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3 years ago

If the volume and moles of a container are held constant, the temperature of a gas is inversely proportional to the pressure.

Chemistry

1 answer:

noname [10]3 years ago

4 0

Answer:

FALSE

Explanation:

Assuming that the gas is ideal

Therefore the gas obeys the ideal gas equation

<h3>Ideal gas equation is </h3><h3>P × V = n × R × T</h3>

where

P is the pressure exerted by the gas

V is the volume occupied by the gas

n is the number of moles of the gas

R is the ideal gas constant

T is the temperature of the gas

Here volume of the gas will be the volume of the container

Given the volume of the container and number of moles of the gas are constant

As R will also be constant, the pressure of the gas will be directly proportional to the temperature of the gas

P ∝ T

∴ Pressure will be directly proportional to the temperature

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For the reaction

sp2606 [1]

Answer:

$n_{H_2}^{equilibrium}=0.393mol$

Explanation:

Hello,

In this case, given the amounts of water and carbon dioxide we should invert the given reaction as hydrogen will be producted rather than consumed:

$H_2O(g) + CO(g)\rightleftharpoons H_2(g) + CO_2(g)$

Consequently, the equilibrium constant is also inverted:

$Kc'=\frac{1}{Kc}=\frac{1}{0.534} =1.87$

In such a way, we can now propose the law of mass action:

$Kc'=\frac{[H_2][CO_2]}{[H_2O][CO]}$

And we can express it in terms of the initial concentrations of the reactants and the change $x$ due to the reaction extent:

$Kc'=\frac{(x)(x)}{([H_2O]_0-x)([CO]_0-x)}=1.87$

Thus, we compute the initial concentration which are same, since equal amount of moles are given:

$[H_2O]_0=[CO]_0=\frac{0.680mol}{70.0L}=0.0097M$

Hence, solving for $x$ by using the quardratic equation or solver, we obtain:

$x_1=0.00561M\\x_2=0.0361M$

For which the correct value is 0.00561M since the other one will produce negative concentrations of water and carbon monoxide at equilibrium. Therefore, the number of moles of hydrogen at equilibrium for the same 70.0-L container turn out:

$n_{H_2}^{equilibrium}=0.00561mol/L*70.0L=0.393mol$

Best regards.

3 0

3 years ago

What is the pOH of a 7.9 x 10(-4)m OH- solution?

aleksley [76]

POH is defined as -log([OH-])

pOH = -log([OH-]) = -log(7.9 x 10^-4) = 3.1

Hope that helps! :)

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Lostsunrise [7]

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