The balanced equation is 2C2H6 + 7O2 → 4CO2 + 6H2O. Therefore, 4 moles of C2H6 require 14 moles of O2, which means 7 moles of O2 for 2 moles of C2H6.

The central carbon atoms in C2H4 are sp2 hybridized, forming a double bond between them.

Carbon in CO2 is sp hybridized, forming two double bonds with oxygen.

The nitrogen atom in NH3 is sp3 hybridized, forming three bonds and one lone pair.

The nitrogen atom in NH3 is sp3 hybridized, forming three N-H bonds and one lone pair.

Strong acid and strong base neutralize each other, resulting in a neutral solution with pH = 7.

Using the formula Ka = [H+]^2 / [HA], where [H+] = 10^(-4) M and [HA] = 0.1 M, we find Ka = (10^-4)^2 / 0.1 = 1 x 10^-5.

Vapor pressure lowering = (n_solute / n_solvent) * P°_solvent = (0.5 / 55.5) * 23.76 = 1.88 mmHg

At STP, 1 mole of gas occupies 22.4 L. Therefore, the molar volume is 22.4 L.

Concentration (M) = moles of solute / liters of solution = 0.5 moles / 1 L = 0.5 M.

Concentration (M) = moles of solute / liters of solution = 0.5 moles / 1 L = 0.5 M.

Molarity (M) = moles of solute / liters of solution = 0.5 moles / 1 L = 0.5 M.

Using the dilution formula M1V1 = M2V2, we have 2 M * 1 L = M2 * 3 L, thus M2 = 2/3 = 0.67 M.

Using the dilution formula M1V1 = M2V2, we have 3 M × 1 L = M2 × 2 L. Thus, M2 = 1.5 M.

Using the dilution formula M1V1 = M2V2, (2 M)(1 L) = M2(3 L) => M2 = 2/3 = 0.67 M.

NaCl dissociates into 2 ions (Na+ and Cl-), so 1 mol of NaCl produces 2 mol of particles in solution.

NaCl dissociates into 2 ions (Na+ and Cl-), so the van 't Hoff factor i = 2.

NaCl dissociates into 2 ions (Na+ and Cl-), so 1 mol of NaCl results in 2 mol of particles.

Volume = moles x volume per mole = 0.5 moles x 22.4 L/mole = 11.2 L.

Volume = moles x volume per mole = 0.5 moles x 22.4 L/mole = 11.2 L.

The freezing point depression is calculated using the formula ΔTf = Kf * m, where Kf for water is 1.86 °C kg/mol.

The freezing point depression is calculated using the formula ΔTf = i * Kf * m. For a non-electrolyte, i = 1, Kf for water = 1.86 °C kg/mol, and m = 1 mol/kg gives ΔTf = 1.86 °C.

The freezing point depression for 1 mole of a non-electrolyte solute in 1 kg of water is -1.86 °C.

Boiling point elevation = i * Kb * m = 1 * 0.512 * 1 = 0.512 °C.

NaCl dissociates into 2 ions (Na+ and Cl-), so the van 't Hoff factor (i) is 2.

Molarity (M) = moles of solute / liters of solution = 1 mole / 1 L = 1 M.

10 g of CaCO3 = 0.1 moles. CaCO3 → CaO + CO2, so 0.1 moles of CO2 = 0.1 * 44 g = 4.4 g.

10 g of Na = 0.43 moles. Na + Cl2 → NaCl, so 0.43 moles of NaCl = 0.43 * 58.5 g = 25.2 g.

The balanced equation is CaCO3 → CaO + CO2. The molar mass of CaCO3 is 100 g/mol and CaO is 56 g/mol. Thus, 10 g of CaCO3 produces (10 g / 100 g/mol) x 56 g/mol = 5.6 g of CaO.

Moles of NaCl = mass (g) / molar mass (g/mol) = 10 g / 58.5 g/mol = 0.171 moles.