Answer
(a) $0.50 \mathrm{mol} \mathrm{O}_{2}=16.00$ $\mathrm{g}$;
(b) By dividing over Avogadro's number, $2.0 \times 10^{23} \mathrm{Cu}$ atoms $=1 / 3$ $\mathrm{mol}$ $\mathrm{Cu} \approx 21$ $\mathrm{g}$;
(c) By dividing over Avogadro's number, $1.0 \times 10^{24} \mathrm{H}_{2} \mathrm{O}$ molecules $\approx$ $1.5$ $\mathrm{mol}$ $\mathrm{H}_{2} \mathrm{O}>27 \mathrm{g}$;
(e) The mass of $1$ $\mathrm{mol}$ $\mathrm{Ne}=20$ $\mathrm {g}$.
So, the greatest mass is (c) & the smallest is (a).
Work Step by Step
(a) $0.50 \mathrm{mol} \mathrm{O}_{2}=16.00$ $\mathrm{g}$;
(b) By dividing over Avogadro's number, $2.0 \times 10^{23} \mathrm{Cu}$ atoms $=1 / 3$ $\mathrm{mol}$ $\mathrm{Cu} \approx 21$ $\mathrm{g}$;
(c) By dividing over Avogadro's number, $1.0 \times 10^{24} \mathrm{H}_{2} \mathrm{O}$ molecules $\approx$ $1.5$ $\mathrm{mol}$ $\mathrm{H}_{2} \mathrm{O}>27 \mathrm{g}$;
(e) The mass of $1$ $\mathrm{mol}$ $\mathrm{Ne}=20$ $\mathrm {g}$.
So, the greatest mass is (c) & the smallest is (a).
