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ira [324]
3 years ago
8

Electron configuration of vanadium-52

Chemistry
1 answer:
Aneli [31]3 years ago
7 0

Answer:

[Ar]4s23d3

Explanation:

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Suppose that in the citric acid cycle, oxalacetate is labeled with 14C in the carboxyl carbon farthest from the keto group. Wher
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Answer:

All of alpha-ketoglutarate formed in the citric acid cycle would contain the radioactive ^{14}C while none of succinate would contain ^{14}C, and all of carbon dioxide released would contain ^{14}C.

Explanation:

When oxaloacetate in the citric acid cycle is labeled with ^{14}C in carboxyl carbon atom which is farthest from keto group, alpha ketoglutarate formed from this oxaloactetate has the full radioactive label. In the next step, the carboxylic group (that contains the ^{14}C) is eliminated in the form of the release of the carbon dioxide and succinate is formed. Succinate thus will not have radioactivity. CO_2 released had all the radioactivity.

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4 years ago
What is the reactant that runs out first in a reaction called?
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limiting reactant

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HELP ASAP: Determine the enthalpy change of the following reaction: CH4 + 4Cl2 -> CCl4+ 4HCl 
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Answer:

A

Explanation:

In this question, we are to calculate the enthalpy of change of the reaction. ΔH

To be able to do that, we use the Hess’ law and it involves the subtraction of the summed heat reaction of the reactants from that of the product.

Thus, mathematically, the enthalpy of change of the reaction would be;

[ΔH(CCl4) + 4 ΔH(HCl)] - [ΔH(CH4) + 4 ΔH(Cl2)]

We can see that we multiplied some heat change by some numbers. This is corresponding to the number of moles of that compound in question in the reaction.

Also, for diatomic gases such as chlorine in the reaction above, the heat of reaction is zero.

Thus, we can have the modified equation as follows;

[ΔH(CCl4) + 4 ΔH(HCl)] - [ΔH(CH4)]

Substituting the values we have according to the question, we have;

-95.98 + 4(-92.3) -(-17.9)

= -95.98 - 369.2 + 17.9

= -447.28 KJ/mol

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