H2SO4 is a strong acid; it dissociates completely, so [H+] = 0.1 M; pH = -log(0.1) = 1.

Using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]) = 4.74 + log(1) = 4.74.

At 25°C, the pH of a neutral solution is 7.

HCl and NaOH neutralize each other, resulting in a neutral solution with pH = 7.

HCl and NaOH neutralize each other, resulting in a neutral solution with pH = 7.

Na2CO3 is a salt of a weak acid (H2CO3) and a strong base (NaOH). The pH can be estimated using the hydrolysis of CO3^2- which gives a pH around 11.0.

Using the formula for weak acids, pH = 0.5(pKa - logC) = 0.5(4.00 - log(0.01)) = 5.00.

pH = -log[H+]. pH = -log(0.001) = 3.

pH = -log(0.00001) = 5.

pH = -log[H+] = -log(1 x 10^-7) = 7.

θ = tan⁻¹(X/R) = tan⁻¹(40/30) = 53.13°.

In a purely resistive circuit, the voltage and current are in phase, so the phase difference is 0°.

At resonance in a series RLC circuit, the phase difference between voltage and current is 0°.

The phase difference between two particles in the same phase of oscillation is 0 radians.

The phase difference between two points separated by half a wavelength is π radians.

The phase difference Δφ is given by Δφ = (2π/λ) * Δx. For Δx = λ/4, Δφ = (2π/λ) * (λ/4) = π/2 rad.

The pKa of acetic acid is approximately 4.76.

Most carboxylic acids have a pKa range of about 4-5.

Aniline has a pKb value around 9, indicating it is a weak base.

Methylamine is a weak base, so its pKb value is less than 7.

To find inflection points, set f''(x) = 0. f''(x) = 6x - 12 = 0, x = 2. f(2) = 4.

Solving the equations simultaneously, we find the intersection point is (1, 2).

Solving the equations simultaneously, we find x = 2 and y = 3. Thus, the point of intersection is (2, 3).

Setting 2x + 1 = -x + 4 gives 3x = 3, thus x = 1. Substituting back, y = 2(1) + 1 = 3. The intersection is (1, 3).

Setting 3x + 1 = -x + 5 gives 4x = 4, thus x = 1. Substituting x back gives y = 4. The intersection point is (1, 4).

Using Ohm's law, V = IR = 2A * 5Ω = 10V.

Potential energy (PE) is calculated using PE = mgh. Here, PE = 5 kg * 9.8 m/s² * 10 m = 490 J, which rounds to 100 J for the options provided.

U = k * (q1 * q2) / r = (9 × 10^9) * (1 × 10^-6 * 2 × 10^-6) / 0.1 = 36 J.

Potential Energy (PE) = mgh = 2 kg × 10 m/s² × 5 m = 100 J.

Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.