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Mashcka [7]
2 years ago
5

What would happen to the position of the equilibrium when the following changes are made to the reaction below?

Chemistry
2 answers:
love history [14]2 years ago
6 0

Answer:

I don't know good luck

Explanation:

crimeas [40]2 years ago
6 0
Any impact on pressure will only affect gas molecules. So in this question the reactant has 0 gas molecules and the products have 1 gas molecule (the O2 molecule).

This means when you decrease the pressure the equilibrium will shift in order to oppose this change.

As you decreased the pressure it will move to the reaction that will increase the pressure back to equilibrium. So if you decrease pressure the equilibrium will shift to the right hand side (forward reaction) to oppose the change in pressure and a higher yield of Hg(l) and O2(g) will be produced.
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Answer:

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Two students (X and Y) measured the diameter of a gold coin four
Eddi Din [679]

Explanation:

To solve this problem, we need to understand the concept of precision and accuracy.

Through this understanding we apply to solve this problem.

  • Precision is the ability to reproduce the same set of values in an experiment.
  • Accuracy is the nearness or closeness of the measured value to the true value.

For example; if the true value is 28.0mm and the following readings were made:

        A  =  24.0mm, 24.0mm, 24.3mm, 23.9mm  This is a precise measurement but not accurate.

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8 0
3 years ago
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tiny-mole [99]

Answer:

K = [H2] [CO] / [HCHO]

Explanation:

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We can obtain the expression for the equilibrium constant for the above equation as follow:

Equilibrium constant, K for a given reaction is the ratio of the concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

Thus, the equilibrium constant, K for the above equation can be written as follow:

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