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2 years ago

A 46.2 L sample of gas exerts 237.4 mm Hg pressure at 139.2 ºC. What volume does the gas have at 702.4 mm Hg and 63.4 ºC?

Chemistry

1 answer:

zvonat [6]2 years ago

5 0

7.111 L is the volume of gas with 702.4 mm Hg and 63.4 ºC.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given data:

$V_1=46.2 L$

$P_1=237.4 mm Hg$

$T_1=139.2 ºC$

$V_2=?$

$P_2=702.4 mm Hg$

$T_2=63.4 ºC$

Using the given below formula:

$\frac{V_1 X P_1}{T_1} = \frac{V_1 X P_1}{T_2}$

$\frac{46.2 L X 237.4 \;mm Hg}{139.2} = \frac{V_2 X 702.4 mm Hg}{63.4 ºC}$

$V_2$ = 7.111 L

Hence, 7.111 L is the volume of gas with 702.4 mm Hg and 63.4 ºC.

Learn more about the ideal gas here:

brainly.com/question/27691721

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4 years ago

Read 2 more answers

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$pK_{b}=4.82$

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The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.805\ g}{82.0343\ g/mol}$

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B + H₂O ⇄ BH⁺ + OH⁻

At t=0 0.4 - -

At t =equilibrium (0.4-x) x x

The expression for dissociation constant is:

$K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}$

$1.5136\times 10^{-5}=\frac {x^2}{0.4-x}$

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³ M

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3 years ago

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Explanation:

Hello there!

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