Question

1. 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)?

known: v = 5.00 ml
unknown: m=\frac{v}{d}
show work:
answer with sig fig:

1. 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?

known:
unknown:
show work:
answer with sig fig:

43 b. convert 42.7 l to milliliters
show work: 42.7l×\frac{0.001 ml}{1 l}
answer: 0.0427 ml

Explanation:

33.

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 $m_{total}=m_{beaker}+m = 87.3\ g+68.0\ g = 155.3\ g$.

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: $V=\frac{50.0\ g}{1.19\times10^{-3}\ g/cm^{3}}\approx4.20\times10^{4}\ cm^{3}$.

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:

$155.3\ g$

40.