q1402 drafts

1-1 Algebra

Exercise \(\PageIndex{3}\)

\(P V=n R T\)

Solve the equation above for:

1. P
2. V
3. n
4. T

Answer a

\[\begin{aligned}
&P V=n R T\\
&\text { Step 1: Divide both sides by V: }\\
&P\left(\frac{V}{V}\right)=\frac{n R T}{V}\\
&\text { Note: }\left(\frac{\mathrm{V}}{\mathrm{V}}\right)=1\\
&P=\frac{n R T}{V}
\end{aligned}\nonumber\]

 

Answer b

\[\begin{aligned}
&P V=n R T\\
&\text { Step 1: Divide both sides by P: }\\
&V\left(\frac{P}{P}\right)=\frac{n R T}{P}\\
&\text { Note: }\left(\frac{\mathrm{P}}{\mathrm{P}}\right)=1\\
&V=\frac{n R T}{P}
\end{aligned}\nonumber\]

 

Answer c

\[\begin{aligned}
&P V=n R T\\
&\text { Step 1: Divide both sides by RT: }\\
&\frac{PV}{RT}=n\left(\frac{RT}{RT}\right)\\
&\text { Note: }\left(\frac{\mathrm{RT}}{\mathrm{RT}}\right)=1\\
&n=\frac{PV}{RT}
\end{aligned}\nonumber\]

Answer d

\[\begin{aligned}
&P V=n R T\\
&\text { Step 1: Divide both sides by nR: }\\
&\frac{PV}{nR}=\left(\frac{nR}{nR}\right)T\\
&\text { Note: }\left(\frac{\mathrm{nR}}{\mathrm{nR}}\right)=1\\
&T=\frac{PV}{nR}
\end{aligned}\nonumber\]

 

Exercise \(\PageIndex{4}\)

\(T_{F}=T_{C}\left(\frac{9}{5}\right)+32\)

Solve the equation above for T_c:

Answer

\(T_{F}=T_{C}\left(\frac{9}{5}\right)+32\)

Step 1: Bring all things with T_C to one side of the eq. and everything else to the other by subtracting 32 from sides and rearranging (note 32-32=0 )

\(T_{C}\left(\frac{9}{5}\right)=T_{F}-32\)

Step 2. multiply both sides by \(\frac{5}{9}\) noting:

\(\left(\frac{9}{5}\right)\left(\frac{5}{9}\right)=\left(\frac{9}{9}\right)\left(\frac{5}{5}\right)=(1)(1)=1\)

\(T_{C}=\left(T_{F}-32\right)\left(\frac{5}{9}\right)\)

 

 

Exercise \(\PageIndex{5}\)

\(q=mc(T_{F}-T_{i})\)

Solve the equation above for T_F:

Answer

Separate T_f from all other variables by dividing both sides by mc and adding T_i to both sides

\[\begin{align} q&=mc(T_{f}-T_{i})\nonumber\\
\frac{q}{mc}&=\frac{mc}{mc}(T_{f}-T_{i})\nonumber\\
\frac{q}{mc}+T_{i}&=T_{f}-T_{i}+T_{i}=T_{f}\nonumber\\
T_{f}&=T_{i}+\frac{q}{mc} \nonumber\end{align} \]

 

 

Exercise \(\PageIndex{6}\)

Solve the following equation for:
 \[ m_Cc_C \Delta T _C =- m_Hc_H \Delta T_H  \nonumber\]

1. c_H
2. c_c
3. T_c
4. T_F

Answer a

\[\begin{aligned}
\text { Divide both sides by } m_{H}\left(T_{F}-T_{H}\right)\\
\frac{m_{c} c_{c}\left(T_{F}-T_{C}\right)}{-m_{H}\left(T_{F}-T_{H}\right)}&=c_{H}\left(\frac{m_{H}\left(T_{F}-T_{H}\right)}{m_{H}\left(T_{F}-T_{H}\right)}\right)\\
c_{H}&=\frac{-m_{C} c_{C}\left(T_{F}-T_{C}\right)}{m_{H}\left(T_{F}-T_{H}\right)}
\end{aligned}\]

Answer b

\[\begin{aligned}
\text { Divide both sides by } m_{C}\left(T_{F}-T_{C}\right)\\
c_{C}\left(\frac{m_{C}\left(T_{F}-T_{C}\right)}{m_{C}\left(T_{F}-T_{C}\right)}\right)&=\frac{-m_{H} c_{H}\left(T_{F}-T_{H}\right)}{m_{C}\left(T_{F}-T_{C}\right)}\\
c_{C}&=\frac{-m_{H} c_{H}\left(T_{F}-T_{H}\right)}{m_{C}\left(T_{F}-T_{C}\right)}
\end{aligned}\]

Answer c

Step 1: Place all terms with T_C on one side of the equal sign and everything else on the other side

\[\begin{aligned}
m_{C} c_{C}\left(T_{F}-T_{C}\right)&=-m_{H}c_{H}\left(T_{F}-T_{H}\right)\\
m_{C} c_{C}T_{F}-m_{C} c_{C}T_{C}&=-m_{H}c_{H}(T_{F}-T_{H})\\
-m_{C} c_{C}T_{C}&=m_{C} c_{C}T_{F}-m_{H}c_{H}(T_{F}-T_{H})
\end{aligned}\nonumber\]

Step 2: Solve for T_C by dividing both sides by \(-m_{C} c_{c}\)

\[\begin{aligned}
\frac{-m_{C} c_{C}T_{C}}{-m_{C} c_{C}}&=\frac{m_{C} c_{C}T_{F}-m_{H}c_{H}(T_{F}-T_{H})}{-m_{C} c_{C}}\\
T_{C}&=T_{F}+\frac{m_{H}c_{H}(T_{F}-T_{H})}{m_{C} c_{C}}
\end{aligned}\nonumber\]

Answer d

Step 1: Place all terms with T_F on one side of the equal sign and everything else on the other side

\[\begin{aligned}
m_{C} c_{C}\left(T_{F}-T_{C}\right)&=-m_{H}c_{H}\left(T_{F}-T_{H}\right)\\
m_{C} c_{C}T_{F}-m_{C} c_{C}T_{C}&=-m_{H} c_{H}T_{F}-m_{H} c_{H}T_{H}\\
m_{C} c_{C}T_{F}+m_{H} c_{H}T_{F}&=m_{C} c_{C}T_{C}+m_{H} c_{H}T_{H}\\
T_{F}(m_{C} c_{C}+m_{H} c_{H})&=m_{C} c_{C}T_{C}+m_{H} c_{H}T_{H}\\
\end{aligned}\nonumber\]

Step 2: Solve for T_C by dividing both sides by \(m_{C} c_{C}+m_{H} c_{H}\)

\[T_{F}=\frac{m_{C} c_{C}T_{C}+m_{H} c_{H}T_{H}}{(m_{C} c_{C}+m_{H} c_{H})}\nonumber\]

 

1-2-Math

 

Exercise \(\PageIndex{1}\)

Consider all of the following values to be measured numbers. Review your rules on significant figures and be sure you can do these without a calculator.

1. 12.56 + 2.4
2. 98.3-89.4
3. 82.0 + 34.4
4. 312.56 X 2.4

Answer a

15.0

Review the Rules for Addition and Subtraction:
The exact answer is 14.96 but 2.4 is only known to the 10ths position, so you truncate the answer there. You round up because the 6 (in the 100ths position) is greater than or equal to 5.

Answer b

8.9

Review the Rules for Addition and Subtraction:
Although both numbers have 3 significant figures the answer has 2 because they are only known to the 10th position.

Answer c

116.4

Review the Rules for Addition and Subtraction:
Although both numbers are known to 3 significant figures the answer has 4 because they are all known to the 10th position.

Answer d

30

Review the Rules for Division and Multiplication:
The answer will have the number of significant digits as the number with the least.

 

Exercise \(\PageIndex{2}\)

Consider all of the following values to be measured numbers. Review your rules on significant figures and be sure you can do these without a calculator.

Solve the following to the correct number of significant figures:

1. \(\frac{198.1}{12.1+198.1}\)
2. \(\frac{12.1}{12.1+198.1}\)
3. \(\frac{4.12}{384}\)
4. 3\(412-0.4\)

Answer a

0.9424

Note the denominator has 4 not 3 sig figs

\(\frac{198.1}{12.1+198.1}=\frac{198.1\text{(4 sig figs)}}{ 210.2 \text{(4 sig figs)}}=0.9424\) final answer has four sig figs

Answer b

0.0576

\(\frac{12.1}{12.1+198.1}=\frac{12.1\text{(3 sig figs)}}{ 210.2 \text{(4 sig figs)}}=0.0576\) final answer has three sig figs

Answer c

0.0107

Answer d

412

1-3 Significant Figures

 

Exercise \(\PageIndex{1}\)

Give the number of significant figures in each measured value.

1. 0.00204 g
2. 20400 g
3. 0.0020400 kg
4. 20103 mL
5. 100 students
6. 0.00001 miles
7. 2.00x10⁴

Answer a

3 sig figs

Answer b

3 sig figs

Answer c

5 sig figs

Answer d

5 sig figs 

 

Answer e

3 sig figs

 

Answer f

1 sig figs

 

Answer g

3 sig figs

 

1-4 Scientific Notation

Exercise \(\PageIndex{1}\)

Convert the following numbers to scientific notation

1. 234.00
2. 0.000100
3. 23470.23
4. 0.0374600

Answer a

2.3400x10²

Answer b

1.00 x 10^-4

Answer c

 2.347034 x 10⁴

Answer d

 3.74600 x 10^-2

 

Exercise \(\PageIndex{2}\)

Express 200 to 2 significant figures

Answer a

2.0 x 10²

1-5 Math

Exercise \(\PageIndex{1}\)

Solve the following:

1. (44.5 + 12.1) X (116 - 104)
2. (32.4 - 41) X (4.867 + 2.295)
3. (0.086 + 0.034) X (1.283 + 0.137)
4. (2 X 10²) X (4 X 10³)
5. 3.18 X 10^-3 + 4.6 X 10^-4
6. 8.4 X 10^-8 + 3.2x10^-3
7. \(\frac{(6.0221367\times 10^{23})(6.62608\times 10^{-34})}{(2.99792458\times 10^{8})(9.6485309\times 10^{4})}\)

Answer a

680
Note that (116-104) = 12 and has two significant figures

Answer b

-60
Note (32.4-41) has one significant figure (but use -8.6 in the calculation).

Answer c

0.170
Note (0.086 + 0.034) has 3 significant figures

Answer d

8 X 10⁵
Note if you get 8 X 10⁶ you are not correctly using your calculator 

Answer e

3.64 X 10^-3
In order to determine the relative precision of each number, you need to bring them to the same power of 10
\[\begin{align} \nonumber & 3.18\times 10^{-3}  \nonumber \\ + \nonumber &0.46\times 10^{-3} \nonumber \\ \hline \nonumber &3.64\times 10^{-3}\nonumber \end{align} \]

Answer f

3.64 X 10^-3
Note, the first number is of the order of 10⁵ smaller than the second and therefor insignificant.

Answer g

1.3795x10^-23
\[\frac{(6.0221367\times 10^{23})(6.62608\times 10^{-34})}{(2.99792458\times 10^{8})(9.6485309\times 10^{4})}\nonumber\\\\
=\frac{(6.0221367)(6.62608)}{(2.99792458)(9.6485309)}\times\frac{(10^{23})(10^{-34})}{(10^{8})( 10^{4})}\nonumber\\\\
=1.3795\times10^{32-34-8-4}\nonumber\]

 

1-6 Percent

Exercise \(\PageIndex{1}\)

You have a solution made by mixing 41.48 g salt with 972 g water

a. What is the mass percent water?

b. What is the mass percent salt?

Answer a

\(\frac{972g}{(972g+41.48g)}(100)=\frac{972g}{(1013.48g)}(100)=95.9\%\)
Note that although the denominator has 4 significant figures, you use them all in the calculation.

Answer b

\(\frac{972g}{(972g+41.48g)}(100)=\frac{972g}{(1013.48g)}(100)=4.093\%\)
Note that although the denominator has 4 significant figures, you use them all in the calculation.

 

Exercise \(\PageIndex{2}\)

An ore sample is 1.67% gold. How much pure gold is in 23.4 g of the ore?

Answer

\[\begin{align}
\%&=\frac{Part}{Whole}(100)=\frac{P}{W}(100)\nonumber\\
P&=\%\frac{W}{100}=1.67\frac{23.4g}{100}=0.39078g\nonumber
\end{align}\nonumber\]g Au in the ore is 0.391g
Note that the 100 in the percent is an exact number and does not influence the significant figures in the final answer.

 

Exercise \(\PageIndex{3}\)

Modern copper deposits tend to be low grade sulfide ores. What quantity of 0.874% copper ore is required to produce 1.00 lb of copper?

Answer

\[\begin{align}
\%&=\frac{Part}{Whole}(100)=\frac{P}{W}(100)\nonumber\\
W&=\frac{P}{\%}(100)\nonumber\\
W&=\frac{1}{0.874}(100)=114lb\nonumber
\end{align}\nonumber\]

 

Exercise \(\PageIndex{4}\)

What mass of sulfur is released upon combustion of 50.0 lbs of low grade coal containing 12.4% sulfur?

Answer

\[\begin{align}
\%&=\frac{Part}{Whole}(100)=\frac{P}{W}(100)\nonumber\\
P&=\frac{\%W}{100}\nonumber\\
P&=\frac{(12.4)(50.0)}{100}=6.20lb\nonumber
\end{align}\nonumber\]