33. a volume of 5.00 ml of mercury is added to a beaker that has a mass of 87.3 g. what is the mass of the…
33. a volume of 5.00 ml of mercury is added to a beaker that has a mass of 87.3 g. what is the mass of the beaker with the added mercury (mercury density = 13.6 g/cm³)?\nknown: v = 5.00 ml\nunknown: m=\frac{v}{d}\nshow work:\nanswer with sig fig:\n\n40. the density of dry air measured at 25°c is 1.19 x 10⁻³ g/cm³. what is the volume of 50.0 g of air?\nknown:\nunknown:\nshow work:\nanswer with sig fig:\n\n43 b. convert 42.7 l to milliliters\nshow work: 42.7l×\frac{0.001 ml}{1 l}\nanswer: 0.0427 ml
Answer
33.
Explanation:
Step1: Convert volume units
Since $1\ mL = 1\ cm^{3}$, a volume of $V = 5.00\ mL$ is also $V = 5.00\ cm^{3}$.
Step2: Use density - mass formula
The density formula is $D=\frac{m}{V}$, so $m = D\times V$. Given $D = 13.6\ g/cm^{3}$ and $V=5.00\ cm^{3}$, then $m=13.6\ g/cm^{3}\times5.00\ cm^{3}=68.0\ g$.
Step3: Find total mass
The mass of the beaker is $m_{beaker}=87.3\ g$. The total mass $m_{total}=m_{beaker}+m = 87.3\ g+68.0\ g = 155.3\ g$.
Answer:
$155.3\ g$
40.
Explanation:
Step1: Identify knowns and unknowns
Known: $D = 1.19\times10^{-3}\ g/cm^{3}$, $m = 50.0\ g$. Unknown: $V$.
Step2: Rearrange density formula
From $D=\frac{m}{V}$, we can solve for $V$ as $V=\frac{m}{D}$.
Step3: Calculate volume
Substitute $m = 50.0\ g$ and $D = 1.19\times10^{-3}\ g/cm^{3}$ into the formula: $V=\frac{50.0\ g}{1.19\times10^{-3}\ g/cm^{3}}\approx4.20\times10^{4}\ cm^{3}$.
Answer:
$4.20\times 10^{4}\ cm^{3}$
43
Explanation:
Step1: Use conversion factor
The conversion factor between liters and milliliters is $1\ L=1000\ mL$ or $\frac{1000\ mL}{1\ L}$.
Step2: Convert volume
To convert $42.7\ L$ to milliliters, we multiply by the conversion factor: $V = 42.7\ L\times\frac{1000\ mL}{1\ L}=42700\ mL$.
Answer:
$42700\ mL$