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choli [55]
3 years ago
6

A container holds 6.4 moles of gas. hydrogen gas makes up 25% of the total moles in the container. if the total pressure is 1.24

atm
Chemistry
2 answers:
klio [65]3 years ago
6 0
The total pressure of the mixture of gases is equal to the sum of the pressure of each gas as if it is alone in the container. The partial pressure of a component of the mixture is said to be equal to the product of the total pressure and the mole fraction of the component in the mixture.

Partial pressure of hydrogen gas = 1.24 atm x .25 = 0.31 atm
Partial pressure of the remaining = 1.24 atm x (1-.25) = 0.93 atm 
Vlad1618 [11]3 years ago
5 0

Answer:

0.31

Explanation:

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The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×
Anuta_ua [19.1K]

Answer:

Kc=~1.49x10^3^4}

Explanation:

We have the reactions:

A: N_2_(_g_) + O_2_(_g_)  2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9

Our <u>target reaction</u> is:

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)

We have NO_(_g_) as a reactive in the target reaction and  NO_(_g_) is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

Then if we add reactions A and B we can obtain the target reaction, so:

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9

For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}

Kc=~1.49x10^+^3^4}

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