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luda_lava [24]
3 years ago
11

A container of hydrogen at 172 kpa was decreased to 85.0 kpa producing a new volume of 765 mL. What was the original volume

Chemistry
1 answer:
labwork [276]3 years ago
6 0

Answer:

\boxed{\text{378 mL}}

Explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

p_{1}V_{1} = p_{2}V_{2}

Data:

\begin{array}{rclrcl}p_{1}& =& \text{172 kPa}\qquad & V_{1} &= & \text{?} \\p_{2}& =& \text{85.0 kPa}\qquad & V_{2} &= & \text{765 mL}\\\end{array}

Calculations:

\begin{array}{rcl}172V_{1} & =& 85.0 \times 765\\172V_{1} & = & 65 025\\V_{1} & = &\textbf{378 mL}\\\end{array}\\\text{The original volume was } \boxed{\textbf{378 mL}}

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