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3 years ago

9

A container is filled with helium gas. It has a volume of 2.25 liters and contains 9.00 moles of helium. How many moles of heliu

m could be held in a 1.85 liter container at the same temperature and pressure?

1 answer:

Darina [25.2K]3 years ago

7 0

Answer: There are 7.4 moles of helium gas present in a 1.85 liter container at the same temperature and pressure.

Explanation:

Given: $V_{1}$ = 2.25 L, $n_{1}$ = 9.0 mol

$V_{2}$ = 1.85 L, $n_{2}$ = ?

Formula used to calculate the moles of helium are as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\$

Substitute the values into above formula as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{2.25 L}{9.0 mol} = \frac{1.85 L}{n_{2}}\\n_{2} = \frac{1.85 L \times 9.0 mol}{2.25 L}\\= 7.4 mol$

Thus, we can conclude that there are 7.4 moles of helium gas present in a 1.85 liter container at the same temperature and pressure.

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Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

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What can be concluded about the atom from knowing that oxygen 18 has an atomic number of 8?

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Explanation:

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$\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}$

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p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C

p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C

Calculations:

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T₁ = ( 1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}$

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We are given:

$t_{1/2}=62.1s$

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$k=\frac{0.693}{62.1}=0.011s^{-1}$

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