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ad-work [718]
3 years ago
8

Consider the following reaction, equilibrium concentrations, and equilibrium constant at

Chemistry
1 answer:
REY [17]3 years ago
5 0

Answer:

Equilibrium concentration of H_{2}O is 12.5 M

Explanation:

Given reaction: C_{2}H_{4}+H_{2}O\rightleftharpoons C_{2}H_{5}OH

Here, K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}

where K_{c} represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations

Here, [C_{2}H_{4}]=0.015M , [C_{2}H_{5}OH]=1.69M and K_{c}=9.0

So, [H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M

Hence equilibrium concentration of H_{2}O is 12.5 M

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In reaction 1 of the Krebs cycle, acetyl‑CoA formed in the pyruvate dehydrogenase reaction condenses with the four‑carbon compound to form <em>citrate </em>with the elimination of coenzyme A. Since the product has three carboxyl groups, this pathway is referred to as the cycle. In reaction 2 of the Krebs cycle, this product then undergoes to form<em> isocitrate. </em>The enzyme is called aconitase because the compound cis‑aconitate is the <em>intermediate product</em> of the reaction. Reaction 3 eliminates CO2 to form the five‑carbon dicarboxylic acid <em>α-cetoglutarate. </em>Oxidation also occurs, with electrons transferred from the substrate to <em>COO-</em> . Consequently, this reaction is an oxidative decarboxylation.

In the image, you can see the reaction 2 in Krebs cycle is a two steps reaction with an intermediate cis-aconitase and a product called isocitrate.

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3 years ago
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B
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3 years ago
Periods on the periodic table are
Roman55 [17]

Answer:

Periods on the periodic table of elements

Explanation:

In the periodic table of elements, there are seven horizontal rows of elements called periods. The vertical columns of elements are called groups, or families. (See also The Periodic Table: Metals, Nonmetals, and Metalloids.) In each period (horizontal row), the atomic numbers increase from left to right.

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Why can water molecules store energy in more ways than neon atoms?
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8 0
2 years ago
Discribe how to prepare 50.00mL of a 3.00M hydrochloric acid solution using 12.0 M HCl.
Leokris [45]

Answer:

The answer to your question is given below.

Explanation:

To prepare 50mL of 3M HCl, we must calculate the volume of the stock solution needed. This can obtained as follow:

Molarity of stock solution (M1) = 12M

Volume of stock solution needed (V1) =?

Molarity of diluted solution (M2) = 3M

Volume of diluted solution (V2) = 50mL

The volume of the stock solution needed can be obtained by using the dilution formula as shown below:

M1V1 = M2V2

12 x V1 = 3 x 50

Divide both side by 12

V1 = (3 x 50)/12

V1 = 12.5mL

The volume of the stock solution needed is 12.5mL

Therefore, to prepare 50mL of 3M HCl, we must measure 12.5mL of the stock solution i.e 12M HCl and then, add water to the mark in a 1L volumetric flask. Now we can measure out 50mL of the solution.

6 0
3 years ago
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