Question

1. to calculate the volume of collected gas at stp, i used combined gas law.

\\( \frac { p _ { 1 } v _ { 1 } } { t _ { 1 } } = \frac { p _ { 2 } v _ { 2 } } { t _ { 2 } } \\)
\\( v _ { 2 } = \frac { p _ { 1 } v _ { 1 } t _ { 2 } } { p _ { 2 } t _ { 1 } } \\)
\\( v _ { 2 } = \frac { 100.13 k p a \times 53.0 m l \times 273 k } { 102.5 k p a \times 294.6 k } \\)
\\( v _ { 2 } \approx \\)

Explanation:

Step1: 代入公式計算

$$V_{2}=\frac{P_{1}V_{1}T_{2}}{P_{2}T_{1}}=\frac{100.13kPa\times53.0mL\times273K}{102.5kPa\times294.6K}$$

Step2: 計算數值

先計算分子：\(100.13\times53.0\times273 = 100.13\times14469 = 1448770.97\)
再計算分母：\(102.5\times294.6= 102.5\times(300 - 5.4)=102.5\times300-102.5\times5.4 = 30750 - 553.5 = 30196.5\)
則 \(V_{2}=\frac{1448770.97}{30196.5}\approx48.0mL\)

Answer:

\(V_{2}\approx48.0mL\)