H3PO4 is a triprotic acid; for a dilute solution, the first dissociation dominates, giving pH ≈ 2.

Using the formula for weak acids, pH = 0.5(pKa - log[C]) where pKa = -log(1.8 x 10^-5) ≈ 4.74. pH = 0.5(4.74 - log(0.05)) ≈ 4.7.

NaCl is a neutral salt, so the pH remains 7.

NH4Cl is a salt of a weak base and strong acid, pH < 7. Calculate using hydrolysis: pH ≈ 4.75.

K2CO3 is a salt of a weak acid (H2CO3) and strong base (KOH), resulting in a basic solution with pH around 10.

K2SO4 is a neutral salt formed from a strong acid (H2SO4) and a strong base (KOH). Therefore, the pH of the solution is 7.

NH4Cl is an acidic salt, pH is approximately 5.1.

Using the formula for salt of weak acid, pH = 14 + 0.5(pKa + logC) = 14 + 0.5(4.74 + log(0.1)) = 9.25.

NaHCO3 is a weak base, pH ≈ 8.00 due to its bicarbonate ion acting as a weak base.

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 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 x = 2 and y = 3. Thus, the point of intersection is (2, 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.