Answer:
1) The value of Kc:
C. remains the same.
2) The value of Qc:
A. is greater than Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium.
4) The concentration of N2 will:
B. decrease.
Explanation:
Hello,
In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:
1) The value of Kc:
C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:
2) The value of Qc:
A. is greater than Kc, since the reaction quotient is:
![Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.
4) The concentration of N2 will:
B. decrease, since less reactant is forming the products.
Best regards.
Answer:
See explanation and image attached
Explanation:
Fischer esterification is a type of reaction used to convert carboxylic acids to ester in the presence of excess alcohol and a strong acid which acts as a catalyst. Another final product formed in the reaction is water.
The mechanism for the fischer esterification of Benzoic acid and C H 3 O H in the presence of HCl as the catalyst is shown in the image attached to this answer.
The final products of the reaction are methyl benzoate, water and H^+ as shown in the image attached.
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.
Answer: Option (3) is the correct answer.
Explanation:
When there is a negative charge on an atom then we add the charge with the number of electrons. Whereas when there is a positive charge on an atom then we subtract the charge from the number of electrons.
Atomic number of chlorine is 17. So, number of electrons present in 
is 17 + 1 = 18 electrons.
Atomic number of cobalt is 27. So, number of electrons present in 
is 27 - 4 = 23 electrons.
Atomic number of iron is 26. So, number of electrons present in 
is 26 - 2 = 24 electrons.
Atomic number of vanadium is 23. So, number of electrons present in V is 23 electrons.
Atomic number of scandium is 21. So, number of electrons present in 
is 21 + 2 = 23 electrons.
Thus, we can conclude that out of the given species, has the greatest number of electrons.
Answer:
6.2moles of Gold
Explanation:
To solve this problem, we are going to use the mole concept approach.
Given that;
Number of atoms of gold is 3.73 x 10²⁴ atoms
Now;
In 1 mole of any substance, we have 6.02 x 10²³ atoms;
So;
If there 6.02 x 10²³ atoms in 1 mole of any substance;
3.73 x 10²⁴ atoms will contain 
= 6.2moles of Gold
