Question

1. convert the following using dimensional analysis. show all work, including units:

a) 1.125 ml to l
b) 0.0085 ng to mg
c) 17.5 g/cm³ to lb/m³
d) 351 m/sec to km/hr
e) 24.5 m³ to in³
f) -17°c to kelvin
g) 42k to celsius

Explanation:

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Part a)

Step1: Identify conversion factor

$1\ \text{L} = 1000\ \text{mL}$ so $\frac{1\ \text{L}}{1000\ \text{mL}} = 1$

Step2: Multiply value by conversion factor

$1.125\ \text{mL} \times \frac{1\ \text{L}}{1000\ \text{mL}}$

Step3: Calculate the result

$1.125 \times \frac{1}{1000}\ \text{L} = 0.001125\ \text{L}$

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Part b)

Step1: List conversion factors

$1\ \text{Mg} = 10^9\ \text{mg}$, $1\ \text{mg} = 10^6\ \text{ng}$, so $\frac{1\ \text{Mg}}{10^{15}\ \text{ng}} = 1$

Step2: Multiply value by conversion factor

$0.0085\ \text{ng} \times \frac{1\ \text{Mg}}{10^{15}\ \text{ng}}$

Step3: Calculate the result

$0.0085 \times 10^{-15}\ \text{Mg} = 8.5 \times 10^{-18}\ \text{Mg}$

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Part c)

Step1: List conversion factors

$1\ \text{lb} = 453.592\ \text{g}$, $1\ \text{m}^3 = 10^6\ \text{cm}^3$

Step2: Set up dimensional analysis

$17.5\ \frac{\text{g}}{\text{cm}^3} \times \frac{1\ \text{lb}}{453.592\ \text{g}} \times \frac{10^6\ \text{cm}^3}{1\ \text{m}^3}$

Step3: Calculate the result

$17.5 \times \frac{10^6}{453.592}\ \frac{\text{lb}}{\text{m}^3} \approx 38580\ \frac{\text{lb}}{\text{m}^3}$

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Part d)

Step1: List conversion factors

$1\ \text{km} = 1000\ \text{m}$, $1\ \text{hr} = 3600\ \text{sec}$

Step2: Set up dimensional analysis

$351\ \frac{\text{m}}{\text{sec}} \times \frac{1\ \text{km}}{1000\ \text{m}} \times \frac{3600\ \text{sec}}{1\ \text{hr}}$

Step3: Calculate the result

$351 \times \frac{3600}{1000}\ \frac{\text{km}}{\text{hr}} = 1263.6\ \frac{\text{km}}{\text{hr}}$

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Part e)

Step1: List conversion factors

$1\ \text{m} = 39.37\ \text{in}$, so $1\ \text{m}^3 = (39.37)^3\ \text{in}^3 \approx 61023.7\ \text{in}^3$

Step2: Multiply value by conversion factor

$24.5\ \text{m}^3 \times \frac{61023.7\ \text{in}^3}{1\ \text{m}^3}$

Step3: Calculate the result

$24.5 \times 61023.7\ \text{in}^3 \approx 1.495 \times 10^6\ \text{in}^3$

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Part f)

Step1: Identify temperature conversion formula

$T(\text{K}) = T(^\circ\text{C}) + 273.15$

Step2: Substitute value into formula

$T(\text{K}) = -17 + 273.15$

Step3: Calculate the result

$T(\text{K}) = 256.15\ \text{K}$

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Part g)

Step1: Identify temperature conversion formula

$T(^\circ\text{C}) = T(\text{K}) - 273.15$

Step2: Substitute value into formula

$T(^\circ\text{C}) = 42 - 273.15$

Step3: Calculate the result

$T(^\circ\text{C}) = -231.15^\circ\text{C}$

Answer:

a) $0.001125\ \text{L}$
b) $8.5 \times 10^{-18}\ \text{Mg}$
c) $\approx 38580\ \text{lb/m}^3$
d) $1263.6\ \text{km/hr}$
e) $\approx 1.495 \times 10^6\ \text{in}^3$
f) $256.15\ \text{K}$
g) $-231.15^\circ\text{C}$