User:B2609/Sandbox

Determining grams of Mg: $$ 2.28 \pm 0.02 \mbox{ cm } \cdot \frac{0.016 \mbox{ g }}{1 \mbox{ cm }} = 0.0365 \pm 0.0003 \mbox{ g } \mbox{ of Mg } $$

Determining moles of H2: $$ \begin{align} 0.0365 \pm 0.82\% \mbox{ g } \cdot \frac{1 \mbox{ mol }}{24.3 \operatorname{g}} & = 0.88695 \pm 0.0003 \operatorname{mol} \mbox{ of Mg } \\ & = 0.88695 \pm 0.82\% \mbox{ mol H}_2 \end{align} $$

Corrected pressure of H2 gas: $$ 102.3 \pm 0.1 \mbox{ kPa } - 2.338 \mbox{ kPa } = 100.0 \pm 0.1 $$

Volume of collected gas at STP, where x = unknown volume: $$ \begin{align} \frac {p_1V_1}{T_1} & = \frac {p_2V_2}{T_2} \\ \frac {100.0 \pm 0.01 \mbox{ kPa } \cdot 0.0444 \pm 0.00002 \mbox{ L }}{293.3 \pm 0.02 \mbox{ K }} & = \frac {101.3 \mbox{ kPa } \cdot x \mbox{ L }}{273.15 \mbox{ K }} \\ x & = \frac {100.0 \pm 0.01 \mbox{ kPa } \cdot 0.0444 \pm 0.00002 \mbox{ L } \cdot 273.15 \mbox{ K }}{293.3 \pm 0.02 \mbox{ K } \cdot 101.3 \mbox{ kPa }} \\ x & = 0.040819 \pm 0.062\% \mbox{ L } \\ x & = 0.0408 \pm 0.062\% \mbox{ L } \end{align} $$

Molar volume of H2: $$ \begin{align} \frac{ 0.88695 \pm 0.82\% \mbox{ mol }}{ 0.0408 \pm 0.062\% \mbox{ L } } & = 21.73897 \pm 0.882\% \frac{\mbox{ mol }}{\mbox{ L }} \\ & = 21.73897 \pm 0.2 \frac{\mbox{ mol }}{\mbox{ L }} \\ & = 21.7 \pm 0.2 \frac{\mbox{ mol }}{\mbox{ L }} \end{align} $$