Question

apply dimensional analysis and convert temperature.

1. 541 k to celsius degrees
2. 350 k to celsius degrees
3. 125°c to kelvin
4. 221°c to kelvin

apply dimensional analysis and solve for density variables.

1. mass = 142.4 g; volume =? ml; density = 0.915 g/ml
2. mass = 121.4 g; volume = 42.4 cm³; density =? g/cm³
3. mass =? g; volume = 124.1 ml; density = 0.821 g/ml
4. mass =? g; volume = 4.51 l; density = 1.15 g/cm³

Explanation:

Step1: Recall temperature conversion formula

The formula to convert Kelvin (K) to Celsius ($^{\circ}C$) is $T_{^{\circ}C}=T_{K}- 273.15$, and to convert Celsius to Kelvin is $T_{K}=T_{^{\circ}C}+273.15$. The formula for density $
ho=\frac{m}{V}$, where $
ho$ is density, $m$ is mass and $V$ is volume.

Step2: Convert 541 K to Celsius

$T_{^{\circ}C}=541 - 273.15=267.85^{\circ}C$

Step3: Convert 350 K to Celsius

$T_{^{\circ}C}=350 - 273.15 = 76.85^{\circ}C$

Step4: Convert 125$^{\circ}C$ to Kelvin

$T_{K}=125 + 273.15=398.15$ K

Step5: Convert 221$^{\circ}C$ to Kelvin

$T_{K}=221+ 273.15 = 494.15$ K

Step6: Solve for volume in density - related problem

Given $
ho=\frac{m}{V}$, then $V=\frac{m}{
ho}$. For $m = 142.4$ g and $
ho=0.915$ g/mL, $V=\frac{142.4}{0.915}\approx155.63$ mL

Step7: Solve for density

$
ho=\frac{m}{V}$, for $m = 121.4$ g and $V = 42.4$ $cm^{3}$, $
ho=\frac{121.4}{42.4}\approx2.86$ g/$cm^{3}$

Step8: Solve for mass

Given $
ho=\frac{m}{V}$, then $m=
ho\times V$. For $
ho = 0.821$ g/mL and $V = 124.1$ mL, $m=0.821\times124.1 = 101.8861$ g

Step9: Convert volume unit and solve for mass

Since $1L = 1000cm^{3}$, $V = 4.51L=4510cm^{3}$. Given $
ho = 1.15$ g/$cm^{3}$, $m=
ho\times V=1.15\times4510 = 5186.5$ g

Answer:

1. $267.85^{\circ}C$
2. $76.85^{\circ}C$
3. $398.15$ K
4. $494.15$ K
5. $155.63$ mL
6. $2.86$ g/$cm^{3}$
7. $101.89$ g
8. $5186.5$ g