Question

a lead ball has a mass of 75.0 g and a density of 11.4 g/cm³. what is the volume of the ball?
855 cm³
0.152 cm³
6.58 cm³
4.82 cm³
question 17 (0.0862 points)
what is the mass percent of oxygen in water?
33.3%
5.60%
88.8%
11.2%

Explanation:

First Question (Volume of Lead Ball)

Step1: Recall density formula

The formula for density is $
ho = \frac{m}{V}$, where $
ho$ is density, $m$ is mass, and $V$ is volume. We need to solve for $V$, so rearrange the formula to $V=\frac{m}{
ho}$.

Step2: Substitute values

Given $m = 75.0\ \text{g}$ and $
ho=11.4\ \text{g/cm}^3$. Substitute into the formula: $V=\frac{75.0\ \text{g}}{11.4\ \text{g/cm}^3}$.

Step3: Calculate the volume

Perform the division: $\frac{75.0}{11.4}\approx 6.58\ \text{cm}^3$.

Step1: Determine water's formula and molar masses

The formula for water is $\ce{H_2O}$. The molar mass of $\ce{H}$ is approximately $1.008\ \text{g/mol}$, and the molar mass of $\ce{O}$ is $16.00\ \text{g/mol}$. The molar mass of $\ce{H_2O}$ is $2\times1.008 + 16.00=18.016\ \text{g/mol}$.

Step2: Calculate mass of oxygen in one mole of water

In one mole of $\ce{H_2O}$, the mass of $\ce{O}$ is $16.00\ \text{g}$ (since there is 1 mole of $\ce{O}$ per mole of $\ce{H_2O}$).

Step3: Calculate mass percent

Mass percent of $\ce{O}$ is $\frac{\text{mass of O in compound}}{\text{molar mass of compound}}\times 100\%=\frac{16.00\ \text{g/mol}}{18.016\ \text{g/mol}}\times 100\%\approx 88.8\%$.

Answer:

6.58 cm³ (the option with 6.58 cm³)

Second Question (Mass Percent of Oxygen in Water)