The correct relationship is ΔG = ΔH - TΔS, where ΔG must be negative for a spontaneous process.

For a process to be spontaneous, the change in Gibbs free energy (ΔG) must be negative.

In a zero-order reaction, the rate is constant and does not depend on the concentration of reactants.

In a zero-order reaction, the rate is constant and does not depend on the concentration of the reactants.

For d orbitals, the azimuthal quantum number l = 2.

For l=2 (d orbital), m_l can take values from -l to +l, which are -2, -1, 0, 1, 2.

For a p orbital, l=1, so m_l can take values -1, 0, +1.

The spin quantum number (m_s) can take values of -1/2 and +1/2.

For a p orbital (l=1), m_l can take values -1, 0, +1.

For l=2 (d orbital), m_l can take values from -2 to +2, which are -2, -1, 0, 1, 2.

According to Raoult's Law, the vapor pressure of the solution is 0.75 * 100 mmHg = 75 mmHg.

Substituting x = 1 into the equation gives 1^2 + 2*1 + k = 0 => 1 + 2 + k = 0 => k = -3.

The discriminant must be non-negative: k^2 - 4*1*9 >= 0 => k^2 >= 36 => k <= -6 or k >= 6.

There are 2 moles of gas on the reactant side (2A).

According to Le Chatelier's principle, increasing the concentration of B will shift the equilibrium to the left, resulting in a decrease in the concentration of A.

Decreasing the volume increases the pressure, and the equilibrium will shift towards the side with fewer moles of gas, which is the right side in this case.

Doubling the volume decreases the pressure, which shifts the equilibrium towards the side with more moles of gas (B and C).

Doubling the volume decreases the pressure, which shifts the equilibrium towards the side with more moles of gas, in this case, the right side.

Decreasing the temperature for an endothermic reaction will shift the equilibrium to the left, favoring the formation of HCl.

Decreasing the temperature for an exothermic reaction will shift the equilibrium to the right, favoring the formation of products.

Adding H2 will shift the equilibrium to the right to produce more NH3.

Increasing the pressure will shift the equilibrium to the right, favoring the formation of NH3, which has fewer moles of gas.

Decreasing the volume increases the pressure, which shifts the equilibrium to the side with fewer moles of gas, favoring the production of NH3.

Adding more I2 increases its concentration, which shifts the equilibrium to the right to produce more HI.

Adding more HI increases the concentration of products, which shifts the equilibrium to the left to form more reactants.

For l=1, m_l can take values -1, 0, 1, which gives 3 possible values.

At equilibrium, 0.5 - 0.2 = 0.3 M of A remains.

At equilibrium, [A] = 0.5 - 0.2 = 0.3 M.

Increasing the concentration of a product shifts the equilibrium to the left to counteract the change.

Using ICE table and Kc expression, we find [NO2] at equilibrium to be 0.2 M.