Answer is: the hydronium ion concentratio is 1.71×10⁻⁷ mol/dm³ and pH<6.76.
The Kw (the ionization constant of water) at 40°C is 2.94×10⁻¹⁴ mol²/dm⁶ or 2.94×10⁻¹⁴ M².
Kw = [H₃O⁺] · [OH⁻].
[H₃O⁺] = [OH⁻] = x.
Kw = x².
x = √Kw.
x = √2.94×10⁻¹⁴ M².
x = [H₃O⁺] = 1.71×10⁻⁷ M; concentration of hydronium ion.
pH = -log[H₃O⁺].
pH = -log(1.71×10⁻⁷ M).
pH = 6.76.
pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity an aqueous solution.
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B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.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.
One chemical change would be baking the muffin.
One physical change would be chopping the fruits to add to the mixture