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:
The Empirical Formular is given as; Ti₆Al₄V
Explanation:
The percent composition of the material is 64.39% titanium, 24.19% aluminum, and 11.42% vanadium.
Elements Titanium Aluminium Vanadium
Percentage 64.39 24.19 11.42
Divide all through by their molar mass
64.39 / 47.87 24.19 / 27 11.42 / 50.94
= 1.345 = 0.896 = 0.224
Divide all though by the smallest number (0.224)
1.345 / 0.224 0.896 / 0.224 0.224 / 0.224
= 6 = 4 = 1
The Empirical Formular is given as; Ti₆Al₄V
Answer:
Q1: 2. Molarity.
Q2: 2.51 x 10⁻³.
Explanation:
<u><em>Q1:</em></u>
- The molarity is defined as the no. of moles of solute in one liter of solution.
- It is can be calculated from the relation: M = (no. of moles of solute)/(Volume in L of the solution).
<u><em>Q2:</em></u>
∵ pH = - log[H₃O⁺]
∴ 2.60 = - log[H₃O⁺]
log[H₃O⁺] = -2.6.
<em>∴ [H₃O⁺] = 2.51 x 10⁻³.</em>