43. glycerol is a viscous liquid often used in cosmetics and soaps. a 3.25 l sample of pure glycerol has a…
43. glycerol is a viscous liquid often used in cosmetics and soaps. a 3.25 l sample of pure glycerol has a mass of 4.10×10³ g. what is the density of glycerol in g cm⁻³?\n44. a supposedly gold nugget is tested to determine its density. it is found to displace 19.3 ml of water and has a mass of 371 grams. could the nugget be made of gold?\n45. ethylene glycol (antifreeze) has a density of 1.11 g cm⁻³.\n a. what is the mass, in g, of 417 ml of this liquid?\n b. what is the volume, in l, of 4.1 kg of this liquid?\n46. acetone (nail polish remover) has a density of 0.7857 g cm⁻³.\n a. what is the mass, in g, of 28.56 ml of acetone?\n b. what is the volume, in ml, of 6.54 g of acetone?\n47. a small airplane takes on 245 l of fuel. if the density of the fuel is 0.803 g ml⁻¹, what mass of fuel has the airplane taken on?\n48. human fat has a density of 0.918 g cm⁻³. how much volume (in cm³) is gained by a person who gains 5.00 kg of pure fat?
Answer
43.
Explanation:
Step1: Convert volume to $cm^3$
$1\ L = 1000\ cm^3$, so $3.25\ L=3.25\times1000 = 3250\ cm^3$
Step2: Calculate density
Density formula is $\rho=\frac{m}{V}$, where $m = 4.10\times 10^{3}\ g$ and $V = 3250\ cm^3$. So $\rho=\frac{4.10\times 10^{3}\ g}{3250\ cm^3}\approx1.26\ g\ cm^{-3}$
Answer:
$1.26\ g\ cm^{-3}$
44.
Explanation:
Step1: Use density formula
The volume of the nugget is equal to the volume of water it displaces, $V = 19.3\ mL=19.3\ cm^3$, and $m = 371\ g$. Using $\rho=\frac{m}{V}$, we have $\rho=\frac{371\ g}{19.3\ cm^3}= 19.2\ g\ cm^{-3}$
Step2: Compare with gold density
The density of gold is approximately $19.3\ g\ cm^{-3}$. Since $19.2\ g\ cm^{-3}$ is close to $19.3\ g\ cm^{-3}$, it could be gold.
Answer:
Yes
45. a.
Explanation:
Step1: Use density - mass - volume relation
Given $\rho = 1.11\ g\ cm^{-3}$, $V=417\ mL = 417\ cm^3$. Using $m=\rho V$, we get $m=1.11\ g\ cm^{-3}\times417\ cm^3 = 462.87\ g$
Answer:
$462.87\ g$
45. b.
Explanation:
Step1: Convert mass to grams
$m = 4.1\ kg=4100\ g$
Step2: Calculate volume
Using $V=\frac{m}{\rho}$, with $\rho = 1.11\ g\ cm^{-3}$ and $m = 4100\ g$, we have $V=\frac{4100\ g}{1.11\ g\ cm^{-3}}\approx3693.7\ cm^3$. Convert to liters: $V=\frac{3693.7\ cm^3}{1000}=3.69\ L$
Answer:
$3.69\ L$
46. a.
Explanation:
Step1: Use mass - density - volume formula
Given $\rho = 0.7857\ g\ cm^{-3}$, $V = 28.56\ mL=28.56\ cm^3$. Using $m=\rho V$, we get $m=0.7857\ g\ cm^{-3}\times28.56\ cm^3\approx22.44\ g$
Answer:
$22.44\ g$
46. b.
Explanation:
Step1: Use volume - density - mass formula
Using $V=\frac{m}{\rho}$, with $m = 6.54\ g$ and $\rho=0.7857\ g\ cm^{-3}$, we have $V=\frac{6.54\ g}{0.7857\ g\ cm^{-3}}\approx8.32\ cm^3 = 8.32\ mL$
Answer:
$8.32\ mL$
47.
Explanation:
Step1: Convert volume to $mL$
$V = 245\ L=245000\ mL$
Step2: Calculate mass
Using $m=\rho V$, with $\rho = 0.803\ g\ mL^{-1}$ and $V = 245000\ mL$, we get $m=0.803\ g\ mL^{-1}\times245000\ mL = 196735\ g=196.735\ kg$
Answer:
$196.735\ kg$
48.
Explanation:
Step1: Convert mass to grams
$m = 5.00\ kg=5000\ g$
Step2: Calculate volume
Using $V=\frac{m}{\rho}$, with $\rho = 0.918\ g\ cm^{-3}$ and $m = 5000\ g$, we have $V=\frac{5000\ g}{0.918\ g\ cm^{-3}}\approx5446.6\ cm^3$
Answer:
$5446.6\ cm^3$