The integral is (5/5)x^5 + C = x^5 + C.

The integral is (6/3)x^3 - 4x + C = 2x^3 - 4x + C.

The integral of e^x is e^x + C.

For a monatomic ideal gas, γ = C_p/C_v = 5/3 = 1.67.

At constant volume, no work is done (W=0), and the heat added equals the change in internal energy (Q=ΔU).

Using t = 1 / (k[A₀]) * (1/[A] - 1/[A₀]), t = 1 / (0.02 * 0.5) * (1/0.25 - 1/0.5) = 25 s.

Using the Arrhenius equation, k = Ae^(-Ea/RT). Calculate k using the given values.

Increasing the temperature generally increases the rate of reaction significantly due to higher kinetic energy.

For a zero-order reaction, t = [A₀] / k. Here, t = 0.5 M / 0.1 M/s = 5 s.

For a zero-order reaction, t = [A₀ - A] / k. Here, t = (0.5 - 0.2) / 0.1 = 3 s.

In an isobaric process, the work done is calculated using the formula W = PΔV, where P is pressure and ΔV is the change in volume.

The Ideal Gas Law (PV = nRT) relates pressure (P), volume (V), and temperature (T) for an ideal gas.

The discriminant must be non-negative: 6² - 4*1*k ≥ 0, which gives k ≤ 9, so the minimum value is -9.

For an exothermic reaction, increasing the temperature shifts the equilibrium to the left, favoring the reactants, as the system tries to absorb the added heat.

The vertex x-coordinate is given by -b/(2a) = -4/(2*-1) = 2.

Set f'(x) = 0. f'(x) = 6x^2 - 18x + 12 = 0. Critical points are x = 2.

Find f'(x) = 6x^2 - 18x + 12, set to 0. Local maxima at x = 2 gives f(2) = 6.

The vertex occurs at x = 2. f(2) = 3(2^2) - 12(2) + 7 = -5.

The x-coordinate of the vertex is given by x = -b/(2a) = 12/(2*3) = 2.

The vertex occurs at x = -b/(2a) = 12/(2*3) = 2. f(2) = 3(2^2) - 12(2) + 9 = 3.

f'(x) = 2x + 2. Setting f'(x) = 0 gives x = -1. f(-1) = 1.

Using the determinant formula, det(E) = 1*(1*0 - 4*6) - 2*(0*0 - 4*5) + 3*(0*6 - 1*5) = 1*(-24) - 2*(-20) - 15 = -24 + 40 - 15 = 1.

Determinant of E = (1*4) - (2*2) = 4 - 4 = 0.

The discriminant must be non-negative: 6² - 4*1*k ≥ 0, thus k ≤ 9. The minimum value of k is -9.

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.

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

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 sum of the roots is 3 + 3 = 6, hence m = 6.

Using the sum of roots: 2 + (-2) = -p, hence p = 0.

For both roots to be negative, p must be greater than 0 and p² > 16. Thus, p < -4.