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3 years ago

Which represents the correct equilibrium constant expression for the reaction below?

Chemistry

2 answers:

charle [14.2K]3 years ago

6 0

Answer:

the answer is D

Explanation:3

ON EDG

Ad libitum [116K]3 years ago

4 0

Answer:

$\large\boxed{\large\boxed{K_c=\dfrac{[Cu^{2+}]}{[Ag^+]^2}}}$

Explanation:

1. Chemical equilibrium equation:

$Cu(s)+2Ag^+(aq)\rightleftharpoons Cu^{2+}+2Ag(s)$

2. Species

In an equilibrium constant expression you do not include the solid substances; only gases and dissolved substances.

The symbol $(s)$ means solid, thus Cu(s) and Ag(s) shall not appear in your equilibrium constant expression.

The symbol $(aq)$ means in aqueous solution, thus the both $Ag^+$ and $Cu^{2+}$ must appear in the equilibrium constant expression.

3. Equilibrium constant expression.

It is the quotient of the product of the concentrations of the species on the right hand side of the equilibrium equation, each raised to its corresponding coefficient, and the product of the concentrations of the species on the left hand side, each raised to its coresponding coefficient.

$K_c=\dfrac{[Cu^{2+}]}{[Ag^+]^2}$

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Answer:

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4 years ago

When potassium carbonate and calcium chromate are mixed, which of the following will pass through a filter?

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3 years ago

How many asymmetric centers are present in the open chain form of the aldohexose D-(-)-gulose?

bulgar [2K]

Answer:

D) 4

Explanation:

If we want to find the asymmetric centers, we must first remember that in an asymmetric center, in the carbon, we will have 4 different types of groups bonded to it.

Therefore we must look for carbons that have 4 different groups. With this in mind, we can look at each carbon.

Carbon 1

In this carbon, we have two bonds with the oxygen, therefore, we dont have 4 different groups. So, this is not an asymmetric center.

Carbon 2

In this carbon, we have an "OH" in the top, an hydrogen in the bottom a carbonyl group in the right, and a CHOH in the left. We have 4 different groups. So, this is an asymmetric center.

Carbon 3

In this carbon, we have an "OH" at the bottom, an hydrogen in the top, a CHOH group in the right (that is bonded to a carbonyl group), and a CHOH in the left. We have 4 different groups. So, this is an asymmetric center.

Carbon 4

In this carbon, we have an "OH" at the top, an hydrogen in the bottom a CHOH group in the right (that is bonded to a CHOH group), and a CHOH in the left. We have 4 different groups. So, this is an asymmetric center.

Carbon 5

In this carbon, we have an "OH" at the bottom, an hydrogen in the top, a CHOH group in the right (that is bonded to a CHOH), and a CH2OH in the left. We have 4 different groups. So, this is an asymmetric center.

Carbon 6

In this carbon, we have two bonds with hydrogens, therefore, we dont have 4 different groups. So, this is not an asymmetric center.

See figure 1

I hope it helps!