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Delvig [45]
3 years ago
10

What is the pressure of a mixture of nitrogen (N2) and oxygen (O2) if partial pressure of N2 is 594 mm Hg and the partial pressu

re of O2 is 165 mm Hg?
Chemistry
1 answer:
ipn [44]3 years ago
5 0

Answer:

Total pressure = 759 mmHg

Explanation:

Given data:

Partial pressure of oxygen = 165 mmHg

Partial pressure of nitrogen = 594 mmHg

Total pressure = ?

Solution:

According to Dalton law of partial pressure,

The total pressure inside container is equal to the sum of partial pressures of individual gases present in container.

Mathematical expression:

P(total) = P₁ + P₂ + P₃+ ............+Pₙ

Total pressure = P(O₂) +  P(N₂)

Total pressure =  165 mmHg + 594 mmHg

Total pressure = 759 mmHg

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Consider 4.00 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.80 L and the temperature is increased to 3
stealth61 [152]

This is an exercise in<u> the General Combined Gas Law</u>.

To start solving this exercise, we obtain the following data:

<h3>Data:</h3>
  • V₁ = 4.00 l
  • P₁ = 365 mmHg
  • T₁ = 20 °C + 273 = 293 K
  • V₂ = 2,80 l
  • T₂ = 30 °C + 273 = 303 K
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We apply the following formula:

  • P₁V₁T₂=P₂V₂T₁     ⇒  General formula

Where:

  • P₁=Initial pressure
  • V₁=Initial volume
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We clear for final pressure (P2)

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{P_{1}V_{1}T_{2}}{V_{2}T_{1}} \ \ \to \ \ \ Formula \end{gathered}$}

We substitute our data into the formula:

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{(365 \ mmHg)(4.00 \not{l})(303 \not{K})}{(2.80 \not{l})(293\not{K})}  \end{gathered}$}

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{442380 \ mmHg}{ 820.4 }  \end{gathered}$}

\boxed{\large\displaystyle\text{$\begin{gathered}\sf P_{2}=539.224 \ mmHg \end{gathered}$}}

Answer: The new canister pressure is 539.224 mmHg.

<h2>{ Pisces04 }</h2>
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The density of carbon tetracoride is 1.58 g/mL. What is the mass of 95.7 mL of carbon tetrachloride?
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Explanation:

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