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B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.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:
pOH is 2.2
Explanation:
The pH scale goes from 0 to 14. As such, the pH and the pOH add up to 14. Subtract your pH from 14 to get your answer.
14 - 11.8 = 2.2
Answer:
10 Litre
Explanation:
Given that ::
v1 = 25L ; n1 = 1.5 mole ; v2 =? ; n2 = (1.5-0.9) = 0.6 mole
Using the relation :
(n2 * v1) / n1 = (n2 * v2) / n2
v2 = (n2 * v1) / n1
v2 = (0.6 mole * 25 Litre) / 1.5 mole
v2 = 15 / 1.5 litre
v2 = 10 Litre
Answer:

Explanation:
Data:
50/50 ethylene glycol (EG):water
V = 4.70 gal
ρ(EG) = 1.11 g/mL
ρ(water) = 0.988 g/mL
Calculations:
The formula for the boiling point elevation ΔTb is

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.
1. Moles of EG

2. Kilograms of water

3. Molal concentration of EG

4. Increase in boiling point

5. Boiling point
Answer:
the atomic mass number for boron is 5
Explanation:
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