I think it's 4 but I am not even 50% sure... But I think that is the answer..
Answer:
A)
.
B)
.
C) 0.9 mol.
D) Increasing both temperature and pressure.
Explanation:
Hello,
In this case, given the information, we proceed as follows:
A)

B) For the calculation of Kc, we rate the equilibrium expression:
![Kc=\frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
Next, since at equilibrium the concentration of ammonia is 0.6 M (0.9 mol in 1.5 dm³ or L), in terms of the reaction extent
, we have:
![[NH_3]=0.6M=2*x](https://tex.z-dn.net/?f=%5BNH_3%5D%3D0.6M%3D2%2Ax)

Next, the concentrations of nitrogen and hydrogen at equilibrium are:
![[N_2]=\frac{1.5mol}{1.5L}-x=1M-0.3M=0.7M](https://tex.z-dn.net/?f=%5BN_2%5D%3D%5Cfrac%7B1.5mol%7D%7B1.5L%7D-x%3D1M-0.3M%3D0.7M)
![[H_2]=\frac{4mol}{1.5L}-3*x=2.67M-0.9M=1.77M](https://tex.z-dn.net/?f=%5BH_2%5D%3D%5Cfrac%7B4mol%7D%7B1.5L%7D-3%2Ax%3D2.67M-0.9M%3D1.77M)
Therefore, the equilibrium constant is:

C) In this case, the equilibrium yield of ammonia is clearly 0.9 mol since is the yielded amount once equilibrium is established.
D) Here, since the reaction is endothermic (positive enthalpy change), one way to increase the yield of ammonia is increasing the temperature since heat is reactant for endothermic reactions. Moreover, since this reaction has less moles at the products, another way to increase the yield is increasing the pressure since when pressure is increased the side with fewer moles is favored.
Best regards.
Answer: Option (A) is the correct answer.
Explanation:
An element is defined as a substance that contains atoms of same type.
As copper wire is a substance that is made up of only atoms of copper. And, all these atoms of copper will be the same. Hence, it is an element.
This means that all the atoms of copper will have same number of protons, neutrons and electrons.
Thus, we can conclude that the statement they are all exactly the same best describes the copper atoms.
Answer:
Back titration is a titration done in reverse; instead of titrating the original sample, a known excess of standard reagent is added to the solution, and the excess is titrated.
I hope it's helpful!