4.5: Miscellaneous Application Problems
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Learning Objectives
In this section, you will learn to apply to concepts for compound interest for savings and annuities to:
- Find the outstanding balance, partway through the term of a loan, of the future payments still remaining on the loan.
- Perform financial calculations in situations involving several stages of savings and/or annuities.
- Find the fair market value of a bond.
- Construct an amortization schedule for a loan.
We have already developed the tools to solve most finance problems. Now we use these tools to solve some application problems.
OUTSTANDING BALANCE ON A LOAN
One of the most common problems deals with finding the balance owed at a given time during the life of a loan. Suppose a person buys a house and amortizes the loan over 30 years, but decides to sell the house a few years later. At the time of the sale, he is obligated to pay off his lender, therefore, he needs to know the balance he owes. Since most long term loans are paid off prematurely, we are often confronted with this problem.
To find the outstanding balance of a loan at a specified time, we need to find the present value \(\mathrm{P}\) of all future payments that have not yet been paid. In this case t does not represent the entire term of the loan. Instead:
- \(t\) represents the time that still remains on the loan
- \(nt\) represents the total number of future payments.
Example \(\PageIndex{1}\)
Mr. Jackson bought his house in 1995, and financed the loan for 30 years at an interest rate of 7.8%. His monthly payment was $1260. In 2015, Mr. Jackson decides to pay off the loan. Find the balance of the loan he still owes.
Solution
The reader should note that the original amount of the loan is not mentioned in the problem. That is because we don't need to know that to find the balance.
The original loan was for 30 years. 20 years have past so there are years still remaining. 12(10) = 120 payments still remain to be paid on this loan.
As for the bank or lender is concerned, Mr. Jackson is obligated to pay $1260 each month for 10 more years; he still owes a total of 120 payments. But since Mr. Jackson wants to pay it all off now, we need to find the present value \(\mathrm{P}\) at the time of repayment of the remaining 10 years of payments of $1260 each month. Using the formula we get for the present value of an annuity, we get
\[\begin{aligned}
\mathrm{P}(1+.078 / 12)^{120} &=\frac{\left.\$ 1260\left[(1+.078 / 12)^{120}-1\right)\right]}{(.078 / 12)} \\
\mathrm{P}(2.17597) &=\$ 227957.85 \\
\mathrm{P} &=\$ 104761.48
\end{aligned} \nonumber \]
to find the outstanding balance of a loan
If a loan has a payment of \(m\) dollars made \(n\) times a year at an interest \(r\), then the outstanding value of the loan when there are \(t\) years still remaining on the loan is given by \(\mathrm{P}\):
\[\mathbf{P}(\mathbf{1}+\mathbf{r} / \mathbf{n})^{\mathbf{n} \mathbf{t}}=\frac{\mathbf{m}\left[(\mathbf{1}+\mathbf{r} / \mathbf{n})^{\mathbf{n} \mathbf{t}}-\mathbf{1} |\right.}{\mathbf{r} / \mathbf{n}} \nonumber \]
IMPORTANT: Note that \(t\) is not the original term of the loan but instead \(t\) is the amount of time still remaining in the future \(nt\) is the number of payments still remaining in the future
If the problem does not directly state the amount of time still remaining in the term of the loan, then it must be calculated BEFORE using the above formula as \(t\) = original term of loan - time already passed since the start date of the loan.
Note that there are other methods to find the outstanding balance on a loan, but the method illustrated above is the easiest.
One alternate method would be to use an amortization schedule, as illustrated toward the end of this section. An amortization schedule shows the payments, interest, and outstanding balance step by step after each loan payment. An amortization schedule is tedious to calculate by hand but can be easily constructed using spreadsheet software.
Another way to find the outstanding balance, that we will not illustrate here, is to find the difference A - B, where
A = the original loan amount (principal) accumulated to the date on which we want to find the outstanding balance (using compound interest formula)
B = the accumulated value of all payments that have been made as of the date on which we want to find the outstanding balance (using formula for accumulated value of an annuity)
In this case we would need do a compound interest calculation and an annuity calculation; we then need to find the difference between them. Three calculations are needed instead of one.
It is a mathematically acceptable way to calculate the outstanding balance. However, it is very strongly recommended that students use the method explained in box above and illustrated in Example \(\PageIndex{1}\), as it is much simpler.
PROBLEMS INVOLVING MULTIPLE STAGES OF SAVINGS AND/OR ANNUITIES
Consider the following situations:
- Suppose a baby, Aisha, is born and her grandparents invest $5000 in a college fund. The money remains invested for 18 years until Aisha enters college, and then is withdrawn in equal semiannual payments over the 4 years that Aisha expects to need to finish college. The college investment fund earns 5% interest compounded semiannually. How much money can Aisha withdraw from the account every six months while she is in college?
- Aisha graduates college and starts a job. She saves $1000 each quarter, depositing it into a retirement savings account. Suppose that Aisha saves for 30 years and then retires. At retirement she wants to withdraw money as an annuity that pays a constant amount every month for 25 years. During the savings phase, the retirement account earns 6% interest compounded quarterly. During the annuity payout phase, the retirement account earns 4.8% interest compounded monthly. Calculate Aisha’s monthly retirement annuity payout.
These problems appear complicated. But each can be broken down into two smaller problems involving compound interest on savings or involving annuities. Often the problem involves a savings period followed by an annuity period. ; the accumulated value from first part of the problem may become a present value in the second part. Read each problem carefully to determine what is needed.
Example \(\PageIndex{2}\)
Suppose a baby, Aisha, is born and her grandparents invest $8000 in a college fund. The money remains invested for 18 years until Aisha enters college, and then is withdrawn in equal semiannual payments over the 4 years that Aisha expects to attend college. The college investment fund earns 5% interest compounded semiannually. How much money can Aisha withdraw from the account every six months while she is in college?
Solution
Part 1: Accumulation of College Savings: Find the accumulated value at the end of 18 years of a sum of $8000 invested at 5% compounded semiannually.
\[\begin{array}{l}
A=\$ 8000(1+.05 / 2)^{(2 \times 18)}=\$ 8000(1.025)^{36}=\$ 8000(2.432535) \\
A=\$ 19460.28
\end{array} \nonumber \]
Part 2: Seminannual annuity payout from savings to put toward college expenses. Find the amount of the semiannual payout for four years using the accumulated savings from part 1 of the problem with an interest rate of 5% compounded semiannually.
\(A\)= $19460.28 in Part 1 is the accumulated value at the end of the savings period. This becomes the present value \(P\)=$19460.28 when calculating the semiannual payments in Part 2.
\[\begin{array}{c}
\$ 19460.28\left(1+\frac{.05}{2}\right)^{2 \times 4}=\frac{m\left[\left(1+\frac{05}{2}\right)^{2 \times 4}-1\right]}{(.05 / 2)} \\
\$ 23710.46=m(8.73612) \\
m=\$ 2714.07
\end{array} \nonumber \]
Aisha will be able to withdraw $2714.07 semiannually for her college expenses.
Example \(\PageIndex{3}\)
Aisha graduates college and starts a job. She saves $1000 each quarter, depositing it into a retirement savings account. Suppose that Aisha saves for 30 years and then retires. At retirement she wants to withdraw money as an annuity that pays a constant amount every month for 25 years. During the savings phase, the retirement account earns 6% interest compounded quarterly. During the annuity payout phase, the retirement account earns 4.8% interest compounded monthly. Calculate Aisha’s monthly retirement annuity payout.
Solution
Part 1: Accumulation of Retirement Savings: Find the accumulated value at the end of 30 years of $1000 deposited at the end of each quarter into a retirement savings account earning 6% interest compounded quarterly.
\[\begin{array}{l}
A=\frac{\$ 1000\left[(1+.06 / 4)^{4 \times 30}-1\right]}{(.06 / 4)} \\
A=\$ 331288.19
\end{array} \nonumber \]
Part 2: Monthly retirement annuity payout: Find the amount of the monthly annuity payments for 25 years using the accumulated savings from part 1 of the problem with an interest rate of 4.8% compounded monthly.
\(A\)= $331288.19 in Part 1 is the accumulated value at the end of the savings period. This amount will become the present value \(\mathrm{P}\) =$331288.19 when calculating the monthly retirement annuity payments in Part 2.
\[\begin{array}{l}
\$ 331288.19(1+.048 / 12)^{12 \times 25}=\frac{m\left[(1+.048 / 12)^{12 \times 25}-1\right]}{(.048 / 12)} \\
\$ 1097285.90=m(578.04483) \\
m=\$ 1898.27
\end{array} \nonumber \]
Aisha will have a monthly retirement annuity income of $1898.27 when she retires.
FAIR MARKET VALUE OF A BOND
Whenever a business, and for that matter the U. S. government, needs to raise money it does it by selling bonds. A bond is a certificate of promise that states the terms of the agreement. Usually the business sells bonds for the face amount of $1,000 each for a stated term , a period of time ending at a specified maturity date.
The person who buys the bond, the bondholder , pays $1,000 to buy the bond.
The bondholder is promised two things: First that he will get his $1,000 back at the maturity date, and second that he will receive a fixed amount of interest every six months.
As the market interest rates change, the price of the bond starts to fluctuate. The bonds are bought and sold in the market at their fair market value .
The interest rate a bond pays is fixed, but if the market interest rate goes up, the value of the bond drops since the money invested in the bond could earn more if invested elsewhere. When the value of the bond drops, we say it is trading at a discount .
On the other hand, if the market interest rate drops, the value of the bond goes up since the bond now yields a higher return than the market interest rate, and we say it is trading at a premium .
Example \(\PageIndex{4}\)
The Orange Computer Company needs to raise money to expand. It issues a 10-year $1,000 bond that pays $30 every six months. If the current market interest rate is 7%, what is the fair market value of the bond?
Solution
The bond certificate promises us two things - An amount of $1,000 to be paid in 10 years, and a semi-annual payment of $30 for ten years. Therefore, to find the fair market value of the bond, we need to find the present value of the lump sum of $1,000 we are to receive in 10 years, as well as, the present value of the $30 semi-annual payments for the 10 years.
We will let P 1 = the present value of the face amount of $1,000
\[P_{1}(1+.07 / 2)^{20}=\$ 1,000 \nonumber \]
Since the interest is paid twice a year, the interest is compounded twice a year and \(nt\) = 2(10)=20
\[\begin{array}{l}
P_{1}(1.9898)=\$ 1,000 \\
P_{1}=\$ 502.56
\end{array} \nonumber \]
We will let P 2 = the present value of the $30 semi-annual payments is
\[\begin{aligned}
\mathrm{P}_{2}(1+.07 / 2)^{20} &=\frac{\$ 30\left[(1+.07 / 2)^{20}-1\right]}{(.07 / 2)} \\
\mathrm{P}_{2}(1.9898) &=848.39 \\
\mathrm{P}_{2} &=\$ 426.37
\end{aligned} \nonumber \]
The present value of the lump-sum $1,000 = $502.56
The present value of the $30 semi-annual payments = $426.37
The fair market value of the bond is P = P
1+
P
2
= $502.56 + $426.37 = $928.93
Note that because the market interest rate of 7% is higher than the bond’s implied interest rate of 6% implied by the semiannual payments, the bond is selling at a discount; its fair market value of $928.93 is less than its face value of $1000.
Example \(\PageIndex{5}\)
A state issues a 15 year $1000 bond that pays $25 every six months. If the current market interest rate is 4%, what is the fair market value of the bond?
Solution
The bond certificate promises two things - an amount of $1,000 to be paid in 15 years, and semi-annual payments of $25 for 15 years. To find the fair market value of the bond, we find the present value of the $1,000 face value we are to receive in 15 years and add it to the present value of the $25 semi-annual payments for the 15 years. In this example, \(nt = 2(15)=30\).
We will let P 1 = the present value of the lump-sum $1,000
\[\begin{array}{l}
\mathrm{P}_{1}(1+.04 / 2)^{30}=\$ 1,000 \\
\mathrm{P}_{1}=\$ 552.07
\end{array} \nonumber \]
We will let P 2 = the present value of the $25 semi-annual payments is
\[\begin{array}{l}
\mathrm{P}_{2}(1+.04 / 2)^{30}=\frac{\$ 25\left[(1+.04 / 2)^{30}-1\right]}{(.04 / 2)} \\
\mathrm{P}_{2}(1.18114)=\$ 1014.20 \\
\mathrm{P}_{2}=\$ 559.90
\end{array} \nonumber \]
The present value of the lump-sum $1,000 = $552.07
The present value of the $30 semi-annual payments = $559.90
Therefore, the fair market value of the bond is
\[P=P_{1} + P_{2}=\$ 552.07+\$ 559.90=\$ 1111.97\nonumber \]
Because the market interest rate of 4% is lower than the interest rate of 5% implied by the semiannual payments, the bond is selling at a premium: the fair market value of $1,111.97 is more than the face value of $1,000.
To summarize:
to find the Fair Market Value of a Bond
Find the present value of the face amount \(\mathrm{A}\) that is payable at the maturity date:
\[\mathbf{A}=\mathbf{P}_{1}(\mathbf{1}+\mathbf{r} / \mathbf{n})^{\mathbf{n} \mathbf{t}} ; \text { solve to find } \mathrm{P}_{1} \nonumber \]
Find the present value of the semiannually payments of $\(m\) over the term of the bond:
\[\mathbf{P}_{2}(\mathbf{1}+\mathbf{r} / \mathbf{n})^{\mathbf{n} \mathbf{t}}=\frac{\mathbf{m}\left[(\mathbf{1}+\mathbf{r} / \mathbf{n})^{\mathbf{n} \mathbf{t}}-\mathbf{1}\right]}{\mathbf{r} / \mathbf{n}} \quad ; \text { solve to find } \mathbf{P}_{2} \nonumber \]
The fair market value (or present value or price or current value) of the bond is the sum of the present values calculated above:
\[\mathrm{P}=\mathbf{P}_{1}+\mathbf{P}_{2} \nonumber \]
AMORTIZATION SCHEDULE FOR A LOAN
An amortization schedule is a table that lists all payments on a loan, splits them into the portion devoted to interest and the portion that is applied to repay principal, and calculates the outstanding balance on the loan after each payment is made.
Example \(\PageIndex{6}\)
An amount of $500 is borrowed for 6 months at a rate of 12%. Make an amortization schedule showing the monthly payment, the monthly interest on the outstanding balance, the portion of the payment contributing toward reducing the debt, and the outstanding balance.
Solution
The reader can verify that the monthly payment is $86.27.
The first month, the outstanding balance is $500, and therefore, the monthly interest on the outstanding balance is
(outstanding balance)(the monthly interest rate) = ($500)(.12/12) = $5
This means, the first month, out of the $86.27 payment, $5 goes toward the interest and the remaining $81.27 toward the balance leaving a new balance of $500 - $81.27 = $418.73.
Similarly, the second month, the outstanding balance is $418.73, and the monthly interest on the outstanding balance is ($418.73)(.12/12) = $4.19. Again, out of the $86.27 payment, $4.19 goes toward the interest and the remaining $82.08 toward the balance leaving a new balance of $418.73 - $82.08 = $336.65. The process continues in the table below.
| Payment # | Payment | Interest | Debt Payment | Balance |
|---|---|---|---|---|
| 1 | $86.27 | $5 | $81.27 | $418.73 |
| 2 | $86.27 | $4.19 | $82.08 | $336.65 |
| 3 | $86.27 | $3.37 | $82.90 | $253.75 |
| 4 | $86.27 | $2.54 | $83.73 | $170.02 |
| 5 | $86.27 | $1.70 | $84.57 | $85.45 |
| 6 | $86.27 | $0.85 | $85.42 | $0.03 |
Note that the last balance of 3 cents is due to error in rounding off.
An amortization schedule is usually lengthy and tedious to calculate by hand. For example, an amortization schedule for a 30 year mortgage loan with monthly payments would have (12)(30)=360 rows of calculations in the amortization schedule table. A car loan with 5 years of monthly payments would have 12(5)=60 rows of calculations in the amortization schedule table. However it would be straightforward to use a spreadsheet application on a computer to do these repetitive calculations by inputting and copying formulas for the calculations into the cells.
Most of the other applications in this section's problem set are reasonably straightforward, and can be solved by taking a little extra care in interpreting them. And remember, there is often more than one way to solve a problem.


13.2: Arrhenius Definition of Acids and Bases
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Learning Objectives
- To recognize a compound as an Arrhenius acid or an Arrhenius base.
- To describe characteristics of acids and bases.
- To write equations of neutralization reactions.
One way to define a class of compounds is by describing the various characteristics its members have in common. In the case of the compounds known as acids, the common characteristics include a sour taste, the ability to change the color of the vegetable dye litmus to red, and the ability to dissolve certain metals and simultaneously produce hydrogen gas. For the compounds called bases, the common characteristics are a slippery texture, a bitter taste, and the ability to change the color of litmus to blue. Acids and bases also react with each other to form compounds generally known as salts.
Although we include their tastes among the common characteristics of acids and bases, we never advocate tasting an unknown chemical!
Chemists prefer, however, to have definitions for acids and bases in chemical terms. The Swedish chemist Svante Arrhenius developed the first chemical definitions of acids and bases in the late 1800s. Arrhenius defined an acid as a compound that increases the concentration of hydrogen ion (H + ) in aqueous solution. Many acids are simple compounds that release a hydrogen cation into solution when they dissolve. Similarly, Arrhenius defined a base as a compound that increases the concentration of hydroxide ion (OH − ) in aqueous solution. Many bases are ionic compounds that have the hydroxide ion as their anion, which is released when the base dissolves in water.
| Acids | Bases | ||
|---|---|---|---|
| Formula | Name | Formula | Name |
| HCl(aq) | hydrochloric acid | NaOH(aq) | sodium hydroxide |
| HBr(aq) | hydrobromic acid | KOH(aq) | potassium hydroxide |
| HI(aq) | hydriodic acid | Mg(OH) 2 (aq) | magnesium hydroxide |
| H 2 S(aq) | hydrosulfuric acid | Ca(OH) 2 (aq) | calcium hydroxide |
| HC 2 H 3 O 2 (aq) | acetic acid | NH 3 (aq) | ammonia |
| HNO 3 (aq) | nitric acid | NaHCO 3 (aq) | sodium bicarbonate |
| HNO 2 (aq) | nitrous acid | CaCO 3 (aq) | calcium carbonate |
| H 2 SO 4 (aq) | sulfuric acid | ||
| H 2 SO 3 (aq) | sulfurous acid | ||
| HClO 3 (aq) | chloric acid | ||
| HClO 4 (aq) | perchloric acid | ||
| HClO 2 (aq) | chlorous acid | ||
| H 3 PO 4 (aq) | phosphoric acid | ||
| H 3 PO 3 (aq) | phosphorous acid | ||
| H 2 CO 3 (aq) | carbonic acid |
Many bases and their aqueous solutions are named using the normal rules of ionic compounds that were presented previously ; that is, they are named as hydroxide compounds. For example, the base sodium hydroxide (NaOH) is both an ionic compound and an aqueous solution. However, aqueous solutions of acids have their own naming rules. The names of binary acids (compounds with hydrogen and one other element in their formula) are based on the root of the name of the other element preceded by the prefix hydro - and followed by the suffix - ic acid . Thus, an aqueous solution of HCl [designated “HCl(aq)”] is called hydrochloric acid, H 2 S(aq) is called hydrosulfuric acid, and so forth. Acids composed of more than two elements (typically hydrogen and oxygen and some other element) have names based on the name of the other element, followed by the suffix - ic acid or -ous acid , depending on the number of oxygen atoms in the acid’s formula. Other prefixes, like per- and hypo- , also appear in the names for some acids. Unfortunately, there is no strict rule for the number of oxygen atoms that are associated with the - ic acid suffix; the names of these acids are best memorized. Table \(\PageIndex{1}\) lists some acids and bases and their names. Note that acids have hydrogen written first, as if it were the cation, while most bases have the negative hydroxide ion, if it appears in the formula, written last.
The name oxygen comes from the Latin meaning “acid producer” because its discoverer, Antoine Lavoisier, thought it was the essential element in acids. Lavoisier was wrong, but it is too late to change the name now.
Example \(\PageIndex{1}\)
Name each substance.
- HF(aq)
- Sr(OH) 2 (aq)
Solution
- This acid has only two elements in its formula, so its name includes the hydro - prefix. The stem of the other element’s name, fluorine, is fluor , and we must also include the - ic acid ending. Its name is hydrofluoric acid.
- This base is named as an ionic compound between the strontium ion and the hydroxide ion: strontium hydroxide.
Exercise \(\PageIndex{1}\)
Name each substance.
- H 2 Se(aq)
- Ba(OH) 2 (aq)
- Answer
-
a. hydroselenic acid
b. barium hydroxide
Notice that one base listed in Table \(\PageIndex{1}\)—ammonia—does not have hydroxide as part of its formula. How does this compound increase the amount of hydroxide ion in aqueous solution? Instead of dissociating into hydroxide ions, ammonia molecules react with water molecules by taking a hydrogen ion from the water molecule to produce an ammonium ion and a hydroxide ion:
\[NH_{3(aq)} + H_2O_{(ℓ)} \rightarrow NH^+_{4(aq)} + OH^−_{(aq)} \label{Eq1} \]
Because this reaction of ammonia with water causes an increase in the concentration of hydroxide ions in solution, ammonia satisfies the Arrhenius definition of a base. Many other nitrogen-containing compounds are bases because they too react with water to produce hydroxide ions in aqueous solution.
Neutralization
As we noted previously, acids and bases react chemically with each other to form salts . A salt is a general chemical term for any ionic compound formed from an acid and a base. In reactions where the acid is a hydrogen ion containing compound and the base is a hydroxide ion containing compound, water is also a product. The general reaction is as follows:
acid + base → water + salt
The reaction of acid and base to make water and a salt is called neutralization . Like any chemical equation, a neutralization chemical equation must be properly balanced. For example, the neutralization reaction between sodium hydroxide and hydrochloric acid is as follows:
\[NaOH{(aq)} + HCl_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(ℓ)} \label{Eq2} \]
with coefficients all understood to be one. The neutralization reaction between sodium hydroxide and sulfuric acid is as follows:
\[2NaOH_{(aq)} + H_2SO_{4(aq)} \rightarrow Na_2SO_{4(aq)} + 2H_2O_{(ℓ)} \label{Eq3} \]
Once a neutralization reaction is properly balanced, we can use it to perform stoichiometry calculations, such as the ones we practiced earlier.
There are a number of examples of acid-base chemistry in everyday life. One example is the use of baking soda, or sodium bicarbonate in baking. NaHCO 3 is a base. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbles of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter “rises.” Baking powder is a combination of sodium bicarbonate, and one or more acid salts that react when the two chemicals come in contact with water in the batter.
\[ HCO_3^- (aq) + H^+ (aq) \rightarrow H_2 CO_3 (aq) \label{4.3.19} \]
\[ H_2 CO_3 (aq) \rightarrow CO_2 (g) + H_2 O(l) \nonumber \]
Example \(\PageIndex{2}\)
Nitric acid [HNO 3 (aq)] can be neutralized by calcium hydroxide [Ca(OH) 2 (aq)].
- Write a balanced chemical equation for the reaction between these two compounds and identify the salt it produces.
- For one reaction, 16.8 g of HNO 3 is present initially. How many grams of Ca(OH) 2 are needed to neutralize that much HNO 3 ?
- In a second reaction, 805 mL of 0.672 M Ca(OH) 2 is present initially. What volume of 0.432 M HNO 3 solution is necessary to neutralize the Ca(OH) 2 solution?
Solution
- Because there are two OH − ions in the formula for Ca(OH) 2 , we need two moles of HNO 3 to provide H + ions. The balanced chemical equation is as follows: Ca(OH) 2 (aq) + 2HNO 3 (aq) → Ca(NO 3 ) 2 (aq) + 2H 2 O(ℓ)
The salt formed is calcium nitrate.
- This calculation is much like the calculations we did in Chapter 6. First we convert the mass of HNO 3 to moles using its molar mass of 1.01 + 14.01 + 3(16.00) = 63.02 g/mol; then we use the balanced chemical equation to determine the related number of moles of Ca(OH) 2 needed to neutralize it; and then we convert that number of moles of Ca(OH) 2 to the mass of Ca(OH) 2 using its molar mass of 40.08 + 2(1.01) + 2(16.00) = 74.10 g/mol.
\(\mathrm{16.8\: g\: HNO_3 \times \dfrac{1\: mol\: HNO_3}{63.02\: g\ HNO_3} \times \dfrac{1\: mol\: Ca(OH)_2}{2\: mol\: HNO_3} \times \dfrac{74.10\: g\: Ca(OH)_2}{1\: mol\: Ca(OH)_2}=9.88\: g\: Ca(OH)_2\: needed}\)
- Having concentration information allows us to employ the skills we developed in Chapter 9. We have two alternative solutions: the multi-step process and the combined one-line process (found at the bottom).
First, we use the concentration and volume data to determine the number of moles of Ca(OH) 2 present. Recognizing that 805 mL = 0.805 L,
\(\mathrm{0.672\: M\: Ca(OH)_2=\dfrac{mol\: Ca(OH)_2}{0.805\: L\: soln}}\)
0.672 M CaOH) 2 × (0.805 L soln) = mol Ca(OH) 2 = 0.541 mol Ca(OH) 2
We combine this information with the proper ratio from the balanced chemical equation to determine the number of moles of HNO 3 needed:
\(\mathrm{0.541\: mol\: Ca(OH)_2 \times \dfrac{2\: mol\: HNO_3}{1\: mol\: Ca(OH)_2}=1.08\: mol\: HNO_3}\)
Now, using the definition of molarity one more time, we determine the volume of acid solution needed:
\(\mathrm{0.432\: M\: HNO_3=\dfrac{1.08\: mol\: HNO_3}{volume\: of\: HNO_3}}\)
\(\mathrm{volume\: of\: HNO_3=\dfrac{1.08\: mol\: HNO_3}{0.432\: M\: HNO_3}=2.50\: L=2.50\times10^3\: mL\: HNO_3}\)
\(\mathrm{0.805\: L\: Ca(OH)_2\: soln \times \dfrac{0.672\: mol\: Ca(OH)_2 }{1\: L\ Ca(OH)_2\: soln } \times \dfrac{2\: mol\: HNO_3}{1\: mol\: Ca(OH)_2} \times \dfrac{1\: L\: HNO_3\: soln}{0.432\: mol\: HNO_3}=2.50\: L\: HNO_3\: soln\: needed}\)
Exercise \(\PageIndex{2}\)
Hydrocyanic acid [HCN(aq)] can be neutralized by potassium hydroxide [KOH(aq)].
- Write a balanced chemical equation for the reaction between these two compounds and identify the salt it produces.
- For one reaction, 37.5 g of HCN is present initially. How many grams of KOH are needed to neutralize that much HCN?
- In a second reaction, 43.0 mL of 0.0663 M KOH is present initially. What volume of 0.107 M HCN solution is necessary to neutralize the KOH solution?
- Answer
-
a. KOH (aq) + HCN (aq) → KCN(aq) + H 2 O(ℓ) KCN is the salt.
b. 77.8 g
c. 0.0266 L or 26.6 mL
Hydrocyanic acid (HCN) is one exception to the acid-naming rules that specify using the prefix hydro- for binary acids (acids composed of hydrogen and only one other element).
Stomach Antacids
Our stomachs contain a solution of roughly 0.03 M HCl, which helps us digest the food we eat. The burning sensation associated with heartburn is a result of the acid of the stomach leaking through the muscular valve at the top of the stomach into the lower reaches of the esophagus. The lining of the esophagus is not protected from the corrosive effects of stomach acid the way the lining of the stomach is, and the results can be very painful. When we have heartburn, it feels better if we reduce the excess acid in the esophagus by taking an antacid. As you may have guessed, antacids are bases. One of the most common antacids is calcium carbonate, CaCO 3 . The reaction,
\[CaCO_3(s)+2HCl(aq)⇌CaCl_2(aq)+H_2O(l)+CO_2(g) \nonumber \]
not only neutralizes stomach acid, it also produces CO 2 ( g ), which may result in a satisfying belch.
Milk of Magnesia is a suspension of the sparingly soluble base magnesium hydroxide, Mg(OH) 2 . It works according to the reaction:
\[Mg(OH)_2(s)⇌Mg^{2+}(aq)+2OH^-(aq) \nonumber \]
The hydroxide ions generated in this equilibrium then go on to react with the hydronium ions from the stomach acid, so that :
\[H_3O^+ + OH^- ⇌ 2H_2O(l) \nonumber \]
This reaction does not produce carbon dioxide, but magnesium-containing antacids can have a laxative effect. Several antacids have aluminum hydroxide, Al(OH) 3 , as an active ingredient. The aluminum hydroxide tends to cause constipation, and some antacids use aluminum hydroxide in concert with magnesium hydroxide to balance the side effects of the two substances.
Example \(\PageIndex{3}\)
Assume that the stomach of someone suffering from acid indigestion contains 75 mL of 0.20 M HCl. How many Tums tablets are required to neutralize 90% of the stomach acid, if each tablet contains 500 mg of CaCO 3 ? (Neutralizing all of the stomach acid is not desirable because that would completely shut down digestion.)
Solution
- Write the balanced chemical equation for the reaction and then decide whether the reaction will go to completion.
- Calculate the number of moles of acid present. Multiply the number of moles by the percentage to obtain the quantity of acid that must be neutralized. Using mole ratios, calculate the number of moles of base required to neutralize the acid.
- Calculate the number of moles of base contained in one tablet by dividing the mass of base by the corresponding molar mass. Calculate the number of tablets required by dividing the moles of base by the moles contained in one tablet.
A. We first write the balanced chemical equation for the reaction:
\(2HCl(aq) + CaCO_3(s) \rightarrow CaCl_2(aq) + H_2CO_3(aq)\)
Each carbonate ion can react with 2 mol of H + to produce H 2 CO 3 , which rapidly decomposes to H 2 O and CO 2 . Because HCl is a strong acid and CO 3 2 − is a weak base, the reaction will go to completion.
B. Next we need to determine the number of moles of HCl present:
\( 75\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .20\: mol\: HCl} {\cancel{L}} \right) = 0. 015\: mol\: HCl \)
Because we want to neutralize only 90% of the acid present, we multiply the number of moles of HCl by 0.90:
\((0.015\: mol\: HCl)(0.90) = 0.014\: mol\: HCl\)
We know from the stoichiometry of the reaction that each mole of CaCO 3 reacts with 2 mol of HCl, so we need
\( moles\: CaCO_3 = 0 .014\: \cancel{mol\: HCl} \left( \dfrac{1\: mol\: CaCO_3}{2\: \cancel{mol\: HCl}} \right) = 0 .0070\: mol\: CaCO_3 \)
C. Each Tums tablet contains
\( \left( \dfrac{500\: \cancel{mg\: CaCO_3}} {1\: Tums\: tablet} \right) \left( \dfrac{1\: \cancel{g}} {1000\: \cancel{mg\: CaCO_3}} \right) \left( \dfrac{1\: mol\: CaCO_3} {100 .1\: \cancel{g}} \right) = 0 .00500\: mol\: CaCO_ 3 \)
Thus we need \(\dfrac{0.0070\: \cancel{mol\: CaCO_3}}{0.00500\: \cancel{mol\: CaCO_3}}= 1.4\) Tums tablets.
Exercise \(\PageIndex{3}\)
Assume that as a result of overeating, a person’s stomach contains 300 mL of 0.25 M HCl. How many Rolaids tablets must be consumed to neutralize 95% of the acid, if each tablet contains 400 mg of NaAl(OH) 2 CO 3 ? The neutralization reaction can be written as follows:
\( NaAl(OH)_2CO_3(s) + 4HCl(aq) \rightarrow AlCl_3(aq) + NaCl(aq) + CO_2(g) + 3H_2O(l) \)
- Answer
-
6.4 tablets
Key Takeaway
- An Arrhenius acid increases the H + ion concentration in water, while an Arrhenius base increases the OH − ion concentration in water.


6.A: Chapter 6 Review Questions
Complete as directed.
Fill in the Blanks:
1. Maximum safe load limit for floors must be posted in all buildings and structures and in the appropriate storage areas. The safe limits must be listed in ________ for all floors except those located on grade.
2. Materials stored inside of buildings under construction shall not be placed within ________ feet of any hoist way or inside floor openings. Such materials shall also be kept at least ________ feet from an exterior wall which does not extend above the top of the material stored.
3. Unless materials are for immediate use, materials shall not be stored on ________ or ________ .
4. When masonry blocks are stored in stacks higher than ________ feet, the stacks shall be tapered back one-half tier above this level.
5. ________ and ________ must be kept clear to provide for ready access and safe movement of material handling equipment or employees.
6. Rigging equipment for material handling shall be ________ prior to use on each shift and as necessary during its use to ensure that it is safe. Defective rigging equipment shall not be used.
7. If the type of installation requires that special hooks, grabs, clamps, etc. must be used, they shall be marked to indicate their maximum safe working loads and they shall be proof-tested prior to their use to ________ of their rated load.
8. ________ is the only permissible method for attaching end fittings to synthetic webbing and to form eyes in the webbing.
Multiple Choice:
9. When lumber is stored, which of the following conditions shall be followed:
a. ________ All nails shall first be removed.
b. ________ It must be stacked on level and solidly supported sills.
c. ________ It must be stable and self-supporting.
d. ________ The piles must not exceed 20 feet provided the lumber to be handled manually, does not exceed 16 feet.
e. ________ All of the above.
True or False:
10. T or F
Materials stored in tiers shall be stacked, racked, blocked, interlocked or otherwise secured to prevent sliding, falling or collapse.


10.20: References
Editors, P. (n.d.). Fetal development: Your baby's journey from conception to birth. Parents.
Retrieved July 21, 2022, from https://www.parents.com/pregnancy/my...tion-to-birth/
Fox, S. I. (2022). Human physiology. McGraw Hill LLC.
Get help and support. Baby Blues Connection. (n.d.). Retrieved July 21, 2022, from
https://www.babybluesconnection.org/
Healthy Moms. Strong Babies. Home | March of Dimes | Healthy Moms. Strong Babies. (n.d.). Retrieved July 21, 2022, from http://www.marchofdimes.com/
Join the cause - international cesarean awareness network. International Cesarean Awareness Network - Education, Support, & Advocacy for Birth Justice & Healing. (2022, July Retrieved July 21, 2022, from http://www.ican-online.org/
Lees, C., Reynolds, K., & McCartan, G. (2007). Pregnancy & birth: Your questions answered. DK.
P, M., & R, A. (2016, October 12). ALLRefer. Retrieved July 21, 2022, from
http://health.allrefer.com/health/fe...ment-info.html
Scanlon, V. C., & Sanders, T. (2019). Essentials of anatomy and physiology. F.A. Davis Company.
Shively, M. J., & Homan, D. P. (2000). Fundamental concepts of human anatomy. Sewing Horse.
Stoppard, M. (1999). Dr Miriam Stoppard's New Parent. Dorling Kindersley.
Sussman, K. (n.d.). Merlot Research Group, Inc.. Merlot Research Group, Inc. Retrieved July 21, 2022, from http://www.MERLOT.com/


1.6: Workplace Hazards
By recognizing and understanding the potential hazards in your work area, you can help prevent accidents from occurring. All accidents are preventable. Both employees and employers must take responsibility for making the workplace safe. Causing injury by failing to follow known safety procedures or knowingly allowing uncontrolled hazardous situations to develop are serious safety violations.
Workplace hazards include things such as slippery floors, loose floor mats, and sharp knives, as well as hazardous materials. It is important for all employees to be aware of hazards, even if they seem obvious. Employers should provide information and training on any safe work procedures related to the job site. Safe work procedures are specific directions for doing a task or operating equipment that may pose a risk or hazard to the worker. Workers should always ask their supervisor if there are any safe work procedures they need to be aware of and/or any written instructions they should be following.
One of the main hazards in any workplace are cleaning products, some of which are everyday products that a person may not regard as hazardous, such as sanitizers and household cleansers. Cleaning products and all other materials that are potentially hazardous are governed by the Workplace Hazardous Materials Information System (WHMIS)
Workplace Hazardous Materials Information System (WHMIS)
The Workplace Hazardous Materials Information System ( WHMIS ) is Canada’s national system of classifying and communicating information about hazardous materials in the workplace. The key components of the system are the classification of materials, cautionary labels and symbols, material safety data sheets ( MSDS ) and workplace education and training programs.
WHMIS was first introduced in 1988, and in 2015 it was changed to include the Globally Harmonized System of Classification and Labelling for chemicals ( GHS ). Currently there is a transition period, until 2018, to allow for suppliers, employers, and workers, to adapt to the new requirements. During this transition period, manufacturers, distributors, and employers will be allowed to comply with either the former WHMIS 1988 requirements or the new WHMIS 2015 requirements to allow for a phased-in approach. [1]
Federal legislation, the Hazardous Products Act HPA [2] establishes which products are regulated under WHMIS and have been traditionally categorized as controlled products within the Controlled Product Regulation (CPR). Changes in 2015 to align with the GHS include a change in terminology, classifying substances as hazardous products , and replacing the CPR with a new Hazardous Product Regulation (HPR) . [3]
Regardless of the changes in terminology, WHMIS covers the sale, distribution, and use of hazardous materials in the workplace.
WorkSafeBC regulations cover the use of hazardous materials in the workplace and identify employers’ responsibilities. Workers who work with hazardous products must know how to handle them safely, which requires being properly trained on their use.
WHMIS categorizes hazardous materials in six hazard classes. These classes are depicted by eight symbols that identify the specific hazards of controlled products. After a controlled product has been classified, the following three WHMIS elements are used to communicate health and safety information:
- WHMIS label
- Material safety data sheets (MSDS)
- WHMIS education and training programs
Hazardous (Controlled) Products
WHMIS defines a hazardous or controlled product as a pure substance or mixture that meets or exceeds criteria for inclusion in one or more of the WHMIS hazard classes. The six classes and eight hazard symbols from WHIMS 1988 are shown in Figure 1, and the symbols from the WHMIS 2015 program, which uses the GHS, are included in Figure 2. The appropriate symbol must appear on the applicable supplier labels.
WHMIS Education and Training Programs
Everyone who uses hazardous materials must be trained to use them properly and to use the personal protective equipment specified in the MSDS. It is the responsibility of the employer to provide WHMIS training as a part of an occupational health and safety program. If you have not received training in WHMIS, ask your employer to schedule a session for you as soon as possible.
Training must:
- Address any WHMIS requirements
- Be reviewed at least annually (more frequently if work conditions or hazards change)
- Provide for the periodic assessment of knowledge by testing or observation


21.6: DC Circuits Containing Resistors and Capacitors
Learning Objectives
By the end of this section, you will be able to:
- Explain the importance of the time constant, τ , and calculate the time constant for a given resistance and capacitance.
- Explain why batteries in a flashlight gradually lose power and the light dims over time.
- Describe what happens to a graph of the voltage across a capacitor over time as it charges.
- Explain how a timing circuit works and list some applications.
- Calculate the necessary speed of a strobe flash needed to “stop” the movement of an object over a particular length.
When you use a flash camera, it takes a few seconds to charge the capacitor that powers the flash. The light flash discharges the capacitor in a tiny fraction of a second. Why does charging take longer than discharging? This question and a number of other phenomena that involve charging and discharging capacitors are discussed in this module.
RC Circuits
An \(RC\) circuit is one containing a resisto r \(R\) and capacitor \(C\). The capacitor is an electrical component that stores electric charge.
Figure shows a simple \(RC\) circuit that employs a DC (direct current) voltage source. The capacitor is initially uncharged. As soon as the switch is closed, current flows to and from the initially uncharged capacitor. As charge increases on the capacitor plates, there is increasing opposition to the flow of charge by the repulsion of like charges on each plate.
In terms of voltage, this is because voltage across the capacitor is given by \(V_c = Q/C\), where \(Q\) is the amount of charge stored on each plate and \(C\) is the capacitance . This voltage opposes the battery, growing from zero to the maximum emf when fully charged. The current thus decreases from its initial value of \(I_9 = \frac{emf}{R}\) to zero as the voltage on the capacitor reaches the same value as the emf. When there is no current, there is no \(IR\) drop, and so the voltage on the capacitor must then equal the emf of the voltage source. This can also be explained with Kirchhoff’s second rule (the loop rule), discussed in Kirchhoff’s Rules , which says that the algebraic sum of changes in potential around any closed loop must be zero.
The initial current is \(I_0 = \frac{emf}{R}\), because all of the \(IR\) drop is in the resistance. Therefore, the smaller the resistance, the faster a given capacitor will be charged. Note that the internal resistance of the voltage source is included in \(R\), as are the resistances of the capacitor and the connecting wires. In the flash camera scenario above, when the batteries powering the camera begin to wear out, their internal resistance rises, reducing the current and lengthening the time it takes to get ready for the next flash.
Voltage on the capacitor is initially zero and rises rapidly at first, since the initial current is a maximum. Figure (b) shows a graph of capacitor voltage versus time (\(t\)) starting when the switch is closed at \(t - 0\). The voltage approaches emf asymptotically, since the closer it gets to emf the less current flows. The equation for voltage versus time when charging a capacitor \(C\) through a resistor \(R\), derived using calculus, is \[V = emf(1 - e^{-t/RC})(charging),\] where \(V\) is the voltage across the capacitor, emf is equal to the emf of the DC voltage source, and the exponential e = 2.718 … is the base of the natural logarithm. Note that the units of \(RC\) are seconds. We define \[\tau = RC,\] where \(\tau\) (the Greek letter tau) is called the time constant for an \(RC\) circuit. As noted before, a small resistance \(R\) allows the capacitor to charge faster. This is reasonable, since a larger current flows through a smaller resistance. It is also reasonable that the smaller the capacitor \(C\), the less time needed to charge it. Both factors are contained in \(\tau = RC\).
More quantitatively, consider what happens when \(t = \tau = RC\). Then the voltage on the capacitor is \[V = emf(1 - e^{-1}) = emf(1 - 0.368) = 0.632 \cdot emf.\]
This means that in the time \(\tau = RC\), the voltage rises to 0.632 of its final value. The voltage will rise 0.632 of the remainder in the next time \(\tau\) It is a characteristic of the exponential function that the final value is never reached, but 0.632 of the remainder to that value is achieved in every time \(\tau\). In just a few multiples of the time constant \(\tau\), then, the final value is very nearly achieved, as the graph in Figure (b) illustrates.
Discharging a Capacitor
Discharging a capacitor through a resistor proceeds in a similar fashion, as Figure illustrates. Initially, the current is \(I_9 - \frac{V_0}{R}\), driven by the initial voltage \(V_0\) on the capacitor. As the voltage decreases, the current and hence the rate of discharge decreases, implying another exponential formula for \(V\). Using calculus, the voltage \(V\) on a capacitor \(C\) being discharged through a resistor \(R\) is found to be \[V = V_0 e^{-t/RC} (discharging).\]
The graph in Figure (b) is an example of this exponential decay. Again, the time constant is \(\tau = RC\). A small resistance \(R\) allows the capacitor to discharge in a small time, since the current is larger. Similarly, a small capacitance requires less time to discharge, since less charge is stored. In the first time interval \(\tau = RC\) after the switch is closed, the voltage falls to 0.368 of its initial value, since \(V = V_0 \cdot e^{-1} = 0.368 V_0\).
During each successive time \(\tau\), the voltage falls to 0.368 of its preceding value. In a few multiples of \(\tau\), the voltage becomes very close to zero, as indicated by the graph in Figure (b).
Now we can explain why the flash camera in our scenario takes so much longer to charge than discharge; the resistance while charging is significantly greater than while discharging. The internal resistance of the battery accounts for most of the resistance while charging. As the battery ages, the increasing internal resistance makes the charging process even slower. (You may have noticed this.)
The flash discharge is through a low-resistance ionized gas in the flash tube and proceeds very rapidly. Flash photographs, such as in Figure , can capture a brief instant of a rapid motion because the flash can be less than a microsecond in duration. Such flashes can be made extremely intense.
During World War II, nighttime reconnaissance photographs were made from the air with a single flash illuminating more than a square kilometer of enemy territory. The brevity of the flash eliminated blurring due to the surveillance aircraft’s motion. Today, an important use of intense flash lamps is to pump energy into a laser. The short intense flash can rapidly energize a laser and allow it to reemit the energy in another form.
Example \(\PageIndex{1}\) : Integrated Concept Problem: Calculating Capacitor Size—Strobe Lights
High-speed flash photography was pioneered by Doc Edgerton in the 1930s, while he was a professor of electrical engineering at MIT. You might have seen examples of his work in the amazing shots of hummingbirds in motion, a drop of milk splattering on a table, or a bullet penetrating an apple (see Figure ). To stop the motion and capture these pictures, one needs a high-intensity, very short pulsed flash, as mentioned earlier in this module.
Suppose one wished to capture the picture of a bullet (moving at \(5 \times 10^2 \, m/s\)) that was passing through an apple. The duration of the flash is related to the \(RC\) time constant \(\tau\). What size capacitor would one need in the \(RC\) circuit to succeed, if the resistance of the flash tube was \(10 \, \Omega\)? Assume the apple is a sphere with a diameter of \(8 \times 10^{-2} \, m\).
Strategy
We begin by identifying the physical principles involved. This example deals with the strobe light, as discussed above. Figure shows the circuit for this probe. The characteristic time \(\tau\) of the strobe is given as \(\tau = RC\).
Solution
We wish to find \(C\), but we don’t know \(\tau\). We want the flash to be on only while the bullet traverses the apple. So we need to use the kinematic equations that describe the relationship between distance \(x\), velocity \(v\), and time \(t\):
\[x = vt \, or \, t - \dfrac{x}{v}.\]
The bullet’s velocity is given as \(5 \times 10^2 \, m\), and the distance \(x\) is \(8 \times 10^{-2} \, m\). The traverse time, then, is
\[t = \dfrac{x}{v} = \dfrac{8.0 \times 10^{-2} \, m}{5.0 \times 10^2 \, m/s} = 1.6 \times 10^{-4} \, s.\]
We set this value for the crossing time \(t\) equal to \(\tau\). Therefore,
\[C = \dfrac{t}{R} = \dfrac{1.6 \times 10^{-4} \, s}{10.0 \, \Omega} = 16 \, \mu F.\]
(Note: Capacitance \(C\) is typically measured in farads, \(F\), defined as Coulombs per volt. From the equation, we see that \(C\) can also be stated in units of seconds per ohm.)
Discussion
The flash interval of \(160 \mu s\) (the traverse time of the bullet) is relatively easy to obtain today. Strobe lights have opened up new worlds from science to entertainment. The information from the picture of the apple and bullet was used in the Warren Commission Report on the assassination of President John F. Kennedy in 1963 to confirm that only one bullet was fired.
RC Circuits for Timing
\(RC\) circuits are commonly used for timing purposes. A mundane example of this is found in the ubiquitous intermittent wiper systems of modern cars. The time between wipes is varied by adjusting the resistance in an \(RC\) circuit. Another example of an \(RC\) circuit is found in novelty jewelry, Halloween costumes, and various toys that have battery-powered flashing lights. (See Figure for a timing circuit.)
A more crucial use of \(RC\) circuits for timing purposes is in the artificial pacemaker, used to control heart rate. The heart rate is normally controlled by electrical signals generated by the sino-atrial (SA) node, which is on the wall of the right atrium chamber. This causes the muscles to contract and pump blood. Sometimes the heart rhythm is abnormal and the heartbeat is too high or too low.
The artificial pacemaker is inserted near the heart to provide electrical signals to the heart when needed with the appropriate time constant. Pacemakers have sensors that detect body motion and breathing to increase the heart rate during exercise to meet the body’s increased needs for blood and oxygen.
Example \ (\PageIndex{2}\): Calculating Time: RC Circuit in a Heart Defibrillator
A heart defibrillator is used to resuscitate an accident victim by discharging a capacitor through the trunk of her body. A simplified version of the circuit is seen in Figure . (a) What is the time constant if an \(8.00 \, \mu F\) capacitor is used and the path resistance through her body is \(1 \times 10^3 \, \Omega\)? (b) If the initial voltage is 10.0 kV, how long does it take to decline to \(5 \times 10^2 \, V\)?
Strategy
Since the resistance and capacitance are given, it is straightforward to multiply them to give the time constant asked for in part (a). To find the time for the voltage to decline to \(5 \times 10^2 \, V\), we repeatedly multiply the initial voltage by 0.368 until a voltage less than or equal to \(5 \times 10^2 \, V\) is obtained. Each multiplication corresponds to a time of \(\tau\) seconds.
Solution for (a)
The time constant \(\tau\) is given by the equation \(\tau = RC\). Entering the given values for resistance and capacitance (and remembering that units for a farad can be expressed as \(s/\Omega\)) gives
\[\tau = RC = (1.00 \times 10^3 \, \Omega)(8.00 \, \mu F) = 8.00 \, ms.\]
Solution for (b)
In the first 8.00 ms, the voltage (10.0 kV) declines to 0.368 of its initial value. That is:
\[V = 0.368 V_0 = 3.680 \times 10^3 \, V \, at \, t = 8.00 \, ms.\]
(Notice that we carry an extra digit for each intermediate calculation.) After another 8.00 ms, we multiply by 0.368 again, and the voltage is
\[V' = 0.368 \, V\]
\[= (0.368)(3.680 \times 10^3 \, V)\]
\[=1.354 \times 10^3 \, V \, at \, t = 16.0 \, ms.\]
Similarly, after another 8.00 ms, the voltage is
\[V" = 0.368 \, V' = (0.368)(1.354 \times 10^3 \, V)\]
\[= 498 \, V \, at \, t = 24.0 \, ms.\]
Discussion
So after only 24.0 ms, the voltage is down to 498 V, or 4.98% of its original value.Such brief times are useful in heart defibrillation, because the brief but intense current causes a brief but effective contraction of the heart. The actual circuit in a heart defibrillator is slightly more complex than the one in Figure , to compensate for magnetic and AC effects that will be covered in Magnetism .
Exercise \(\PageIndex{1}\)
When is the potential difference across a capacitor an emf?
- Answer
-
Only when the current being drawn from or put into the capacitor is zero. Capacitors, like batteries, have internal resistance, so their output voltage is not an emf unless current is zero. This is difficult to measure in practice so we refer to a capacitor’s voltage rather than its emf. But the source of potential difference in a capacitor is fundamental and it is an emf.
PHET EXPLORATIONS: CIRCUIT CONSTRUCTION KIT (DC ONLY)
An electronics kit in your computer! Build circuits with resistors, light bulbs, batteries, and switches. Take measurements with the realistic ammeter and voltmeter. View the circuit as a schematic diagram, or switch to a life-like view.
Section Summary
- An \(RC\) circuit is one that has both a resistor and a capacitor.
- The time constant \(\tau\) for an \(RC\) circuit is \(\tau = RC\).
-
When an initially uncharged \((V_0=0\) at \(t=0)\) capacitor in series with a resistor is charged by a DC voltage source, the voltage rises, asymptotically approaching the emf of the voltage source; as a function of time,
\[V=emf(1−e^{−t/RC})(charging).\] - Within the span of each time constant \(τ\), the voltage rises by 0.632 of the remaining value, approaching the final voltage asymptotically.
-
If a capacitor with an initial voltage \(V_0\) is discharged through a resistor starting at \(t=0\), then its voltage decreases exponentially as given by
\[V=V_0e^{−t/RC}(discharging)\]. - In each time constant τ, the voltage falls by 0.368 of its remaining initial value, approaching zero asymptotically.
Glossary
- RC circuit
- a circuit that contains both a resistor and a capacitor
- capacitor
- an electrical component used to store energy by separating electric charge on two opposing plates
- capacitance
- the maximum amount of electric potential energy that can be stored (or separated) for a given electric potential


24.12: Chapter 12 (Abdominal Assessment)
Answer Key to Chapter 12 Learning Activities
- A – Nausea, B – Vomiting, D – Bloating
- A – Bowel Sounds
- B – There are hypoactive bowel sounds in all quadrants, C – Firmness is palpated in left lower quadrant
-
SAMPLE NOTE:
D (SO): Patient reports ongoing constipation without bowel movement for greater than 24 hours. The patient reports intermittent nausea that is increasing in frequency and has had one episode of vomiting. The patient’s abdomen appears bloated and bowel sounds are hypoactive in all quadrants. There is notable firmness present in the left lower quadrant with palpation. The patient denies tenderness.
A: Provider updated regarding patient’s bowel status and assessment. Order received to administer Milk of Magnesia 30 mL PO times one dose. Medication was administered at 0800.
R (P). Patient reported having a large formed brown stool at 1100. Stool is noted to be soft and brown. Patient reports relief from nausea. Bowel sounds are noted in all quadrants. Will continue to monitor bowel status.
Nurse Signature, Credentials, Date/Time
Answers to interactive elements are given within the interactive element.


8: Introduction
Environmental History is about looking at the past as if the environment matters. American History is about looking at the past of not only the United States, but of both the American continents. This wider view is especially important when we realize that people occupied the Americas for over 15,000 years before Europeans arrived and that when the came to the Americas, Europeans focused their interest for centuries on areas that are not part of the current United States. As we get closer to the present, we will focus more on the U.S., but we’ll try to remind ourselves from time to time that we’re not the only nation in the Americas by considering how other nations have experienced and affected the environment.
Not that long ago the world seemed so big and human actions seemed comparatively so small that it was easy to ignore our effect on the world around us. Today we understand that people have a big impact on our surroundings. And we’re becoming aware that our environment has played an important part in our individual lives and in the growth of our cultures. Instead of being just a neutral backdrop, the environment is now recognized as a powerful shaper of human choices. That is, history.
By the environment, I mean everything around us. The natural world, but also the manmade world. Often it’s difficult to draw a distinct line between those two. We intuitively feel that tree-lined suburbs are more natural than the city and that the wilderness is even more pristine. But if we look closer we often find the suburban trees were part of a developer’s design, and even wilderness areas are special human-made places that have been deliberately protected or even rehabilitated so they resemble our idea of pristine nature.
The environment is also more than just the green part of the world. Even the most sophisticated urbanite depends for her food, water, and energy on elements of a much wider environment than she may be aware of. Nor are we talking only about environmentalists. The most jaded materialist depends on the environment just as much as the most dewy-eyed idealist.
If environment is a surprisingly complicated word, so is history.  These are the civics and social studies lessons we are all so familiar with. Historians recognize that many of these stories are not as true as we’d like to think. A lot of our histories have changed quite a bit over time, as have our reasons for telling them. History isn’t just data about the past, it’s the stories we tell about the past.
So where does that leave us? American Environmental History, for the purposes of this book, means looking at our past with special attention to how our surroundings influenced our opportunities, our choices, and our actions; and to how our actions reshaped our surroundings.
There are two basic elements of looking at a past that includes the environment. First, we pay close attention to the physical world. For example, the shape of the American continents and the fact that they are connected to each other but cut off from the rest of the world has influenced American cultures from their prehistoric beginnings to the present. Landforms, waterways, natural resources, and climate have all been important factors in how our society has developed and how it continues to change. Second, we remember that how we think about the environment, like how we think about the past, has changed over time; and that those changes affect our current choices and actions. You don’t have to dig too deeply into current headlines to find debates raging over coal mining, oil pipelines, depleted fisheries, mass extinctions, and climate change. Our beliefs about the world around us and our role in it are important elements of the competing positions taken by advocates on either side of these important issues.
Our goal is to deal with both events and ideas. But realistically, this is a survey of American Environmental History stretching from prehistory to the present. We’re going to spend most of our time on events, especially because the environmental elements of many key events in American History are not well understood. Too often our stories about America’s past devote all their attention to political debates and to the ideologies of elite leaders or philosophies of founding fathers, and these tales are told against a neutral background that isn’t part of the story at all. The main goal of this text is to show that it’s really not possible to understand our past while ignoring the environment.
In a sense, that simple goal is also one of the ways this American Environmental History text differs from the academic field. When historians study Environmental History, we typically spend a lot of our time drilling into the details and rarely step back to talk about the big picture. As a result, there are plenty of books available about particular issues such as the Dust Bowl, Colonial America, Agriculture, or our changing perceptions of Wilderness. I assign chapters from some of these texts to my students. But we never had a textbook for my college course, because there wasn’t a survey text that covered the whole, big picture.
Until now. This book is based on the content of the course I taught for several years at the University of Massachusetts in Amherst and currently teach at Bemidji State University. It’s designed to give you an overview of American Environmental History and includes suggestions for further reading and chapter supplements that discuss some of those detailed studies I mentioned. There’s a lot of material to cover. I’ve tried to arrange it so it’s both concise and comprehensive. So welcome and let’s get started!


8.1: Social Emotional Theories of Development
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Erik Erikson- Industry vs. Inferiority
Erik Erikson proposed that we are motivated by a need to achieve competence in certain areas of our lives. As we’ve learned in previous chapters, Erikson’s psychosocial theory has eight stages of development over the lifespan, from infancy through late adulthood. At each stage, there is a conflict, or task, that we need to resolve. Successful completion of each developmental task results in a sense of competence and a healthy personality. Failure to master these tasks leads to feelings of inadequacy.
During the elementary school stage (ages 6-12), children face the task of Industry versus Inferiority . Children begin to compare themselves to their peers to see how they measure up.
They either develop a sense of pride and accomplishment in their schoolwork, sports, social activities, and family life, or they feel inferior and inadequate when they don’t measure up.2
According to Erikson, children in middle childhood are very busy or industrious. They are constantly doing, planning, playing, getting together with friends, achieving. This is a very active time and a time when they are gaining a sense of how they measure up when compared with friends. Erikson believed that if these industrious children can be successful in their endeavors, they will get a sense of confidence for future challenges. If not, a sense of inferiority can be particularly haunting during middle childhood.3
Sigmund Freud - Psychoanalytic Theory
The great psychoanalyst Sigmund Freud (1856–1939) focused on unconscious, biological forces that he felt shape individual personality. Freud (1933) thought that the personality consists of three parts: the id, the ego, and the superego. The id is the selfish part of the personality and consists of biological instincts that all babies have, including the need for food and, more generally, the demand for immediate gratification. As babies get older, they learn that not all their needs can be immediately satisfied and thus develop the ego, or the rational part of the personality. As children get older still, they internalize society’s norms and values and thus begin to develop their superego, which represents society’s conscience. If a child does not develop normally and the superego does not become strong enough, the individual is more at risk for being driven by the id to commit antisocial behavior.4
Lawrence Kohlberg’s Stages of Moral Development
Kohlberg (1963) built on the work of Piaget and was interested in finding out how our moral reasoning changes as we get older. He wanted to find out how people decide what is right and what is wrong. Just as Piaget believed that children’s cognitive development follows specific patterns, Kohlberg (1984) argued that we learn our moral values through active thinking and reasoning, and that moral development follows a series of stages. Kohlberg's six stages are generally organized into three levels of moral reasons. To study moral development, Kohlberg looked at how children (and adults) respond to moral dilemmas. One of Kohlberg’s best knownmoral dilemmas is the Heinz dilemma:
In Europe, a woman was near death from a special kind of cancer. There was one drug that the doctors thought might save her. It was a form of radium that a druggist in the same town had recently discovered. The drug was expensive to make but the druggist was charging ten times what the drug cost him to make. He paid $200 for the radium and charged $2,000 for a small dose of the drug. The sick woman’s husband, Heinz, went to everyone he knew to borrow the money but he could only get together about $1,000, about half of what the drug cost. He told the druggist that his wife was dying and asked him to sell it cheaper or let him pay later. But the druggist said: “No, I discovered the drug and I’m going to make money from it.” Heinz got desperate and broke into the man’s store to steal the drug for his wife. Should the husband have done that? (Kohlberg, 1969, p. 379) 6
Level One - Preconventional Morality
In stage one, moral reasoning is based on concepts of punishment. The child believes that if the consequence for an action is punishment, then the action was wrong. In the second stage, the child bases his or her thinking on self-interest and reward ("You scratch my back, I'll scratch yours"). The youngest subjects seemed to answer based on what would happen to the man as a result of the act. For example, they might say the man should not break into the pharmacy because the pharmacist might find him and beat him. Or they might say that the man should break in and steal the drug and his wife will give him a big kiss. Right or wrong, both decisions were based on what would physically happen to the man as a result of the act. This is a self-centered approach to moral decision-making. He called this most superficial understanding of right and wrong preconventional morality . Preconventional morality focuses on self-interest. Punishment is avoided and rewards are sought. Adults can also fall into these stages,particularly when they are under pressure.
Level Two - Conventional Morality
Those tested who based their answers on what other people would think of the man as a result of his act were placed in Level Two. For instance, they might say he should break into the store, then everyone would think he was a good husband, or he should not because it is against the law. In either case, right and wrong is determined by what other people think. In stage three, the person wants to please others. At stage four, the person acknowledges the importance of social norms or laws and wants to be a good member of the group or society. A good decision is one that gains the approval of others or one that complies with the law. This he called conventional morality , people care about the effect of their actions on others . Some older children, adolescents, and adults use this reasoning.
Level Three, post-conventional morality, is not included because it focuses on adolescence and adulthood. However, it is in the table below if you’d like an overview of Level Three - Stages 5 and 6.
Preconventional Morality (young children)
| Stage | Description |
| Stage 1 |
Focus is on self-interest and punishment is avoided. The man shouldn’t steal the drug, as he may get caught and go to jail. |
| Stage 2 | Rewards are sought. A person at this level will argue that the man should steal the drug because he does not want to lose his wife who takes care of him. |
Conventional Morality (older children, adolescents, most adults)
| Stage | Description |
| Stage 3 | Focus is on how situational outcomes impact others and wanting to please and be accepted. The man should steal the drug because that is what good husbands do. |
| Stage 4 |
People make decisions based on laws or formalized rules. The man should obey the law because stealing is a crime. |
Post Conventional Morality (rare in adolescents, a few adults)
| Stage | Description |
| Stage 5 | Individuals employ abstract reasoning to justify behaviors. The man should steal the drug because laws can be unjust and you have to consider the whole situation. |
| Stage 6 | Moral behavior is based on self-chosen ethical principles. The man should steal the drug because life is more important than property. |
Although research has supported Kohlberg’s idea that moral reasoning changes from an early emphasis on punishment and social rules and regulations to an emphasis on more general ethical principles, as with Piaget’s approach, Kohlberg’s stage model is probably too simple. For one, people may use higher levels of reasoning for some types of problems but revert to lower levels in situations where doing so is more consistent with their goals or beliefs (Rest, 1979). Second, it has been argued that the stage model is particularly appropriate for Western, rather than non-Western, samples in which allegiance to social norms, such as respect for authority, may be particularly important (Haidt, 2001). In addition, there is frequently little correlation between how we score on the moral stages and how we behave in real life. Perhaps the most important critique of Kohlberg’s theory is that it may describe the moral development of males better than it describes that of females (Jaffee & Hyde, 2000). 7
Contributors and Attributions
2. Psychology - 9.2: Lifespan Theories by CNX Psychology is licensed under CC BY 4.0
3. Sociology: Brief Edition – Explaining Socialization by Steven E. Barkan is licensed under CC BY-NC-SA 3.0
4. Sociology: Brief Edition – Explaining Socialization by Steven E. Barkan is licensed under CC BY-NC-SA 3.0
6. Psychology - 9.2: Lifespan Theories by CNX Psychology is licensed under CC BY 4.0
7. Lifespan Development: A Psychological Perspective by Martha Lally and Suzanne Valentine-French is licensed under CC BY-NC-SA 3.0


18.2: Relationship Between Solubility and Ksp
- Quantitatively related \(K_{sp}\) to solubility
Considering the relation between solubility and \(K_{sp}\) is important when describing the solubility of slightly ionic compounds. However, this article discusses ionic compounds that are difficult to dissolve; they are considered "slightly soluble" or "almost insoluble." Solubility product constants (\(K_{sq}\)) are given to those solutes, and these constants can be used to find the molar solubility of the compounds that make the solute. This relationship also facilitates finding the \(K_{sq}\) of a slightly soluble solute from its solubility.
Introduction
Recall that the definition of solubility is the maximum possible concentration of a solute in a solution at a given temperature and pressure. We can determine the solubility product of a slightly soluble solid from that measure of its solubility at a given temperature and pressure, provided that the only significant reaction that occurs when the solid dissolves is its dissociation into solvated ions, that is, the only equilibrium involved is:
\[\ce{M}_p\ce{X}_q(s)⇌p\mathrm{M^{m+}}(aq)+q\mathrm{X^{n−}}(aq)\]
In this case, we calculate the solubility product by taking the solid’s solubility expressed in units of moles per liter (mol/L), known as its molar solubility .
We began the chapter with an informal discussion of how the mineral fluorite is formed. Fluorite, \(\ce{CaF2}\), is a slightly soluble solid that dissolves according to the equation:
\[\ce{CaF2}(s)⇌\ce{Ca^2+}(aq)+\ce{2F-}(aq)\nonumber \]
The concentration of Ca 2 + in a saturated solution of CaF 2 is 2.1 × 10 –4 M ; therefore, that of F – is 4.2 × 10 –4 M , that is, twice the concentration of \(\ce{Ca^{2+}}\). What is the solubility product of fluorite?
Solution
First, write out the K sp expression, then substitute in concentrations and solve for K sp :
\[\ce{CaF2(s) <=> Ca^{2+}(aq) + 2F^{-}(aq)} \nonumber\]
A saturated solution is a solution at equilibrium with the solid. Thus:
\[\begin{align*} K_\ce{sp} &= \ce{[Ca^{2+}][F^{-}]^2} \\[4pt] &=(2.1×10^{−4})(4.2×10^{−4})^2 \\[4pt] &=3.7×10^{−11}\end{align*}\]
As with other equilibrium constants, we do not include units with K sp .
In a saturated solution that is in contact with solid Mg(OH) 2 , the concentration of Mg 2 + is 3.7 × 10 –5 M . What is the solubility product for Mg(OH) 2 ?
\[\ce{Mg(OH)2}(s)⇌\ce{Mg^2+}(aq)+\ce{2OH-}(aq)\nonumber\]
- Answer
-
2.0 × 10 –13
The K sp of copper(I) bromide, \(\ce{CuBr}\), is 6.3 × 10 –9 . Calculate the molar solubility of copper bromide.
Solution
The solubility product constant of copper(I) bromide is 6.3 × 10 –9 .
The reaction is:
\[\ce{CuBr}(s)⇌\ce{Cu+}(aq)+\ce{Br-}(aq)\nonumber\]
First, write out the solubility product equilibrium constant expression:
\[K_\ce{sp}=\ce{[Cu+][Br- ]}\nonumber\]
Create an ICE table (as introduced in the chapter on fundamental equilibrium concepts), leaving the \(\ce{CuBr}\) column empty as it is a solid and does not contribute to the K sp :
At equilibrium:
\[ \begin{align*} K_\ce{sp} &=\ce{[Cu+][Br- ]} \\[4pt] 6.3×10^{−9} &=(x)(x)=x^2 \\[4pt] x&=\sqrt{(6.3×10^{−9})}=7.9×10^{−5} \end{align*}\]
Therefore, the molar solubility of \(\ce{CuBr}\) is 7.9 × 10 –5 M .
Finding the Solubility of a Salt: https://youtu.be/98BuldrICXM
Summary
Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium. Equilibrium is the state at which the concentrations of products and reactant are constant after the reaction has taken place. The solubility product constant (\(K_{sp}\)) describes the equilibrium between a solid and its constituent ions in a solution. The value of the constant identifies the degree to which the compound can dissociate in water. The higher the \(K_{sp}\), the more soluble the compound is. \(K_{sq}\) is defined in terms of activity rather than concentration because it is a measure of a concentration that depends on certain conditions such as temperature, pressure, and composition. It is influenced by surroundings. \(K_{sp}\) is used to describe the saturated solution of ionic compounds. (A saturated solution is in a state of equilibrium between the dissolved, dissociated, undissolved solid, and the ionic compound.)
Contributors and Attributions
-
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>).


6.0: Introduction
Nurses are responsible for consistently providing patient care that adheres to current evidence-based practices and meets professional standards and guidelines. There are effective scientific and practical control methods to prevent the development of infection and the transmission of diseases in a healthcare setting. Nurses provide healing care to patients of all ages and stages of life. While they are able to foster the emotional aspects of healing, clinical safety must be at the forefront of decisions to make sure both patients and healthcare providers remain safe. While it is widely understood that practices such as handwashing and covering a sneeze are important in everyday life, nurses hold a deeper responsibility to prevent and control infection against the most dangerous infectious agents.


3.2: Phenotype Analysis
Background Information
Most cells making up the human body contain a nucleus. Inside of the nucleus are chromosomes that are responsible for transmitting hereditary information from cell to cell in cell division and from parent to offspring in reproduction. Chromosomes are thread-like structures that consist of tightly coiled DNA (deoxyribonucleic acid), the genetic material, and associated proteins. DNA essentially serves as the blueprints for life. DNA is organized into genes. Genes (i.e., "packets of DNA") control the structure and function of all cells, as well as large-scale processes such as development and reproduction. Most genes encode for a specific protein. Proteins play a variety of different essential roles in the body, including acting as enzymes, allowing communication between cells, immune functions, regulation of molecules into and out of cells, contraction of muscles and many more.
Humans have 23 pairs of chromosomes (46 total chromosomes) in the nucleus of all somatic cells (non-sex cells). However, the sex cells (sperm cells and eggs) contain only one copy of each chromosome. Human reproduction requires a sperm cell to fuse with an egg during fertilization to create the zygote. The zygote contains 23 pairs of chromosomes. Twenty-two pairs of the chromosomes are autosomes, containing genes that are unrelated to sex.  A single gene on the Y chromosome, SRY (sex-determining region Y), triggers an embryo to develop male physical characteristics.
Chromosomes that are the same length and contain the same genes are called homologous chromosomes. Humans have 22 pairs of homologous chromosomes, receiving one of each pair from the mother and one of each pair from the father. Each chromosome contains thousands of individual genes. There are thought to be about 20,000 genes in human DNA. Interestingly, the number of genes in human DNA is not appreciably different from the number of genes in chimpanzees or mice.
Some human traits are determined by a single gene, or mostly determined by a single gene. Different versions of a gene allow for variations in a population. The different versions of a gene are called alleles. For example, earlobe shape is controlled by a single gene. One allele for earlobe shape causes individuals to have free earlobes and another allele causes people to have attached earlobes. Humans have two alleles for each gene (receiving one allele from our mom and one allele from our dad). If both alleles are the same version then this is referred to as homozygous. When an individual has two different alleles for the same gene this is referred to as heterozygous. The alleles that an individual has are referred to as the genotype. But, the outward expression of the alleles (what we see) is the phenotype. The phenotype that is expressed in heterozygous individuals is the dominant phenotype and is caused by the dominant allele. The recessive allele is the phenotype that is not seen in the phenotype of heterozygous individuals. The dominant allele is generally represented by a capital letter, while the recessive allele is generally represented by a lowercase letter. The alleles for earlobe shape are E and e. The dominant allele, E, determines free earlobes. Homozygous dominant (EE) and heterozygous (Ee) individuals have the free earlobe phenotype. But, homozygous recessive (ee) individuals have the recessive attached earlobe phenotype.
Most traits are determined by more than one gene. For example, skin color and height are determined by many genes. A few human phenotypes however, are determined by a single gene. We will explore some of these single gene traits in the laboratory.
Materials needed
- A group of students to observe phenotypes
Exercise 1: Phenotype Analysis
- Use a lab partner to help you determine your phenotype for the traits listed.
- Complete the table below. Use two alleles per trait for the genotype. If you exhibit the dominant phenotype, use a dash to represent the second allele.
Example :
B_ genotype for the phenotype of brown eyes (dash indicates second allele could be B or b which means a genotype of BB or Bb)
bb genotype for the phenotype of blue eyes
Interlocking fingers
Interlock fingers. Observe which thumb is on top (right or left). The tendency to place the left thumb on top is due to a dominant allele (I) and the genotype is I- (either II or I-). The right thumb on top is determined by the ii genotype.
Ear lobes
The dominant allele (E) results in the phenotype of free earlobes. The recessive allele (e) is for attached earlobes.
Widow’s peak
Widow’s peak occurs when the hairline forms a distinct point in the center of the forehead. Widow’s peak is controlled by a dominant allele (W).
Tongue curling
A dominant allele (T) gives the individual the ability to curl the tongue in a U-shape.
Hitch hiker’s thumb
A person homozygous recessive for this trait (hh) can bend the last (distal) thumb joint back to about a 90 degree angle. Those with the H allele cannot.
PTC tasting
If you can taste the bitterness of PTC, you have the dominant allele (P). Place the PTC paper on your tongue for a few seconds. If you cannot taste anything, you do not possess the dominant allele.
Straight hair
A person with the dominant allele (H) has straight hair. Homozygous recessive individuals have curly hair.
Handedness
If you are right-handed then you have the dominant allele (R). The recessive allele (r) encodes left-handedness.
Mid-digital hair
A person with the dominant allele (H) has hair on the middle joint of any finger. People with no hair on the middle finger joint are homozygous recessive.
Number of students in class_________
| Trait and alleles | My phenotype is: (dominant or nondominant) | My genotype is (X_ or xx) | # in class with dominant phenotype | % of class with dominant phenotype |
|---|---|---|---|---|
| Interlocking fingers (I or i) | ||||
| Earlobes (E or e) | ||||
| Widow’s peak (W or w) | ||||
| Tongue curling (T or t) | ||||
| Hitchhiker’s thumb (H or h) | ||||
| PTC tasting (P or p) | ||||
| Straight hair (H or h) | ||||
| Handedness (R or r) | ||||
| Mid digital hair (h or h) |


1.17: What is Reflective Thinking?
- Understand the connection between reflection and learning
Reflecting on Learning
According to the old saying from Confucius, “To know what you know and what you do not know, that is true knowledge.” Reflection is an act that helps you better determine your own true knowledge about something. For many students, acknowledging their limitations or what they don’t understand about certain material can be key to real learning. Without such practices, we may be faking our way through learning, which isn’t really learning at all. Inversely, we should strive to better and honestly understand our strengths, as well as our growth and progress. You have probably had friends who are good at a particular school subject, and when they are asked how they know so much about it or how they do so well at it, they say something like “It’s just easy for me” or “I’ve always been good at it.” But those answers oversimplify what learning is. If those students reflected on the question more, they might find interesting answers that help them more deeply consider themselves as learners. Writing can help us do that kind of deeper reflection.
While reflection certainly is a mental act, many people write to help make sense of their thoughts. And writing about your own writing is a deeply metacognitive practice that can help you more broadly as a learner.
Many people may view reflection as thinking about an event that has happened, but that is only part of it. Reflection can be about something that happened in the past, is happening now, or will/might happen in the future. For example, as a college student, you might reflect on a math test you took last month, how your current homework assignment is puzzling you, and how a math class might impact you next semester. It is important to see reflection as a continuing process because it is a valuable part of learning and growing in all disciplines and in all aspects of life.
Thinking about your own writing and being aware of your own strengths and weaknesses as a writer can help you improve. See if you can guess at the following types of knowledge types you may review while reflecting.
https://lumenlearning.h5p.com/content/1290965000592956998/embed
Reflecting Across the Disciplines
If reflection seems like an act unique to your writing class, think about other courses and majors and how we see instances of reflection. Consider health fields and professions. Early in their educations, future nurses, doctors, and other health professionals are taught the importance of reflection in their work. People such as nurses and doctors regularly reflect on the condition of their patients, the care they provide, potential treatments, and future care. Writing allows health care professionals, for example, to keep a reflective log of patients (think about why nurses and doctors rely on those charts and patient records so much). This type of thinking requires practice, which is why reflection is part of the courses these students take. For most people, it is not enough to just reflect in the mind.
Contributors and Attributions
- What is Reflective Thinking?. Authored by : Guy Krueger. Provided by : University of Mississippi. License : CC BY-SA: Attribution-ShareAlike
- Reflection as a Continuum. Authored by : Guy Krueger. Provided by : University of Mississippi. License : CC BY-SA: Attribution-ShareAlike
- Process of Reflective Writing. Authored by : Karen Forgette. Provided by : University of Mississippi. License : CC BY-SA: Attribution-ShareAlike
- Reflective Writing. Provided by : SkillsTeamHullUni. Located at : https://www.youtube.com/watch?v=QoI67VeE3ds&feature=emb_logo . License : Other . License Terms : Standard YouTube License
- Meme: I don't know what I'm writing about. License : All Rights Reserved . License Terms : Fair Use