Sodium has an atomic number of 11, and its electron configuration is 1s² 2s² 2p⁶ 3s¹. The outermost electrons are in n=3.

Potassium (K) has an atomic number of 19, and its electron configuration ends in 4s, so the outermost electron has n=4.

Potassium has the electronic configuration [Ar] 4s1, so the outermost electron is in n=4.

The ground state of hydrogen corresponds to n=1.

Potassium has an atomic number of 19, and its outermost electrons are in the n=4 shell.

Sodium has an atomic number of 11, and its outermost electrons are in the n=3 shell.

The universal gas constant R is 0.0821 L·atm/(K·mol).

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

The van 't Hoff factor (i) is equal to the number of particles the solute dissociates into. For a strong electrolyte that dissociates into 3 ions, i = 3.

The van 't Hoff factor (i) for glucose is 1, as it does not dissociate in solution.

For a non-electrolyte solute, the van 't Hoff factor (i) is 1, as it does not dissociate into ions.

Vapor pressure = P°_solvent * (n_solvent / (n_solvent + n_solute)) = 23.76 * (55.5 / (55.5 + 0.1)) = 22.88 mmHg

Using Raoult's law, the vapor pressure of the solution = (moles of solvent / total moles) * P0 = (10 / 11) * P0 = 0.909 P0.

Vapor pressure lowering = (moles of solute / total moles) * P0 = (1 / (1 + 3)) * P0 = 0.25 P0, so vapor pressure = P0 - 0.25 P0 = 0.75 P0.

Using Raoult's law, P_solution = X_solvent * P°_solvent = (9/10) * 100 mmHg = 90 mmHg.

Using Raoult's law, P_solution = X_solvent * P°_solvent. X_solvent = 3/(1+3) = 0.75, so P_solution = 0.75 * 100 mmHg = 75 mmHg.

Using Raoult's law, the vapor pressure of the solution = (moles of solvent / total moles) * P0 = (3 / (3 + 1)) * P0 = 0.75 P0.

The vapor pressure of a solution containing a non-volatile solute is lower than that of the pure solvent due to the presence of solute particles.

At STP, 1 mole of an ideal gas occupies 22.4 liters.

At STP, 1 mole of an ideal gas occupies 22.4 liters.

At STP, 1 mole of gas occupies 22.4 L. Therefore, 4 moles occupy 4 x 22.4 L = 89.6 L.

Using the formula M = moles/volume, Volume = moles/M = 0.5 moles / 1 M = 0.5 L.

At STP, 1 mole of an ideal gas occupies 22.4 liters.

At STP (Standard Temperature and Pressure), 1 mole of an ideal gas occupies 22.4 liters.

Volume = moles x volume per mole = 4 moles x 22.4 L/mole = 89.6 L.

Volume = moles of solute / molarity = 4 moles / 2 M = 2 L.

C2H5OH + 3O2 → 2CO2 + 3H2O. 2 moles of C2H5OH produce 4 moles of CO2. Volume = 4 * 22.4 L = 89.6 L.

Volume percent = (volume of solute / total volume) x 100 = (30 mL / 150 mL) x 100 = 20%.

Volume percent = (volume of solute / total volume) × 100 = (30 mL / (30 mL + 70 mL)) × 100 = 30%.

The work done by a gas during an isothermal expansion is W = nRT ln(Vf/Vi).