Kp is expressed in terms of partial pressures, so for the reaction, Kp = (P_NH3)^2 / (P_N2)(P_H2)^3.

Kp is expressed in terms of the partial pressures of the gases involved in the reaction. For this reaction, Kp = (P_NH3)^2 / (P_N2)(P_H2)^3.

Decreasing the volume increases the pressure, and the equilibrium will shift towards the side with fewer moles of gas. Here, it shifts to the right, producing more C.

The equilibrium constant Kc is given by the expression Kc = [C]^3 / ([A]^2[B]), where the concentrations are raised to the power of their coefficients.

Decreasing the volume increases the pressure, and according to Le Chatelier's principle, the equilibrium will shift towards the side with fewer moles of gas, which is the right side in this case.

For an exothermic reaction, increasing the temperature shifts the equilibrium to the left, favoring the reactants.

Removing SO3 will shift the equilibrium to the right to produce more SO3, according to Le Chatelier's principle.

Increasing the volume decreases the pressure, and according to Le Chatelier's principle, the equilibrium will shift to the side with more moles of gas, which is the left side in this case.

The equilibrium constant Kp is expressed as Kp = (P_B^2 * P_C) / (P_A^3).

Decreasing the volume increases the pressure. According to Le Chatelier's principle, the equilibrium will shift towards the side with fewer moles of gas. If there are more moles of gas on the left side, the equilibrium will shift to the left.

According to Le Chatelier's principle, increasing the concentration of a reactant will shift the equilibrium to the right.

If 1 is included, we can choose from the remaining elements {2, 3, 4}, which has 2^3 = 8 subsets.

The number of ways to choose 2 elements from 4 is given by the combination formula C(4,2) = 6.

The subsets with exactly 2 elements are {a, b}, {a, c}, and {b, c}. Total = 3.

A · B = 1*0 + 0*1 + 0*0 = 0.

A · B = 2*4 + 3*5 = 8 + 15 = 23.

A · B = 3*1 + (-2)*4 + 1*(-2) = 3 - 8 - 2 = -7.

Setting 1^3 - 3(1) + b = 2(1) + 1 gives b = 2.

F2 has a bond order of 1, while C2 has a bond order of 2.

He2 has a bond order of 0, as it has equal bonding and antibonding electrons.

The molecular orbital diagram of F2 is similar to that of O2, with the same energy level arrangement.

O2 and F2 both have a bond order of 2.

To ensure differentiability at x = -1, we find f'(-1) exists. Setting a = 0 ensures the derivative is defined.

The function is a polynomial and is differentiable for all real numbers, hence any value of a works.

Setting the derivative f'(1) = 0 gives a = 1 for differentiability.

f(x) is a polynomial and is differentiable for all a, hence any value of a works.

Setting f'(1) = 0 gives a = 1, ensuring differentiability at that point.

Setting f'(1) = 0 gives a = 1 for differentiability at x = 1.

Setting a = 1 gives continuity at x = 0.

Setting 3(2) + a = 4(2) - 1 gives a = 1.