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-Dominant- [34]
3 years ago
9

For the reaction CO+2H2=CH3OH at 700 K, equilibrium concentrations are [H2]=0.072 M, [CO]= 0.020M, and [CH3OH]= 0.030 M. Calcula

te K
Chemistry
1 answer:
bekas [8.4K]3 years ago
3 0

The balanced equation for the reaction is

CO(g) + 2H₂(g) ⇄ CH₃OH(g)

 

The given concentrations are at equilibrium state. Hence we can use them directly in calculation with the expression for the equilibrium constant, k. expression for k can be written as

   k = [CH₃OH(g)] / [CO(g)] [H₂<span>(g) ]²

</span>[H₂<span>]=0.072 M
[CO]= 0.020M
[CH</span>₃OH]= 0.030 M

 

From substitution,

   k = 0.030 M / 0.020 M x (0.072 M)²

   k = 289.35 M⁻²

<span>
Hence, equilibrium constant for the given reaction at 700 K is 289.35 M</span>⁻².

<span> </span>

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Please explain how to find each answer solution.
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Explanation:

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Option E is correct ✔

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₁₆S = 1s² 2s² 2p⁶ 3s² 3p⁴ or [Ne] 3s² 3p⁴

₁₈S²⁻ = 1s² 2s² 2p⁶ 3s² 3p⁶ or [Ne] 3s² 3p⁶

Therefore,

Option E is correct ✔

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Therefore,

Option D is correct ✔

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Option B is correct ✔

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And here Group 13 element is Al

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Option B is correct ✔

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3 years ago
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Answer:

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MnSO₄ <------> Mn²⁺ + SO₄²⁻

Therefore, all we have to do is calculate the moles of Mn in both solutions, do the sum and then, calculate the concentration with the new volume:

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moles MnSulf = 0.339 * 0.0125 = 0.0042 moles

the total moles are:

moles of Mn²⁺ = 0.0138 + 0.0042 = 0.018 moles

Finally the concentration: 12.5 + 28.3 = 40.8 mL or 0.0408 L

M = 0.018 / 0.0408

M = 0.441 M

This would be the final concentration of the manganese after the mixing of the two solutions

7 0
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