Using Formal Charges to Distinguish between Lewis Structures
As an example of how formal charges can be used to determine the most stable Lewis structure for a substance, we can compare two possible structures for CO 2 . Both structures conform to the rules for Lewis electron structures.
CO 2
1. C gets more bonds when neutral, so it is the central atom.
2. C has 4 valence electrons and each O has 6 valence electrons, for a total of 16 valence electrons.
3. Placing one electron pair between the C and each O gives O–C–O, with 12 electrons left over.
4. Dividing the remaining electrons between the O atoms gives three lone pairs on each atom:
This structure has an octet of electrons around each O atom but only 4 electrons around the C atom.
5. No electrons are left for the central atom.
6. To give the carbon atom an octet of electrons, we can convert two of the lone pairs on the oxygen atoms to bonding electron pairs. There are, however, two ways to do this. We can either take one electron pair from each oxygen to form a symmetrical structure or take both electron pairs from a single oxygen atom to give an asymmetrical structure:
Both Lewis electron structures give all three atoms an octet. How do we decide between these two possibilities? The formal charges for the two Lewis electron structures of CO2 are as follows:
Both Lewis structures have a net formal charg e of zero, but the structure on the right has a 1+ charge on the more electronegative atom (O). Thus the symmetrical Lewis structure on the left is predicted to be more stable, and it is, in fact, the structure observed experimentally. Remember, though, that formal charges do not represent the actual charges on atoms in a molecule or ion. They are used simply as a bookkeeping method for predicting the most stable Lewis structure for a compound.
The thiocyanate ion (\(\ce{SCN^{−}}\)), which is used in printing and as a corrosion inhibitor against acidic gases, has at least two possible Lewis electron structures. Draw two possible structures, assign formal charges on all atoms in both, and decide which is the preferred arrangement of electrons.
Given: chemical species
Asked for: Lewis electron structures, formal charges, and preferred arrangement
Strategy:
A Use the step-by-step procedure to write two plausible Lewis electron structures for SCN − .
B Calculate the formal charge on each atom using Equation \ref{2.3.1}.
C Predict which structure is preferred based on the formal charge on each atom and its electronegativity relative to the other atoms present.
Solution:
A Possible Lewis structures for the SCN − ion are as follows:
B We must calculate the formal charges on each atom to identify the more stable structure. If we begin with carbon, we notice that the carbon atom in each of these structures shares four bonding pairs, the number of bonds typical for carbon, so it has a formal charge of zero. Continuing with sulfur, we observe that in (a) the sulfur atom shares one bonding pair and has three lone pairs and has a total of six valence electrons. The formal charge on the sulfur atom is therefore 6 - (6 + 2/2) = 1−. In (b), the sulfur atom has a formal charge of 0. In (c), the sulfur atom has a formal charge of 1+. Continuing with the nitrogen, we observe that in (a) the nitrogen atom shares three bonding pairs and has one lone pair and has a total of 5 valence electrons. The formal charge on the nitrogen atom is therefore 5 - (2 + 6/2) = 0. In (b), the nitrogen atom has a formal charge of 1−. In (c), the nitrogen atom has a formal charge of 2−.
C Which structure is preferred? Structure (b) is preferred because the negative charge is on the more electronegative atom (N), and it has lower formal charges on each atom as compared to structure (c): 0, 1− versus 1+, 2−.
Salts containing the fulminate ion (\(\ce{CNO^{−}}\)) are used in explosive detonators. Draw three Lewis electron structures for \(\ce{CNO^{−}}\) and use formal charges to predict which is more stable. (Note: \(\ce{N}\) is the central atom.)
- Answer
-
The second structure is predicted to be the most stable.
Exercises
- Draw the Lewis structure of each of these molecules: \(\ce{CH3^{+}}\), \(\ce{NH2^{-}}\), \(\ce{CH3^{-}}\), \(\ce{NH4^{+}}\), \(\ce{BF4^{-}}\). In each case, use the method of calculating formal charge described to satisfy yourself that the structures you have drawn do in fact carry the charges shown.
Answer
2.3: Formal Charges
2.3 Exercises
Questions
Q2.3.1
Give the formal charges for all non-hydrogen atoms in the following molecules: \(\ce{BH4^{-}}\), \(\ce{H2O}\), \(\ce{CH3O^{-}}\)
Solutions
S2.3.1
BH 4 − (B = −1)
H 2 O (O = 0)
CH 3 O − (C = 0, O = −1)


14.19: Understanding Domain Aggregation
Learning Outcomes
- Discuss domain aggregation
There are twelve aggregate functions that extract and group data from a larger data set.
- DSum
- DAvg
- DCount
- DLookup
- DMin and DMax
- DFirst and DLast
- DStDev
- DStDevP
- DVar
- DvarP
These functions all have the same structure:
FunctionName( “ [Field Name] “,” [Dataset Name] “, “[Criteria] “ )
FunctionName is the name of the domain aggregation function being invoked; Field Name refers to the field containing the data in question; DataSet Name refers to the table in play; and, Criteria limits the inclusion of data per the scope necessary.
These functions can provide everything from a total of the values listed to an average of those values to the standard deviation of the values listed. The Help menu has detailed information about the functions.
Contributors and Attributions
- Understanding Domain Aggregation. Authored by : Robert Danielson. Provided by : Lumen Learning. License : CC BY: Attribution


3.3: First Aid and Personal Protective Equipment
First Things First
Every workplace and therefore every employer must render first aid to an injured employee. Depending on the severity of the injury, i.e. if the employer emergency plan is activated, then the duty of care is to get the employee to an emergency facility as soon as possible. When the workplace is a construction site or some distance away from an urgent care or hospital emergency room the employer must have the capacity to provide immediate first aid commensurate to the injury and then proceed if necessary to offsite care for medical treatment. The employer must ensure the availability of medical personnel for advice and consultation no matter the work location.
Construction job site accidents and injuries as a result of tool and equipment failure and/or incorrect use are common. Cuts and punctures from sharp objects, contusions from blunt objects or impacts, burns from open flame torches and hot pipes, splashing of chemicals or debris to the eyes, and electrical shock are just a few of the common injuries associated with construction and maintenance trades.
As construction and maintenance work offer the potential for many types of traumatic and life threatening injuries, workers in skilled trades should be aware of the hazards and be prepared to respond in the event of an injury. Since many construction and maintenance industry tasks may be performed by a single person, often isolated from others, it is recommended that construction industry workers receive First Aid and Cardiopulmonary Resuscitation (CPR), and Occupational Safety and Health Administration (OSHA) 10 or 30 Hour Safety Training. Training enables industry trades-persons to better assess workplace hazards and respond to them appropriately, whether an incident involves oneself, a coworker, or others on the job site. In person, hands-on First Aid/CPR training can be found through local health and welfare organizations, educational institutions (credit or non-credit), and medical providers. OSHA in person courses can be found at local educational institutions (credit or non-credit), and in online formats through various educational institutions and commercial providers.
The First Aid Kit
First aid supplies are required to be easily accessible under paragraph Sec. 1926.50 (d)(1). An example of the minimal contents of a generic first aid kit is described in American National Standard (ANSI) Z308.1-1978 "Minimum Requirements for Industrial Unit-Type First-aid Kits". The contents of the kit listed in the ANSI standard should be adequate for small work sites. When larger operations or multiple operations are being conducted at the same location, employers should determine the need for additional first aid kits at the worksite, additional types of first aid equipment and supplies and additional quantities and types of supplies and equipment in the first aid kits.
If it is reasonably anticipated employees will be exposed to blood or other potentially infectious materials (OPIM) while using first-aid supplies, employers shall provide personal protective equipment (PPE). Appropriate PPE includes gloves, gowns, face shields, masks and eye protection.
Personal Protective Equipment (PPE)
What is personal protective equipment?
Personal protective equipment , commonly referred to as "PPE", is equipment worn to minimize exposure to hazards that cause serious workplace injuries and illnesses. These injuries and illnesses may result from contact with chemical, radiological, physical, electrical, mechanical, or other workplace hazards. Personal protective equipment may include items such as gloves, safety glasses and shoes, earplugs or muffs, hard hats, respirators, or coveralls, vests and full body suits.
What can be done to ensure proper use of personal protective equipment?
All personal protective equipment should be safely designed and constructed, and should be maintained in a clean and reliable fashion. It should fit comfortably, encouraging worker use. If the personal protective equipment does not fit properly, it can make the difference between being safely covered or dangerously exposed. When engineering, work practice, and administrative controls are not feasible or do not provide sufficient protection, employers must provide personal protective equipment to their workers and ensure its proper use. Employers are also required to train each worker required to use personal protective equipment to know:
- When it is necessary
- What kind is necessary
- How to properly put it on, adjust, wear and take it off
- The limitations of the equipment
- Proper care, maintenance, useful life, and disposal of the equipment
If PPE is to be used, a PPE program should be implemented. This program should address the hazards present; the selection, maintenance, and use of PPE; the training of employees; and monitoring of the program to ensure its ongoing effectiveness.


16.1: Data versus Information
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Learning Objectives
- Distinguish between data and information.
- Define information system (IS) and identify the tasks of the information systems manager.
By the time the company took the plunge and committed $100 million to marketing-related information technology (IT), Caesars had been collecting and storing data about customers for almost a decade. “While the company thought it important to collect customer information,” recalls a senior marketing executive, “the problem was we had millions of customers to collect information on, but we had no systematic way of turning it into a marketing decision. We didn’t know what to do with it.” In other words, Caesars was collecting a lot of data but not necessarily any information . So what’s the difference?
As an example, suppose that you want to know how you’re doing in a particular course. So far, you’ve taken two 20-question multiple-choice tests. On the first, you got questions 8, 11, and 14 wrong; on the second, you did worse, missing items 7, 15, 16, and 19. The items that you got wrong are merely data—unprocessed facts. What’s important is your total score. You scored 85 on the first exam and 80 on the second. These two numbers constitute information—data that have been processed, or turned into some useful form. Knowing the questions that you missed simply supplied you with some data for calculating your scores.
Now let’s fast-forward to the end of the semester. At this point, in addition to taking the two tests, you’ve written two papers and taken a final. You got a 90 and 95 on the papers and a 90 on the final. You now have more processed data, but you still want to organize them into more useful information. What you want to know is your average grade for the semester. To get the information you want, you need yet more data—namely, the weight assigned to each graded item. Fortunately, you’ve known from day one that each test counts 20 percent, each paper 10 percent, and the final exam 40 percent. A little math reveals an average grade of 87.
Though this is the information you’re interested in, it may be mere data to your instructor, who may want different information: an instructor who intends to scale grades, for example, will want to know the average grade for the entire class. You’re hoping that the class average is low enough to push your average of 87 up from a B+ to an A– (or maybe even an A—it doesn’t hurt to hope for the best). The moral of the story is that what constitutes information at one stage can easily become data at another: or, one person’s information can be another person’s data.
As a rule, you want information; data are good only for generating the information. So, how do you convert data into information that’s useful in helping you make decisions and solve problems? That’s the question we’ll explore in the next section.


1.3E: Exercises
Practice Makes Perfect
Simplify Expressions with Absolute Value
In the following exercises, fill in \(<,>,\) or \(=\) for each of the following pairs of numbers.
1.
5. \(|15−7|−|14−6|\)
- Answer
-
0
6. \(|17−8|−|13−4|\)
7. \(18−|2(8−3)|\)
- Answer
-
8
8. \(15−|3(8−5)|\)
9. \(18−|12−4(4−1)+3|\)
- Answer
-
15
10. \(27−|19+4(3−1)−7|\)
11. \(10−3|9−3(3−1)|\)
- Answer
-
1
12. \(13−2|11−2(5−2)|\)
Add and Subtract Integers
In the following exercises, simplify each expression.
13. ⓐ \(−7+(−4)\)
ⓑ \(−7+4\)
ⓒ \(7+(−4).\)
- Answer
-
ⓐ \(−11\) ⓑ \(−3\) ⓒ \(3\)
ⓑ \(−5+9\)
ⓒ \(5+(−9)\)
15. \(48+(−16)\)
- Answer
-
32
16. \(34+(−19)\)
17. \(−14+(−12)+4\)
- Answer
-
\(-22\)
18. \(−17+(−18)+6\)
19. \(19+2(−3+8)\)
- Answer
-
\(29\)
20. \(24+3(−5+9)\)
21. ⓐ \(13−7\)
ⓑ \(−13−(−7)\)
ⓒ \(−13−7\)
ⓓ \(13−(−7)\)
- Answer
-
ⓐ 6 ⓑ −6 ⓒ −20 ⓓ 20
22. ⓐ \(15−8\)
ⓑ \(−15−(−8)\)
ⓒ \(−15−8\)
ⓓ \(15−(−8)\)
23. \(−17−42\)
- Answer
-
\(-59\)
24. \(−58−(−67)\)
25. \(−14−(−27)+9\)
- Answer
-
22
26. \(64+(−17)−9\)
27. ⓐ \(44−28\) ⓑ \(44+(−28)\)
- Answer
-
ⓐ 16 ⓑ 16
28. ⓐ \(35−16\) ⓑ \(35+(−16)\)
29. ⓐ \(27−(−18)\) ⓑ \(27+18\)
- Answer
-
ⓐ 45 ⓑ 45
30. ⓐ \(46−(−37)\) ⓑ \(46+37\)
31. \((2−7)−(3−8)\)
- Answer
-
0
32. \((1−8)−(2−9)\)
33. \(−(6−8)−(2−4)\)
- Answer
-
4
34. \(−(4−5)−(7−8)\)
35. \(25−[10−(3−12)]\)
- Answer
-
6
36. \(32−[5−(15−20)]\)
Multiply and Divide Integers
In the following exercises, multiply or divide.
37. ⓐ \(−4⋅8\)
ⓑ \(13(−5)\)
ⓒ \(−24÷6\)
ⓓ \(−52÷(−4)\)
- Answer
-
ⓐ \(−32\) ⓑ \(−65\) ⓒ \(−4\) ⓓ \(13\)
ⓑ \(9(−7)\)
ⓒ \(35÷(−7)\)
ⓓ \(−84÷(−6)\)
ⓑ \(−180÷15\)
ⓒ \(3(−13)\)
ⓓ \(−1(−14)\)
- Answer
-
ⓐ \(−4\) ⓑ \(−12\) ⓒ \(−39\) ⓓ \(14\)
ⓑ \(−192÷12\)
ⓒ \(9(−7)\)
ⓓ \(−1(−19)\)
Simplify and Evaluate Expressions with Integers
In the following exercises , simplify each expression.
41. ⓐ \((−2)^6\) ⓑ \(−2^6\)
- Answer
-
ⓐ \(64\) ⓑ \(−64\)
42. ⓐ \((−3)^5\) ⓑ \(−3^5\)
43. \(5(−6)+7(−2)−3\)
- Answer
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\(−47\)
44. \(8(−4)+5(−4)−6\)
45. \(−3(−5)(6)\)
- Answer
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\(90\)
46. \(−4(−6)(3)\)
47. \((8−11)(9−12)\)
- Answer
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\(9\)
48. \((6−11)(8−13)\)
49. \(26−3(2−7)\)
- Answer
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\(41\)
50. \(23−2(4−6)\)
51. \(65÷(−5)+(−28)÷(−7)\)
- Answer
-
\(-9\)
52. \(52÷(−4)+(−32)÷(−8)\)
53. \(9−2[3−8(−2)]\)
- Answer
-
\(-29\)
54. \(11−3[7−4(−2)]\)
55. \(8−|2−4(4−1)+3|\)
- Answer
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\(1\)
56. \(7−|5−3(4−1)−6|\)
57. \(9−3|2(2−6)−(3−7)|\)
- Answer
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\(-3\)
58. \(5−2|2(1−4)−(2−5)|\)
59. \((−3)^2−24÷(8−2)\)
- Answer
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\(5\)
60. \((−4)^2−32÷(12−4)\)
In the following exercises , evaluate each expression.
61. \(y+(−14)\) when ⓐ \(y=−33\) ⓑ \(y=30\)
- Answer
-
ⓐ \(−47\) ⓑ \(16\)
62. \(x+(−21)\) when ⓐ \(x=−27\) ⓑ \(x=44\)
63. \((x+y)^2\) when \(x=−3\) and \(y=14\)
- Answer
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\(121\)
64. \((y+z)^2\) when \(y=−3\) and \(z=15\)
65. \(9a−2b−8\) when \(a=−6\) and \(b=−3\)
- Answer
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\(-56\)
66. \(7m−4n−2\) when \(m=−4\) and \(n=−9\)
67. \(3x^2−4xy+2y^2\) when \(x=−2\) and \(y=−3\)
- Answer
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\(6\)
68. \(4x^2−xy+3y^2\) when \(x=−3\) and \(y=−2\)
Translate English Phrases to Algebraic Expressions
In the following exercises, translate to an algebraic expression and simplify if possible.
69. the sum of 3 and −15, increased by 7
- Answer
-
\((3+(−15))+7;−5\)
70. the sum of \(−8\) and \(−9\), increased by \(23\)
ⓑ subtract \(11\) from \(−25\)
- Answer
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ⓐ \(10−(−18);28\)
ⓑ\(−25−11;−36\)
ⓑ subtract \(−6\) from \(−13\)
73. the quotient of \(−6\) and the sum of \(a\) and \(b\)
- Answer
-
\(\dfrac{−6}{a+b}\)
74. the product of \(−13\) and the difference of \(c\) and \(d\)
Use Integers in Applications
In the following exercises, solve.
75. Temperature On January 15, the high temperature in Anaheim, California, was \(84°\). That same day, the high temperature in Embarrass, Minnesota, was \(−12°\). What was the difference between the temperature in Anaheim and the temperature in Embarrass?
- Answer
-
\(96^\circ\)
76. Temperature On January 21, the high temperature in Palm Springs, California, was \(89°\), and the high temperature in Whitefield, New Hampshire, was \(−31°\). What was the difference between the temperature in Palm Springs and the temperature in Whitefield?
77. Football On the first down, the Chargers had the ball on their 25-yard line. On the next three downs, they lost 6 yards, gained 10 yards, and lost 8 yards. What was the yard line at the end of the fourth down?
- Answer
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21
78. Football On the first down, the Steelers had the ball on their 30-yard line. On the next three downs, they gained 9 yards, lost 14 yards, and lost 2 yards. What was the yard line at the end of the fourth down?
79. Checking Account Mayra has $124 in her checking account. She writes a check for $152. What is the new balance in her checking account?
- Answer
-
\(−\$ 28\)
80. Checking Account Reymonte has a balance of \(−$49\) in his checking account. He deposits $281 to the account. What is the new balance?
Writing Exercises
81. Explain why the sum of −8 and 2 is negative, but the sum of 8 and −2 is positive.
- Answer
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Answers will vary.
82. Give an example from your life experience of adding two negative numbers.
83. In your own words, state the rules for multiplying and dividing integers.
- Answer
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Answers will vary.
84. Why is \(−4^3=(−4)^3\)?
Self Check
ⓐ After completing the EXAMPLEs, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ After reviewing this checklist, what will you do to become confident for all objectives?


3.6: Capillary Electrophoresis
Capillary electrophoresis (CE) encompasses a family of electrokinetic separation techniques that uses an applied electric field to separate out analytes based on their charge and size. The basic principle is hinged upon that of electrophoresis, which is the motion of particles relative to a fluid (electrolyte) under the influence of an electric field. The founding father of electrophoresis, Arne W. K. Tiselius (Figure \(\PageIndex{1} a \) ), first used electrophoresis to separate proteins, and he went on to win a Nobel Prize in Chemistry in 1948 for his work on both electrophoresis and adsorption analysis. However, it was Stellan Hjerten (Figure \(\PageIndex{1} b\) ) who worked under Arne W. K. Tiselius, who pioneered work in CE in 1967, although CE was not well recognized until 1980 when James W. Jorgenson (Figure \(\PageIndex{1} c \) ) and Krynn D. Lukacs published a series of papers describing this new technique.
Instrument Overview
The main components of CE are shown in Figure \(\PageIndex{2}\). The electric circuit of the CE is the heart of the instrument.
Injection Methods
The samples that are studied in CE are mainly liquid samples. A typical capillary column has an inner diameter of 50 μm and a length of 25 cm. Because the column can only contain a minimal amount of running buffer, only small sample volumes can be tested (nL to μL). The samples are introduced mainly by two injection methods: hydrodynamic and electrokinetic injection. The two methods are displayed in Table \(\PageIndex{1}\) A disadvantage of electrokinetic injection is that the composition of the injected sample may not be the same as the composition of the original sample. This is because the injection method is dependent on the electrophoretic and electroosmotic mobility of the species in the sample. However, both injection methods depend on the temperature and the viscosity of the solution. Hence, it is important to control both parameters when a reproducible volume of sample injections is desired. It is advisable to use internal standards instead of external standards when performing quantitative analysis on the samples as it is hard to control both the temperature and viscosity of the solution.
| Injection Methods | Working Principle |
|---|---|
| Hydrodynamic Injection | The sample vial is enclosed in a chamber with one end of fixed capillary column immersed in it. Pressure is then applied to the chamber for a fixed period so that the sample can enter the capillary. After the sample, has been introduced, the capillary is withdrawn and then re-immersed into the source reservoir and separation takes place. |
| Electrokinetic Injection | The sample is enclosed in a chamber with one end of capillary column immersed in it with an electrode present. The electric field is applied, and the samples enter the capillary. After the sample, has been introduced, the capillary is withdrawn and then re-immersed into the source reservoir and separation takes place. |
Column
After the samples have been injected, the capillary column is used as the main medium to separate the components. The capillary column used in CE shares the same characteristics as the capillary column used in gas chromatography (GC); however, the most critical components of the CE column are:
- the inner diameter of the capillary,
- the total length of the capillary,
- the length of the column from the injector to the detector.
Solvent Buffer
The solvent buffer carries the sample through the column. It is crucial to employ a good buffer as a successful CE experiment is hinged upon this. CE is based on the separation of charges in an electric field. Therefore, the buffer should either sustain the pre-existing charge on the analyte or enable the analyte to obtain a charge, and it is important to consider the pH of the buffer before using it.
Applied Voltage (kV)
The applied voltage is important in the separation of the analytes as it drives the movement of the analyte. It is important that it is not too high as it may become a safety concern.
Detectors
Analytes that have been separated after the applying the voltage can be detected by many detection methods. The most common method is UV-visible absorbance. The detection takes place across the capillary with a small portion of the capillary acting as the detection cell. The on-tube detection cell is usually made optically transparent by scraping off the polyimide coating and coating it with another optically transparent material so that the capillary would not break easily. For species that do not have a chromophore, a chromophore can be added to the buffer solution. When the analyte passes by, there would be a decrease in signal. This decreased signal will correspond to the amount of analyte present. Other common detection techniques employable in CE are fluorescence and mass spectrometry (MS).
Theory
In CE, the sample is introduced into the capillary by the above-mentioned methods. A high voltage is then applied causing the ions of the sample to migrate towards the electrode in the destination reservoir, in this case, the cathode. Sample components migration and separation are determined by two factors, electrophoretic mobility and electroosmotic mobility.
Electrophoretic Mobility
The electrophoretic mobility, \(μ_{ep}\), is inherently dependent on the properties of the solute and the medium in which the solute is moving. Essentially, it is a constant value, that can be calculated as given by \ref{1} where \(q\) is the solute`s charge, \(η\) is the buffer viscosity and \(r\) is the solute radius.
\[ \mu _{ep} = \dfrac{q}{6\pi \eta r} \label{1} \]
The electrophoretic velocity, \(v_{ep}\), is dependent on the electrophoretic mobility and the applied electric field, \(E\) (\ref{2}).
\[ \nu _{ep} = \mu _{ep} E \label{2} \]
Thus, when solutes have a larger charge to size ratio the electrophoretic mobility and velocity will increase. Cations and the anion would move in opposing directions corresponding to the sign of the electrophoretic mobility with is a result of their charge. Thus, neutral species that have no charge do not have an electrophoretic mobility.
Electroosmotic Mobility
The second factor that controls the migration of the solute is the electroosmotic flow. With zero charge, it is expected that the neutral species should remain stationary. However, under normal conditions, the buffer solution moves towards the cathode as well. The cause of the electroosmotic flow is the electric double layer that develops at the silica solution interface.
At pH more than 3, the abundant silanol (-OH) groups present on the inner surface of the silica capillary, de-protonate to form negatively charged silanate ions (-SiO - ). The cations present in the buffer solution will be attracted to the silanate ions and some of them will bind strongly to it forming a fixed layer. The formation of the fixed layer only partially neutralizes the negative charge on the capillary walls. Hence, more cations than anions will be present in the layer adjacent to the fixed layer, forming the diffuse layer. The combination of the fixed layer and diffuse layer is known as the double layer as shown in Figure \(\PageIndex{3}\). The cations present in the diffuse layer will migrate towards the cathode, as these cations are solvated the solution will also flow with it, producing the electroosmotic flow. The anions present in the diffuse layer are solvated and will move towards the anode. However, as there are more cations than anions the cations will push the anions together with it in the direction of the cathode. Hence, the electroosmotic flow moves in the direction of the cathode.
The electroosmotic mobility, μ eof , is described by \ref{3} where ξ is the zeta potential, ε is the buffer dielectric constant and η is the buffer viscosity. The electroosmotic velocity, v eof , is the rate at which the buffer moves through the capillary is given by \ref{4} .
\[ \mu _{eof} \ =\ \frac{\zeta \varepsilon }{4\pi \eta } \label{3} \]
\[ \nu _{eof}\ =\ \mu _{eof}E \label{4} \]
Zeta Potential
The zeta potential, ξ, also known as the electrokinetic potential is the electric potential at the interface of the double layer. Hence, in our case, it is the potential of the diffuse layer that is at a finite distance from the capillary wall. Zeta potential is mainly affected and directly proportional to two factors:
- The thickness of the double layer. A higher concentration of cations possibly due to an increase in the buffer`s ionic strength would lead to a decrease in the thickness of the double layer. As the thickness of the double layer decreases, the zeta potential would decrease that results in the decrease of the electroosmotic flow.
- The charge on the capillary walls. A greater density of the silanate ions corresponds to a larger zeta potential. The formation of silanate ions is pH dependent. Hence, at pH less than 2 there is a decrease in the zeta potential and the electroosmotic flow as the silanol exists in its protonated form. However, as the pH increases, there are more silanate ions formed causing an increase in zeta potential and hence, the electroosmotic flow.
Order of Elution
Electroosmotic flow of the buffer is generally greater than the electrophoretic flow of the analytes. Hence, even the anions would move to the cathode as illustrated in Figure \(\PageIndex{4}\). Small, highly charged cations would be the first to elute before larger cations with lower charge. This is followed by the neutral species which elutes as one band in the middle. The larger anions with low charge elute next and lastly, the highly charged small anion would have the longest elution time. This is clearly portrayed in the electropherogram in Figure \(\PageIndex{5}\).
Optimizing the CE Experiment
There are several components that can be varied to optimize the electropherogram obtained from CE. Hence, for any given setup certain parameters should be known:
- the total length of the capillary (L),
- the length the solutes travel from the start to the detector (l),
- the applied voltage (V).
Reduction in Migration Time, t mn
To shorten the analysis time, a higher voltage can be used or a shorter capillary tube can be used. However, it is important to note that the voltage cannot be arbitrarily high as it will lead to joule heating. Another possibility is to increase μ eof by increasing pH or decreasing the ionic strength of the buffer, \ref{5} .
\[ t_{mn} \ =\ \frac{1\ L}{(\mu _{ep} \ +\ \mu_{eof}) V } \label{5} \]
Efficiency
In chromatography, the efficiency is given by the number of theoretical plates, N. In CE, there exist a similar parameter, \ref{6} where D is the solute`s diffusion coefficient. Efficiency increase s with an increase in voltage applied as the solute spends less time in the capillary there will be less time for the solute to diffuse. Generally, for CE, N will be very large.
\[ N\ =\frac{1^{2}}{2Dt_{mn}} = \frac{\mu _{tot} V l}{2DL} \label{6} \]
Resolution Between Two Peaks
The resolution between two peaks, R, is defined by \ref{7} where Δv is the difference in velocity of two solutes and ṽ is the average velocity of two solutes.
\[ R= \frac{\sqrt{N} }{4} \times \frac{\Delta v}{ \tilde{\nu } } \label{7} \]
Substituting the equation by N gives \ref{8}
\[ R\ = 0.177(\mu _{ep,1} \ -\ \mu _{ep,2}) \sqrt{ \frac{V}{D(\nu _{av} + \mu _{eof})} } \label{8} \]
Therefore, increasing the applied voltage, V, will increase the resolution. However, it is not very effective as a 4-fold increase in applied voltage would only give a 2-fold increase in resolution. In addition, increase in N, the number of theoretical plates would result in better resolution.
Selectivity
In chromatography, selectivity, α, is defined as the ratio of the two retention factors of the solute. This is the same for CE, \ref{9} , where t 2 and t 1 are the retention times for the two solutes such that, α is more than 1.
\[ \alpha =\frac{t_{2}}{t_{1}} \label{9} \]
Selectivity can be improved by adjusting the pH of the buffer solution. The purpose is to change the charge of the species being eluted.
Comparison Between CE and HPLC
CE unlike High-performance liquid chromatography (HPLC) accommodates many samples and tends to have a better resolution and efficiency. A comparison between the two methods is given in Table \(\PageIndex{2}\).
| CE | HPLC |
|---|---|
| Wider selection of analyte to be analyzed | Limited by the solubility of the sample |
| Higher efficiency, no stationary mass transfer term as there is no stationary phase | Efficiency is lowered due to the stationary mass transfer term (equilibration between the stationary and mobile phase) |
| Electroosmotic flow profile in the capillary is flat as a result no band broadening. Better peak resolution and sharper peaks | Rounded laminar flow profile that is common in pressure driven systems such as HPLC. Resulting in broader peaks and lower resolution |
| Can be coupled to most detectors depending on application | Some detectors require the solvent to be changed and prior modification of the sample before analysis |
| Greater peak capacity as it uses a very large number of theoretical plates, N | The peak capacity is lowered as N is not as large |
| High voltages are used when carrying out the experiment | No need for high voltage |
Micellar Electrokinetic Chromatography
CE allows the separation of charged particles, and it is mainly compared to ion chromatography. However, no separation takes place for neutral species in CE. Thus, a modified CE technique named micellar electrokinetic chromatography (MEKC) can be used to separate neutrals based on its size and its affinity to the micelle. In MEKC, surfactant species is added to the buffer solution at a concentration at which micelles will form. An example of a surfactant is sodium dodecyl sulfate (SDS) as seen in Figure \(\PageIndex{6}\)
Neutral molecules are in dynamic equilibrium between the bulk solution and interior of the micelle. In the absence of the micelle the neutral species would reach the detector at t 0 but in the presence of the micelle, it reaches the detector at t mc , where t mc is greater than t 0 . The longer the neutral molecule remains in the micelle, the longer it's migration time. Thus small, non-polar neutral species that favor interaction with the interior of the micelle would take a longer time to reach the detector than a large, polar species. Anionic, cationic and zwitter ionic surfactants can be added to change the partition coefficient of the neutral species. Cationic surfactants would result in positive micelles that would move in the direction of electroosmotic flow. This enables it to move faster towards the cathode. However, due to the fast migration, it is possible that insufficient time is given for the neutral species to interact with the micelle resulting in poor separation. Thus, all factors must be considered before choosing the right surfactant to be used. The mechanism of separation between MEKC and liquid chromatography is the same. Both are dependent on the partition coefficient of the species between the mobile phase and stationary phase. The main difference lies in the pseudo stationary phase in MEKC, the micelles. The micelle which can be considered the stationary phase in MEKC moves at a slower rate than the mobile ions.
Quantum dots (QD) are semiconductor nanocrystals that lie in the size range of 1-10 nm, and they have different electrophoretic mobility due to their varying sizes and surface charge. CE can be used to separate and characterize such species, and a method to characterize and separate CdSe QD in the aqueous medium has been developed. The QDs were synthesized with an outer layer of trioctylphosphine (TOP, Figure \(\PageIndex{7} a\)) and trioctylphosphine oxide (TOPO, Figure \(\PageIndex{7} b\)), making the surface of the QD hydrophobic. The background electrolyte solution used was SDS, in order to make the QDs soluble in water and form a QD-TOPO/TOP-SDS complex. Different sizes of CdSe were used and the separation was with respect to the charge-to-mass ratio of the complexes. It was concluded from the study that the larger the CdSe core (i.e., the larger the charge-to-mass ratio) eluted out last. The electropherogram from the study is shown in Figure \(\PageIndex{8}\) from which it is visible that good separation had taken place by using CE. Laser-induced fluorescence detection was used, the buffer system was SDS, and the pH of the system set up was fixed at 6.5. The pH is highly important in this case as the stability of the system and the separation is dependent on it.