Skip to main content
Chemistry LibreTexts

2: Mathematical Fundamentals

  • Page ID
    257701
  • Mathematical Fundamentals

    Exercise \(\PageIndex{1a}\)

    Define the following:

    1. Commutative property of addition
    2. Associative property of addition
    3. Distributive property of multiplication
    Answer a

    The order of addition does not matter:  A+B=B+A

    Answer b

    the sum of three or more numbers is the same, no matter how you add them.  (A+B)+C=(B+C)+A=(C+A)+B...

    Answer c

    The product of multiplying the sum of two numbers by a third number is the same as the sum of multiplying both numbers by the third and summing the two products.

    A(B+C)=AB+AC

    Algebra review

     

    Exercise \(\PageIndex{2}\)

    Solve For \(A\) for each of the following equations:

    a. \(A(X+Y)=Z\)

    b. \((AX+Y)=Z\)

    c. \((AX+AY)=Z\)

    d. \((AX+Y)Z=1\)

    e. \(\frac{A}{X+Y}=Z\)

    f. \(\frac{1}{AX+AY}=Z\)

    g.\(\frac{1}{X+AY}=Z\)

    h.\(\frac{A}{X+Y}=ZA+3\)

    i.\(\frac{A}{X+Y}=ZA+D\)

    j.\(\frac{A}{X+Y}=ZA+D\)

    Answer a

    \(A(X+Y)=Z\\\frac{A(X+Y)}{X+Y}=\frac{Z}{X+Y}\\\frac{A\cancel{(X+Y)}}{\cancel{X+Y}}=\frac{Z}{X+Y}\\A=\frac{Z}{X+Y}\)

    Answer b

    \((AX+Y)=Z\\(AX+Y)-Y=Z-Y\\(AX+\cancel{Y})-\cancel{Y}=Z-Y\\\frac{AX}{X}=\frac{Z-Y}{X}\\\frac{A\cancel{X}}{\cancel{X}}=\frac{Z-Y}{X}\\A=\frac{Z-Y}{X}\)

    Answer c

    \((AX+AY)=Z\\A(X+Y)=Z\\\frac{A(X+Y)}{(X+Y)}=\frac{Z}{X+Y}\\\frac{A\cancel{(X+Y)}}{\cancel{(X+Y)}}=\frac{Z}{X+Y}\\A=\frac{Z}{X+Y}\)

    Answer d

    \((AX+Y)Z=1\\\frac{(AX+Y)Z}{Z}=\frac{1}{Z}\\\frac{(AX+Y)\cancel{Z}}{\cancel{Z}}=\frac{1}{Z}\\AX+Y-Y=\frac{1}{Z}-Y\\AX+\cancel{Y}-\cancel{Y}=\frac{1}{Z}-Y\\\frac{AX}{X}=\frac{\frac{1}{Z}-Y}{X}\\\frac{A\cancel{X}}{\cancel{X}}=\frac{\frac{1}{Z}-Y}{X}\\A=\frac{\frac{1}{Z}-Y}{X}\)

    Answer e

    \(\frac{A}{X+Y}=Z\\\frac{A}{X+Y}(X+Y)=Z(X+Y)\\\frac{A}{\cancel{X+Y}}\cancel{(X+Y)}=Z(X+Y)\\A=Z(X+Y)\)

    Answer f

    \(\frac{1}{AX+AY}=Z\\AX+AY=\frac{1}{Z}\\A(X+Y)=\frac{1}{Z}\\\frac{A(X+Y)}{(X+Y)}=\frac{1}{Z(X+Y)}\\\frac{A\cancel{(X+Y)}}{\cancel{(X+Y)}}=\frac{1}{Z(X+Y)}\\A=\frac{1}{Z(X+Y)}\)

    Answer g

    \(\frac{1}{X+AY}=Z\\X+AY=\frac{1}{Z}\\X+AY-X=\frac{1}{Z}-X\\\cancel{X}+AY\cancel{-X}=\frac{1}{Z}-X\\\frac{AY}{Y}=\frac{\frac{1}{Z}-X}{Y}\\\frac{A\cancel{Y}}{\cancel{Y}}=\frac{\frac{1}{Z}-X}{Y}\\A=\frac{\frac{1}{Z}-X}{Y}\)

    Answer h

    \(\frac{A}{X+Y}=ZA+3\\\frac{A}{X+Y}(X+Y)=(ZA+3)(X+Y)\\\frac{A}{\cancel{(X+Y)}}\cancel{(X+Y)}=(ZA+3)(X+Y)\\A=ZA(X+Y)+3(X+Y)\\A-ZA(X+Y)=ZA(X+Y)-ZA(X+Y)+3(X+Y)\\A-ZA(X+Y)=\cancel{ZA(X+Y)}-\cancel{ZA(X+Y)}+3(X+Y)\\    A(1-ZA(X+Y))=3(X+Y)\\\frac{A(1-ZA(X+Y))}{(1-ZA(X+Y))}=\frac{3(X+Y)}{(1-ZA(X+Y))}\\\frac{A\cancel{(1-ZA(X+Y))}}{\cancel{(1-ZA(X+Y))}}=\frac{3(X+Y)}{(1-ZA(X+Y))}\\A=\frac{3(X+Y)}{1-Z(X+Y)}\)

    Answer i

    \(\frac{A}{X+Y}=ZA+D\\\frac{A}{X+Y}(X+Y)=(ZA+D)(X+Y)\\\frac{A}{\cancel{(X+Y)}}\cancel{(X+Y)}=(ZA+D)(X+Y)\\A=ZA(X+Y)+D(X+Y)\\A-ZA(X+Y)=ZA(X+Y)-ZA(X+Y)+D(X+Y)\\A-ZA(X+Y)=\cancel{ZA(X+Y)}-\cancel{ZA(X+Y)}+D(X+Y)\\    A(1-ZA(X+Y))=D(X+Y)\\\frac{A(1-ZA(X+Y))}{(1-ZA(X+Y))}=\frac{D(X+Y)}{(1-ZA(X+Y))}\\\frac{A\cancel{(1-ZA(X+Y))}}{\cancel{(1-ZA(X+Y))}}=\frac{D(X+Y)}{(1-ZA(X+Y))}\\A=\frac{D(X+Y)}{1-Z(X+Y)}\)

    Answer j

    \(\frac{A+P}{X+Y}=ZA+D\\\frac{A+P}{X+Y}(X+Y)=(ZA+D)(X+Y)\\\frac{A+P}{\cancel{(X+Y)}}\cancel{(X+Y)}=(ZA+D)(X+Y)\\A+P=ZA(X+Y)+D(X+Y)\\A+P-P=ZA(X+Y)+D(X+Y)-P\\A+\cancel{P}-\cancel{P}=ZA(X+Y)+D(X+Y)-P\\A-ZA(X+Y)=ZA(X+Y)-ZA(X+Y)+D(X+Y)-P\\A-ZA(X+Y)=\cancel{ZA(X+Y)}-\cancel{ZA(X+Y)}+D(X+Y)-P\\    A(1-ZA(X+Y))=D(X+Y)-P\\\frac{A(1-ZA(X+Y))}{(1-ZA(X+Y))}=\frac{D(X+Y)-P}{(1-ZA(X+Y))}\\\frac{A\cancel{(1-ZA(X+Y))}}{\cancel{(1-ZA(X+Y))}}=\frac{D(X+Y)-P}{(1-ZA(X+Y))}\\A=\frac{D(X+Y)-P}{1-Z(X+Y)}\)  

     

     

    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 Tc:

    Answer

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

    Step 1: Bring all things with TC 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 TF:

    Answer

    Separate Tf from all other variables by dividing both sides by mc and adding Ti 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. cH
    2. cc
    3. Tc
    4. TF
    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 TC 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 TC 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 TF 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 TC by dividing both sides by \(m_{C} c_{C}+m_{H} c_{H}\)

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

     

     

    Numbers

    This section was just an introduction to the 4 ways commonly used to express numbers, and each way will be gone over in turn, and that is where practice problems will exist.

    Exercise \(\PageIndex{3a}\)

    Identified the following as measured, defined or a counted numbers, and if there is uncertainty.

    1. mass  = 3 kg water
    2. 3 feet = 1 yard
    3. 3 eggs
    4. 3 dozen eggs
    Answer a

    measured, there is uncertainty

    Answer b

    defined, there is no uncertainty

    Answer c

    counted, there is no uncertainty

    Answer d

    counted and defined, a dozen is defined as 12 eggs, so there are 36 eggs, and this is an exact number with no uncertainty

    Units of Measurement

    Exercise \(\PageIndex{4a}\)

    Give the SI Base Unit for each of the following

    1. time
    2. mass
    3. quantity
    4. temperature
    5. length
    Answer a

    second (s)

    Answer b

    kilogram (k)

    Answer c

    mole (n)

    Answer d

    kelvin (K)

    Answer e

    meter (m)

     

    Measured Numbers & Significant Digits

    Exercise \(\PageIndex{5a}\)

    Give the number of significant figures in each measured value.

    1. 5.87
    2. 0.031
    3. 52.90
    4. 0.2001
    5. 500
    Answer a

    3 (see section 2.4)

    Answer b

    2 (see see section 2.4)

    Answer c

    4 (see section 2.4)

    Answer d

    4 (see section 2.4)

    Answer e

    1(see section 2.4),  note, some texts will say it is not determined.  That is, you do not know. We will use the convention that trailing zeros are not significant if there is not a decimal, but if your instructor uses a different convention, play the game and use the rules your class uses.

    Exercise \(\PageIndex{5b}\)

    Give the number of significant figures in each measured value.

    1. 0.0053
    2. 2300
    3. 32.00
    4. 34.483
    5. 3450000
    Answer a

    2 (see section 2.4)

    Answer b

    2 (see section 2.4)

    Answer c

    4 (see section 2.4)

    Answer d

    5 (see section 2.4)

    Answer e

    3(see section 2.4)

     

    Exercise \(\PageIndex{5c}\)

    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
     

     

    Exercise \(\PageIndex{5d}\)

    Solve the following problems:

    1. \(3.03860 +3.7+36.8 \)
    2. \(95.7 +4.41+0.111\)
    Answer a

    \(\begin{align} \nonumber & 3.03860  \\ \nonumber &3.7 \\ + \;\; \nonumber 3 &6.8 \\ \hline  4 &3.5386 \end{align} \)

    The answer is 43.5 because 3.7 is only known to the 10ths position (see section 2.5.3)

    Answer b

    \(\begin{align} \nonumber 9 & 5.7  \\ \nonumber & 4.4 \\ + \;\;  \nonumber &0.111 \\ \hline  10 &0.211 \end{align} \)

    The answer is 100.2 because 95.7 is only known to the 10ths position (see section 2.5.3)

     

    Exercise \(\PageIndex{5e}\)

    Solve the following problems:

    1. \( \frac{(3.03860)(3.7)}{36.8} \)
    2. \(\frac{95.7}{(4.41)(0.111)}\)
    Answer a

    0.3055=0.31

    round up because digit after truncated 5 is odd. (see section 2.5.3)

    Answer b

    195.5016=195

    round down because digit after truncated 5 is even (see section 2.5.3)

     

    Exercise \(\PageIndex{5f}\)

    Solve the following problems:

    1. \( \frac{(3.03860+8.7)}{36.8} \)
    2. \(\frac{95.7}{(9.4+0.7)}\)
    Answer a

    0.318983696=0.319

    Answer b

    9.475247525=9.47

    Round the truncated 5 down as it was followed by an even number)

     

    Exercise \(\PageIndex{5g}\)

    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 \(\PageIndex{5h}\)

    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

    Scientific Notation

    Exercise \(\PageIndex{6a}\)

    Solve the following problems:

    1. \begin{equation}
      3.03860 \times 10^{32}+3.7 \times 10^{30}+3.68 \times 10^{31}
      \end{equation}
    2. \begin{equation}
      3.03860 \times 10^{-32}+3.7 \times 10^{-30}+3.68 \times 10^{-31}
      \end{equation}
    Answer a

    \( \begin{align}\nonumber & 3.03860 \times 10^{32}\\\nonumber &0.037 \times 10^{32}\\\nonumber+& 0.368\times 10^{32}\\ \hline & 3.4436 \times 10^{32} \end{align}\)

    The answer is \(3.444\times 10^{32}\) because two of the least precise value was known to the thousandths (see section 2.6)

    Answer b

    \( \begin{align}\nonumber & 3.7 \times 10^{-30}\\\nonumber &0.368 \times 10^{-30}\\\nonumber+& 0.0303860\times 10^{-30}\\ \hline & 4.098386\times 10^{-30} \end{align}\)

    The answer is \(4.1\times 10^{-30}\) because two of the least precise value was known to the thousandths (see section 2.6)

    Exercise \(\PageIndex{6b}\)

    Solve the following problems:

    1. \begin{equation}
      \frac{\left(9.7 \times 10^{583}\right)\left(7.339 \times 10^{98}\right)}{\left(5.432 \times 10^{-645}\right)\left(3.446 \times 10^{-4484}\right)}
      \end{equation}
    2. \begin{equation}
      \frac{\left(9.7 \times 10^{583}\right)\left(7.339 \times 10^{-98}\right)}{\left(5.432 \times 10^{-645}\right)\left(3.446 \times 10^{4484}\right)}
      \end{equation}
    Answer a

    \(\frac{(9.7\times10^{583})(7.339\times10^{98})}{(5.432\times10^{-645})(3.446\times10^{-4484})}\\(\frac{9.7\times7.339}{5.432\times3.446})(10^{583+98-(-645)-(-4484)})\\3.80306\times10^{5810}\)

    The answer is \(3.8\times 10^{5810}\) (see section 2.6)

    Answer b

    \(\frac{(9.7\times10^{583})(7.339\times10^{-98})}{(5.432\times10^{-645})(3.446\times10^{4484})}\\(\frac{9.7\times7.339}{5.432\times3.446})(10^{583-98-(-645)-(-4484)})\\3.80306\times10^{5614}\)

    The answer is \(3.8\times 10^{5614}\)  (see section 2.6)

    Exercise \(\PageIndex{6c}\)

    Solve the following problems:

    1. \begin{equation}
      \frac{\left(9.7 \times 10^{-583}\right)\left(7.339 \times 10^{-98}\right)}{\left(5.432 \times 10^{645}\right)\left(3.446 \times 10^{4484}\right)}
      \end{equation}
    2. \begin{equation}
      \frac{\left(9.70 \times 10^{5}+6.3 \times 10^{4}\right)\left(3.33 \times 10^{5}\right)}{\left(5.00 \times 10^{45}\right)\left(3.20 \times 10^{-5}\right)}
      \end{equation}
    Answer a

    \(\frac{(9.7\times10^{-583})(6.3\times10^{-98})}{(5.432\times10^{645})(3.446\times10^{4484})}\\(\frac{9.7\times7.339}{5.432\times3.446})(10^{-583-98-645-4484})\\3.80306\times10^{-5810}\)

    The answer is \(3.8\times 10^{-5810}\) (see section 2.6)

    Answer b

    \(\frac{(9.70\times10^5\;+\;6.30\times10^4)(3.33\times10^5)}{(5.00\times10^{45})(3.20\times10^{-5})}\\\frac{(9.70\times10^5\;+\;0.63\times10^5)(3.33\times10^5)}{(5.00\times10^{45})(3.20\times10^{-5})}\\\frac{(10.33\times10^5)(3.33\times10^5)}{(5.00\times10^{45})(3.20\times10^{-5})}\\(\frac{10.33\times3.33}{5.00\times3.20})(10^{5+5-45-(-5)})\\2.1499\times10^{-30}\)

    The answer is \(2.15\times 10^{-30}\) (see section 2.6)

     

    Exercise \(\PageIndex{6d}\)

    Solve the following problems:

    1. \begin{equation}
      6.13833 \times 10^{52}+6.9 \times 10^{50}+6.77 \times 10^{51}
      \end{equation}
    2. \begin{equation}
      \frac{\left(1.7 \times 10^{984}\right)\left(7.469 \times 10^{998}\right)}{\left(5.432 \times 10^{-645}\right)\left(3.446 \times 10^{-4484}\right)}
      \end{equation}
    Answer a

    \( \begin{align}\nonumber & 6.13833 \times 10^{52}\\\nonumber &0.068 \times 10^{52}\\\nonumber+& 0.677\times 10^{52}\\ \hline & 6.88333 \times 10^{52} \end{align}\)

    The answer is \(6.883\times 10^{52}\) because two of the added numbers had a least precise value was known to the thousandths (see section 2.6)

    Answer b

    (\(\frac{1.7\times7.469}{5.432\times3.446})(10^{984+998-(-645)-(-4484)})\\0.678322693\times10^{7111}\)

    The answer is \(6.8\times 10^{7110}\)  (see section 2.6)

     

    Exercise \(\PageIndex{6e}\)

    Solve the following problems:

    1. \begin{equation}
      \frac{\left(2.27 \times 10^{577}\right)\left(7.6 \times 10^{498}\right)}{\left(5.432 \times 10^{-945}\right)\left(3.446 \times 10^{-44484}\right)}
      \end{equation}
    2. \begin{equation}
      4.238 \times 10^{-22}+7.896 \times 10^{-20}+9.99 \times 10^{-21}
      \end{equation}
    Answer a

    (\(\frac{2.27\times7.6}{5.432\times3.446})(10^{577+498-(-945)-(-44484)})\\0.9216465783\times10^{46504}\)

    The answer is \(9.2\times 10^{46503}\) (see section 2.6)

    Answer b

    \( \begin{align}\nonumber & 7.896 \times 10^{-20}\\\nonumber &0.04238 \times 10^{-20}\\\nonumber+& 0.999\times 10^{-20}\\ \hline & 8.89738\times 10^{-20} \end{align}\)

    The answer is \(8.897\times 10^{-20}\) because two of the least precise value was known to the thousandths (see section 2.6)

     

     

    Exercise \(\PageIndex{6f}\)

    Solve the following problems:

    1. \begin{equation}
      4.03807 \times 10^{42}+3.8 \times 10^{40}+6.28 \times 10^{41}
      \end{equation}
    2. \begin{equation}
      \frac{\left(1.77 \times 10^{1577}\right)\left(8.3 \times 10^{1984}\right)}{\left(5.32 \times 10^{-1945}\right)\left(3.46 \times 10^{-44}\right)}
      \end{equation}
    Answer a

    \( \begin{align}\nonumber & 4.03807 \times 10^{42}\\\nonumber &0.038 \times 10^{42}\\\nonumber+& 0.628\times 10^{42}\\ \hline & 4.7047 \times 10^{42} \end{align}\)

    The answer is \(4.705\times 10^{42}\) (see section 2.6)

    Answer b

    (\(\frac{1.77\times8.3}{5.32\times3.46})(10^{1577+1984-(-1945)-(-44)})\\0.7981116085\times10^{5550}\)

    The answer is \(8.0\times 10^{5549}\)  (see section 2.6)

     

    SI Prefixes

    Exercise \(\PageIndex{7a}\)

    Give the name of the prefix and the quantity indicated by the following symbols that are used with SI base units.

    1. c
    2. d
    3. G
    4. k
    Answer a

    centi-, × 10−2

    Answer b

    deci-, × 10−1

    Answer c

    Giga-, × 109

    Answer d

    kilo-, × 103

    Exercise \(\PageIndex{7b}\)

    Give the name of the prefix and the quantity indicated by the following symbols that are used with SI base units.

    1. m
    2. n
    3. p
    4. T
    Answer a

    milli-, × 10−3

    Answer b

    nano-, × 10−9

    Answer c

    pico-, × 10−12

    Answer d

    tera-, × 1012

     

    Unit Conversions

    Exercise \(\PageIndex{8a}\)

    How many dm are in an object which is 2.45 feet long?

    12in = 1 ft

    Answer

    \(2.450 f t\left(\frac{12 i n}{f t}\right)\left(\frac{2.54 c m}{i n}\right)\left(\frac{10^{1} d m}{10^{2} c m}\right)=7.4676 d m=7.468 d m\) 

    Exercise \(\PageIndex{8b}\)

    How many gallons are in a cuboid can that is 2.05 ft wide, 12.4 inches long and 1.30 yards deep?

    4 qt = 1 gal, 1.057qt=1L, 12in = 1 ft, 3 ft = 1 yd

    Answer

    \(2.05 f t(12.4 i n)(1.30 y d)\left(\frac{3 f t}{y d}\right)\left(\frac{12 i n}{f t}\right)^{2}\left(\frac{2.54 c m}{i n}\right)^{3}\left(\frac{1 m L}{1 c m^{3}}\right)\left(\frac{1 L}{1000 m L}\right)\left(\frac{1.057 q t}{1 L}\right)\left(\frac{g a l}{4 q t}\right)=61.8 g a l\)

    Exercise \(\PageIndex{8c}\)

    How many square inches are in a rectangle which is 3.24ft by 2.67 yd?

    Answer

    \(3.24 f(2.67 y d)\left(\frac{3 f t}{y d}\right)\left(\frac{12 i n}{f t}\right)^{2}=3737.1456 i n^{2}=3740 i n^{2}\)

    Exercise \(\PageIndex{8d}\)

    How many cubic inches are in 1.00 mL?

    Answer

    \(1.00 m l\left(\frac{c m^{3}}{1 m l}\right)\left(\frac{i n}{2.54 c m}\right)^{3}=0.0610 i n^{3}\)

    Exercise \(\PageIndex{8e}\)

    How many cubic feet are in 1.00 TL?

    Answer

    \(1.00 T l\left(\frac{10^{3} m L}{10^{-12} T L}\right)\left(\frac{1 c m^{3}}{1 m L}\right)\left(\frac{i n}{2.54 c m}\right)^{3}\left(\frac{f t}{12 i n}\right)^{3}=3.53 \times 10^{10} f^{3}\)

    Exercise \(\PageIndex{8f}\)

    How many kL are in a cube which is 3.24x1021mm on each side?

    Answer

    \(\left(3.24 \times 10^{21} \mathrm{mm}\right)^{3}\left(\frac{10^{2} \mathrm{cm}}{10^{3} \mathrm{mm}}\right)^{3}\left(\frac{\mathrm{mL}}{1 \mathrm{cm}^{3}}\right)\left(\frac{10^{-3} \mathrm{kL}}{10^{3} \mathrm{mL}}\right)=3.40 \times 10^{55} \mathrm{kL}\)

    Exercise \(\PageIndex{8g}\)

    How many square inches are in a rectangle which is 1.75x1011mm by 7.34x10-13km

    Answer

    \(1.75 \times 10^{11} \mathrm{mm}\left(7.34 \times 10^{-13} \mathrm{km}\right)\left(\frac{\mathrm{cm}}{10 \mathrm{mm}}\right)\left(\frac{10^{2} \mathrm{cm}}{10^{-3} \mathrm{km}}\right)\left(\frac{\mathrm{in}}{2.54 \mathrm{cm}}\right)^{2}=199 \mathrm{in}^{2}\)

    Exercise \(\PageIndex{8h}\)

    How many mL are in an cuboid that is 2.00x1021mm by 3.00x102yard by 5.00x1018ft?

    Answer

    \(2.00 x 10^{21} m m\left(3.00 \times 10^{2} y d\right)\left(5.00 x 10^{18} f\right)\left(\frac{3 f t}{y d}\right)\left(\frac{12 i n}{f t}\right)^{2}\left(\frac{1 c m}{10 m m}\right)\left(\frac{2.54 c m}{1 i n}\right)^{2}\left(\frac{m L}{1 c m^{3}}\right)=8.36 x 10^{44} m L\)

    Exercise \(\PageIndex{8i}\)

    How many liters is the volume of 5.873 kg of Mercury? Use 13.6 g/mL as the density of Mercury

    Answer

    \(5.873kg\left(\frac{1000 g}{1 kg}\right)\left(\frac{1 mL}{13.6g}\right)\left(\frac{1 L}{1000mL}\right)=0.431L\)

    Exercise \(\PageIndex{8j}\)

    How many gallons is the volume of 14.0 kg of Gallium? Use 5.9 g/mL as the density of Gallium

    Answer

    \(14.0kg\left(\frac{1000 g}{1 kg}\right)\left(\frac{1 mL}{5.9g}\right)\left(\frac{1 L}{1000mL}\right)\left ( \frac{1.057qt}{1 L} \right )\left ( \frac{1gal}{4qt} \right )=0.62703gal=0.63gal\)

    Exercise \(\PageIndex{8k}\)

    How many pounds would 3.789 gallons of water weight if at a given temperature the water has a density of 0.96g/mL?

    Answer

    \(\left ( 3.789gal \right )\left ( \frac{4qt}{1gal} \right )\left ( \frac{1L}{1.057qt} \right )\left ( \frac{1000mL}{1L} \right )\left ( \frac{0.96g}{mL} \right )\left ( \frac{1lb}{453.6g} \right )= 5.392lb= 5.4lb\)

    Exercise \(\PageIndex{8l}\)

    How many pounds would 135 liters of water weight if at a given temperature the water has a density of 0.99g/mL?

    Answer

    \(\left ( 135L \right )\left ( \frac{1000mL}{1L} \right )\left ( \frac{0.99g}{mL} \right )\left ( \frac{1lb}{453.6g} \right )= 294.64lb= 295lb\)

    Exercise \(\PageIndex{8m}\)

    You have a rectangular metal surface that is 3.81X1023in long by 4.22x10-2yards wide that you wish to make a platinum catalyst with. To do this you need to coat the surface with platinum and wish to use the minimum amount of platinum.  Using the process of chemical vapor deposition you will make a layer that is the thickness of a single layer of platinum atoms, which have a depth of 175 pm. The current price is $908/oz, 28.3495g=1 ounce and the density of platinum is 24.15 g/ml. What would be the cost of the platinum to make this surface?

    Answer

    \[3.81x10^{23}\left ( 0.0422yd \right )\left ( 175pm \right )
    \left ( \frac{36in}{yd} \right )
    \left ( \frac{2.54cm}{in} \right )^2
    \left ( \frac{10^2cm}{10^{12}pm} \right )
    \left ( \frac{1ml}{cm^3} \right )
    \left ( \frac{24.15g}{ml} \right )
    \left ( \frac{1 oz}{28.3495g} \right )
    \left ( \frac{\$908}{oz} \right ) \\ =\$5.06806x10^{19}=\$5.07x10^{19} \nonumber\]

     

    Fractions and Percent

    Exercise \(\PageIndex{9a}\)

    What mass of Water do you need to add to 20.6g of salt to make a solution that is 25.0% Salt?

    Hint: Solve this in terms of fractions, where fraction of salt = \(\frac{\%salt}{100}\)

    Answer

    fs=fraction salt, ms=mass salt, mw=mass water,  if percent salt = 25.0%, the fs= 0.250

    \(f_{s}=\frac{m_{s}}{m_{s}+m_{w}} \\ \; \\  \text{solve for }m_w \\ \; \\
    f_{s}\left(m_{s}+m_{w}\right)=m_{s} \\ \; \\
    f_{s} m_{s}+f_{s} m_{w}=m_{s} \\ \; \\
    f_{s} m_{w}=m_{s}-f_{s} m_{s}=m_{s}\left(1-f_{s}\right) \\ \; \\
    m_{w}=\frac{\left(1-f_{s}\right)}{f_{s}} m_{s}=\frac{(1-.250)}{.250} 20.6 g=\frac{.750}{.250} 20.6 g=61.8 g
    \)

    Exercise \(\PageIndex{9b}\)

    What mass of Water do you need to add to 47.5g of salt to make a solution that is 5.0% Salt?

    Hint: Solve this in terms of fractions, where fraction of salt = \(\frac{\%salt}{100}\)

    Answer

    fs=fraction salt, ms=mass salt, mw=mass water,  if percent salt = 5.0%, the fs= 0.050

    \(f_{s}=\frac{m_{s}}{m_{s}+m_{w}} \\ \; \\  \text{solve for }m_w \\ \; \\
    f_{s}\left(m_{s}+m_{w}\right)=m_{s} \\ \; \\
    f_{s} m_{s}+f_{s} m_{w}=m_{s} \\ \; \\
    f_{s} m_{w}=m_{s}-f_{s} m_{s}=m_{s}\left(1-f_{s}\right) \\ \; \\
    m_{w}=\frac{\left(1-f_{s}\right)}{f_{s}} m_{s}=\frac{(1-.050)}{.050} 47.5 g=\frac{.950}{.050} 47.5 g=902.5 g = 902 g
    \)

    Exercise \(\PageIndex{9c}\)

    What mass of Water do you need to add to 0.84g of salt to make a solution that is 32.0% Salt?

    Hint: Solve this in terms of fractions, where fraction of salt = \(\frac{\%salt}{100}\)

    Answer

    fs=fraction salt, ms=mass salt, mw=mass water,  if percent salt = 30.0%, the fs= 0.320

    \(f_{s}=\frac{m_{s}}{m_{s}+m_{w}} \\ \; \\  \text{solve for }m_w \\ \; \\
    f_{s}\left(m_{s}+m_{w}\right)=m_{s} \\ \; \\
    f_{s} m_{s}+f_{s} m_{w}=m_{s} \\ \; \\
    f_{s} m_{w}=m_{s}-f_{s} m_{s}=m_{s}\left(1-f_{s}\right) \\ \; \\
    m_{w}=\frac{\left(1-f_{s}\right)}{f_{s}} m_{s}=\frac{(1-.320)}{.320} 0.84 g=\frac{.680}{.320} 0.84 g=1.785 g = 1.8 g
    \)

    Exercise \(\PageIndex{9d}\)

    Consider Mixing 55.4 g of table salt (NaCl) with 102 g of water

    1. What is the percent salt?
    2. What is the percent water?
    Answer a

    \( \% \; salt =  \frac{m_{salt}}{m_{salt}+m_{water}}(100)  = \frac{55.4g}{55.4g+102g}(100) = \frac{55.4g}{157.4g}(100)=35.196 \%=35.2\%  \)

    Answer b

    \( \% \; water =  \frac{m_{water}}{m_{salt}+m_{water}}(100)  = \frac{102g}{55.4g+102g}(100) = \frac{102g}{157.4g}(100)=64.803 \%  =64.8 \% \)

    Exercise \(\PageIndex{9e}\)

    Consider Mixing 15.2 g of table salt (NaCl) with 87 g of water

    1. What is the percent salt?
    2. What is the percent water?
    Answer a

    \( \% \; salt =  \frac{m_{salt}}{m_{salt}+m_{water}}(100)  = \frac{15.2g}{15.2g+87g}(100) = \frac{15.2g}{102.2g}(100)=14.872 \%=14.9\%  \)

    Answer b

    \( \% \; water =  \frac{m_{water}}{m_{salt}+m_{water}}(100)  = \frac{87g}{15.2g+87g}(100) = \frac{87g}{102.2g}(100)=85.127 \%  = 85 \% \)

    Exercise \(\PageIndex{9f}\)

    Consider Mixing 4.7 g of table salt (NaCl) with 25.0 g of water

    1. What is the percent salt?
    2. What is the percent water?
    Answer a

    \( \% \; salt =  \frac{m_{salt}}{m_{salt}+m_{water}}(100)  = \frac{4.7g}{4.7g+25.0g}(100) = \frac{4.7g}{29.7g}(100)=15.824 \%=16\%  \)

    Answer b

    \( \% \; water =  \frac{m_{water}}{m_{salt}+m_{water}}(100)  = \frac{25.0g}{4.7g+25.0g}(100) = \frac{25.0g}{29.7g}(100)=84.175 \%  = 84.2 \% \)

    Exercise \(\PageIndex{9g}\)

    Consider Mixing 55.4 g of table salt (NaCl) with 102 g of water

    1. What is the fraction salt?
    2. What is the fraction water?
    Answer a

    \( \text{fraction salt = } \frac{m_{salt}}{m_{salt}+m_{water}}  = \frac{55.4g}{55.4g+102g} = \frac{55.4g}{157.4g}=0.3519695=0.352 \nonumber \)

    Answer b

    \( \text{fraction water = } \frac{m_{water}}{m_{salt}+m_{water}}  = \frac{102g}{55.4g+102g} = \frac{102g}{157.4g}=0.6480305 = 0.648 \nonumber \)

    Exercise \(\PageIndex{9h}\)

    Consider Mixing 15.2 g of table salt (NaCl) with 87 g of water

    1. What is the fraction salt?
    2. What is the fraction water?
    Answer a

    \( \text{fraction salt = } \frac{m_{salt}}{m_{salt}+m_{water}}  = \frac{15.2g}{15.2g+87g} = \frac{15.2g}{102.2g}=0.14872798=0.149 \nonumber \)

    Answer b

    \( \text{fraction water = } \frac{m_{water}}{m_{salt}+m_{water}}  = \frac{87g}{15.2g+87g} = \frac{87g}{102.2g}=0.85127202 = 0.85 \nonumber \)

    Exercise \(\PageIndex{9i}\)

    Consider Mixing 4.7 g of table salt (NaCl) with 25.0 g of water

    1. What is the fraction salt?
    2. What is the fraction water?
    Answer a

    \( \text{fraction salt = } \frac{m_{salt}}{m_{salt}+m_{water}}  = \frac{4.7g}{4.7g+25.0g} = \frac{4.7g}{29.7g}=0.15824916=0.16 \nonumber \)

    Answer b

    \( \text{fraction water = } \frac{m_{water}}{m_{salt}+m_{water}}  = \frac{25.0g}{4.7g+25.0g} = \frac{25.0g}{29.7g}=0.84175084 = 0.842 \nonumber \)

     

    Graphing and Temperature Conversion

    Exercise \(\PageIndex{10a}\)

    Convert the following:

    1. 37.0 ºC to _____ K
    2. 331 K to _____ ºC
    Answer a

    310.2 K

    \[(37.0+273.15)=310.2K\]

    (see section 1B.1.5)

    Answer b

    57.9 ºC

    \begin{equation}
    (331-273.15)=57.9^{\circ} \mathrm{C}
    \end{equation}

    (see section 1B.1.5)

    Exercise \(\PageIndex{10b}\)

    Convert the following:

    1. 57.8 ºC to _____ ºF
    2. -129 ºF to _____ ºC
    Answer a

    136 ºF

    \begin{equation}
    [(57.8+40) 1.8]-40=136^{\circ} \mathrm{F}
    \end{equation}

    (see section 1B.1.5)

    Answer b

    -89.4 ºC

    \begin{equation}
    \left(\frac{(-129+40)}{1.8}\right)-40=-89.4^{\circ} \mathrm{C}
    \end{equation}

    (see section 1B.1.5)

    Exercise \(\PageIndex{10c}\)

    Convert the boiling temperature of gold, 2966 °C, into degrees Fahrenheit and Kelvin.

    Answer

    \(\left [\left (2966^{\circ}C+40  \right ) 1.8  \right ]-40=5371^{\circ}F\)

    \(2966 ^{\circ} C +273.15 = 3239K\)

    (see section 1B.1.5)

    Exercise \(\PageIndex{10d}\)

    Convert the temperature of scalding water, 54 °C, into degrees Fahrenheit and Kelvin.

    Answer

    \(\left [\left (54^{\circ}C+40  \right ) 1.8  \right ]-40=130^{\circ}F\)

    \(54 ^{\circ} C +273.15 = 330\)

    130 °F, 330 K

    (see section 1B.1.5)

    Exercise \(\PageIndex{10e}\)

    Convert the temperature of the coldest area in a freezer, −10 °F, to degrees Celsius and Kelvin.

    Answer

    \( [(\frac{10 ^\circ F\;+\;40)}{1.6}]-40\;=\;-23 ^{\circ} C \)

    \( [(\frac{10 ^\circ F\;+\;40)}{1.6}]-40\;+\;273.15\;=\;250\;K\)

    −23 °C, 250 K

    (see section 1B.1.5)

    • Was this article helpful?