The central atom in ozone (O3) is sp2 hybridized, forming a resonance structure.

Phosphorus in PCl5 is dsp3 hybridized, allowing it to form five bonds.

The equation factors to (x + 2)(x + 2) = 0, giving a double root x = -2.

For a first-order reaction, the half-life is independent of the initial concentration. Therefore, it remains 10 minutes.

For a first-order reaction, the half-life is given by the expression t1/2 = 0.693/k.

For a first-order reaction, the half-life is independent of the initial concentration.

A reversible process must be quasi-static, meaning it occurs infinitely slowly.

ΔG = ΔH - TΔS = 200 kJ - 400 K * 0.5 kJ/K = 200 kJ - 200 kJ = 0 kJ.

The rate of disappearance of A is equal to the rate of formation of B, multiplied by the stoichiometric coefficients. Here, it is 1.0 mol/L/s.

If doubling the concentration of A doubles the rate, the reaction is first order with respect to A.

The correct relationship at constant temperature and pressure is ΔG = ΔH - TΔS.

According to Le Chatelier's principle, if the concentration of products increases, the equilibrium will shift to the left.

At equilibrium, the change in Gibbs free energy (ΔG) is zero, indicating that the system is at maximum entropy.

If ΔG° is negative, the equilibrium constant K is greater than 1.

A positive ΔG° indicates that the reaction is non-spontaneous in the forward direction, thus spontaneous in the reverse.

A positive ΔG° indicates that the reaction is non-spontaneous in the forward direction under standard conditions.

A positive ΔG° indicates that the reaction is non-spontaneous in the forward direction.

For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant, given by Rate = k[A].

To find the temperature at which the reaction becomes spontaneous, set ΔG = 0: 0 = ΔH - TΔS. Thus, T = ΔH/ΔS = (100,000 J)/(200 J/K) = 500 K.

To find the temperature at which the reaction becomes spontaneous, set ΔG = 0: 0 = ΔH - TΔS, thus T = ΔH/ΔS = 100,000 J / 200 J/K = 500 K.

To find the temperature at which the reaction becomes spontaneous, set ΔG = 0: 0 = ΔH - TΔS, thus T = ΔH/ΔS = (50,000 J/mol) / (100 J/mol·K) = 500 K.

For a reversible process, the change in entropy is given by ΔS = Q/T, where Q is the heat exchanged and T is the temperature.

For a reversible process, the change in entropy (ΔS) is given by ΔS = Q/T, where Q is the heat absorbed and T is the temperature.

The change in entropy (ΔS) for a reversible process is given by ΔS = Q/T, where Q is the heat absorbed and T is the temperature.

For a reversible process, the change in entropy of the universe is zero, as the system and surroundings are in equilibrium.

The efficiency of a Carnot engine is given by η = 1 - (T2/T1), where T1 is the temperature of the hot reservoir and T2 is the temperature of the cold reservoir.

According to Raoult's Law, the vapor pressure of the solution is P_A = X_A * P_A^0 = 0.6 * 100 mmHg = 60 mmHg.

For a solution to obey Raoult's Law, the interactions between solute and solvent must be similar to those in the pure components.

For a spontaneous process, the total entropy change of the universe (system + surroundings) must be positive.

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