Combined gas law, Dalton’s law and Downward displacement, Graham’s Law, Gas Stoichiometry, Coulombs Law, and Gibbs Free energy Quiz

Step 1: Write the combined gas law equation:

\(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)

Step 2: Substitute the known values:

\(\frac{2.00 \cdot 3.00}{300} = \frac{P_2 \cdot 1.50}{450}\)

Step 3: Solve for \(P_2\):

\(\frac{6.00}{300} = \frac{1.50P_2}{450}\)

\(0.02 = \frac{1.50P_2}{450}\)

\(P_2 = \frac{0.02 \cdot 450}{1.50} = 6.00\,\text{atm}\)

Final Answer: The final pressure is \(6.00\,\text{atm}\).

Step 1: Write Dalton's law equation:

\(P_{\text{total}} = P_{\text{oxygen}} + P_{\text{nitrogen}} + P_{\text{carbon dioxide}}\)

Step 2: Substitute the partial pressures:

\(P_{\text{total}} = 0.60 + 0.30 + 0.10\)

\(P_{\text{total}} = 1.00\,\text{atm}\)

Final Answer: The total pressure is \(1.00\,\text{atm}\).

Step 1: Write Graham's law equation:

\(\frac{\text{Rate of hydrogen}}{\text{Rate of oxygen}} = \sqrt{\frac{M_{\text{oxygen}}}{M_{\text{hydrogen}}}}\)

Step 2: Substitute the molar masses:

\(\frac{\text{Rate of hydrogen}}{\text{Rate of oxygen}} = \sqrt{\frac{32}{2}}\)

\(\frac{\text{Rate of hydrogen}}{\text{Rate of oxygen}} = \sqrt{16} = 4\)

Final Answer: Hydrogen gas diffuses \(4\,\text{times faster}\) than oxygen gas.

Step 1: Write the balanced chemical equation:

\(\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\)

Step 2: Calculate moles of \(\text{CH}_4\):

\(\text{Molar mass of CH}_4 = 12 + (4 \times 1) = 16\,\text{g/mol}\)

\(\text{Moles of CH}_4 = \frac{10.0}{16} = 0.625\,\text{mol}\)

Step 3: Use the mole ratio to find moles of \(\text{CO}_2\):

\(\text{Moles of CO}_2 = 0.625 \times 1 = 0.625\,\text{mol}\)

Step 4: Convert moles of \(\text{CO}_2\) to volume at STP:

\(\text{Volume} = \text{Moles} \times 22.4\,\text{L/mol}\)

\(\text{Volume} = 0.625 \times 22.4 = 14.0\,\text{L}\)

Final Answer: \(14.0\,\text{L}\) of \(\text{CO}_2\) are produced.

Step 1: Write Coulomb's law equation:

\(F = k \frac{|q_1 q_2|}{r^2}\)

Step 2: Substitute the known values:

\(F = (8.99 \times 10^9) \frac{(2.0 \times 10^{-6})(3.0 \times 10^{-6})}{(0.05)^2}\)

\(F = (8.99 \times 10^9) \frac{6.0 \times 10^{-12}}{0.0025}\)

\(F = (8.99 \times 10^9) \times 2.4 \times 10^{-9}\)

\(F = 21.6\,\text{N}\)

Final Answer: The force is \(21.6\,\text{N}\).

Step 1: Write the Gibbs free energy equation:

\(\Delta G = \Delta H - T \Delta S\)

Step 2: Convert \(\Delta S\) to \(\text{kJ/mol}\,\text{K}\):

\(\Delta S = \frac{-200}{1000} = -0.200\,\text{kJ/mol}\,\text{K}\)

Step 3: Substitute the values into the equation:

\(\Delta G = -150 - (298 \times -0.200)\)

\(\Delta G = -150 + 59.6\)

\(\Delta G = -90.4\,\text{kJ/mol}\)

Final Answer: \(\Delta G = -90.4\,\text{kJ/mol}\)

Step 1: Write Boyle’s Law as part of the Combined Gas Law:

\(P_1 V_1 = P_2 V_2\)

Step 2: Substitute the known values:

\((1.20)(2.50) = P_2(1.00)\)

Step 3: Solve for \(P_2\):

\(P_2 = \frac{1.20 \times 2.50}{1.00} = 3.00\,\text{atm}\)

Final Answer: The final pressure is \(3.00\,\text{atm}\).

Step 1: Calculate the total pressure:

\(P_{\text{total}} = P_{\text{hydrogen}} + P_{\text{helium}} + P_{\text{argon}}\)

\(P_{\text{total}} = 0.40 + 0.50 + 0.10 = 1.00\,\text{atm}\)

Step 2: Calculate the fraction of the pressure due to helium:

\(\text{Fraction of helium} = \frac{P_{\text{helium}}}{P_{\text{total}}}\)

\(\text{Fraction of helium} = \frac{0.50}{1.00} = 0.50\)

Final Answer: Helium accounts for \(50\%\) of the total pressure.

Step 1: Write Graham's Law equation:

\(\frac{\text{Rate of gas A}}{\text{Rate of oxygen}} = \sqrt{\frac{M_{\text{oxygen}}}{M_{\text{gas A}}}}\)

Step 2: Rearrange to solve for \(M_{\text{gas A}}\):

\(M_{\text{gas A}} = \frac{M_{\text{oxygen}}}{\left(\frac{\text{Rate of gas A}}{\text{Rate of oxygen}}\right)^2}\)

Step 3: Substitute the known values (\(M_{\text{oxygen}} = 32\,\text{g/mol}\)):

\(M_{\text{gas A}} = \frac{32}{(1.5)^2} = \frac{32}{2.25} = 14.22\,\text{g/mol}\)

Final Answer: The molar mass of the unknown gas is \(14.22\,\text{g/mol}\).

Step 1: Write the balanced chemical equation:

\(\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3\)

Step 2: Calculate moles of \(\text{N}_2\):

\(\text{Molar mass of N}_2 = 28\,\text{g/mol}\)

\(\text{Moles of N}_2 = \frac{20.0}{28} = 0.714\,\text{mol}\)

Step 3: Use the mole ratio to find moles of \(\text{NH}_3\):

\(\text{Moles of NH}_3 = 0.714 \times 2 = 1.428\,\text{mol}\)

Step 4: Convert moles of \(\text{NH}_3\) to volume at STP:

\(\text{Volume} = \text{Moles} \times 22.4\,\text{L/mol}\)

\(\text{Volume} = 1.428 \times 22.4 = 31.97\,\text{L}\)

Final Answer: The volume of \(\text{NH}_3\) produced is \(31.97\,\text{L}\).

Step 1: Write the combined gas law equation:

\(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)

Step 2: Rearrange to solve for \(V_2\):

\(V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}\)

Step 3: Substitute the known values:

\(V_2 = \frac{2.50 \times 5.00 \times 350}{1.50 \times 300}\)

\(V_2 = \frac{4375}{450} = 9.72\,\text{L}\)

Final Answer: The new volume is \(9.72\,\text{L}\).

Step 1: Write the formula for mole fraction:

\(\text{Mole Fraction of Oxygen} = \frac{P_{\text{oxygen}}}{P_{\text{total}}}\)

Step 2: Substitute the known values:

\(\text{Mole Fraction of Oxygen} = \frac{0.40}{1.60}\)

\(\text{Mole Fraction of Oxygen} = 0.25\)

Final Answer: The mole fraction of oxygen is \(0.25\).

Step 1: Write Graham's Law equation:

\(\frac{\text{Rate of helium}}{\text{Rate of unknown gas}} = \sqrt{\frac{M_{\text{unknown gas}}}{M_{\text{helium}}}}\)

Step 2: Rearrange to solve for \(M_{\text{unknown gas}}\):

\(M_{\text{unknown gas}} = M_{\text{helium}} \times \left(\frac{\text{Rate of helium}}{\text{Rate of unknown gas}}\right)^2\)

Step 3: Substitute the known values (\(M_{\text{helium}} = 4.00\,\text{g/mol}\)):

\(M_{\text{unknown gas}} = 4.00 \times (2.0)^2 = 4.00 \times 4 = 16.00\,\text{g/mol}\)

Final Answer: The molar mass of the unknown gas is \(16.00\,\text{g/mol}\).

Step 1: Write the balanced chemical equation:

\(\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3\)

Step 2: Calculate moles of \(\text{NH}_3\):

\(\text{Molar mass of NH}_3 = 14 + (3 \times 1) = 17\,\text{g/mol}\)

\(\text{Moles of NH}_3 = \frac{10.0}{17} = 0.588\,\text{mol}\)

Step 3: Use the mole ratio to find moles of \(\text{H}_2\):

\(\text{Moles of H}_2 = 0.588 \times \frac{3}{2} = 0.882\,\text{mol}\)

Step 4: Convert moles of \(\text{H}_2\) to volume at STP:

\(\text{Volume} = \text{Moles} \times 22.4\,\text{L/mol}\)

\(\text{Volume} = 0.882 \times 22.4 = 19.75\,\text{L}\)

Final Answer: \(19.75\,\text{L}\) of \(\text{H}_2\) are needed.

Step 1: Write Coulomb's Law equation:

\(F = k \frac{|q_1 q_2|}{r^2}\)

Step 2: Substitute the known values:

\(F = (8.99 \times 10^9) \frac{(3.0 \times 10^{-6})(2.0 \times 10^{-6})}{(0.20)^2}\)

\(F = (8.99 \times 10^9) \frac{6.0 \times 10^{-12}}{0.04}\)

\(F = (8.99 \times 10^9) \times 1.5 \times 10^{-10}\)

\(F = 1.35\,\text{N}\)

Final Answer: The electrostatic force is \(1.35\,\text{N}\).

Step 1: Write the combined gas law equation:

\(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)

Step 2: Rearrange to solve for \(T_2\):

\(T_2 = \frac{P_2 V_2 T_1}{P_1 V_1}\)

Step 3: Substitute the known values:

\(T_2 = \frac{(0.80)(9.00)(400)}{(1.20)(6.00)}\)

\(T_2 = \frac{2880}{7.20} = 400\,\text{K}\)

Final Answer: The new temperature is \(400\,\text{K}\).

Step 1: Calculate the mole fraction of oxygen gas:

\(\text{Mole Fraction of Oxygen} = \frac{P_{\text{oxygen}}}{P_{\text{total}}} = \frac{0.20}{(0.60 + 0.20 + 0.40)}\)

\(\text{Mole Fraction of Oxygen} = \frac{0.20}{1.20} = 0.1667\)

Step 2: Use the mole fraction to calculate the new partial pressure:

\(P_{\text{oxygen}} = \text{Mole Fraction} \times P_{\text{total}}\)

\(P_{\text{oxygen}} = 0.1667 \times 2.00 = 0.333\,\text{atm}\)

Final Answer: The partial pressure of oxygen gas is \(0.333\,\text{atm}\).

Step 1: Write Graham's Law equation:

\(\frac{\text{Rate of hydrogen}}{\text{Rate of gas}} = \sqrt{\frac{M_{\text{gas}}}{M_{\text{hydrogen}}}}\)

Step 2: Rearrange to solve for \(M_{\text{gas}}\):

\(M_{\text{gas}} = M_{\text{hydrogen}} \times \left(\frac{\text{Rate of hydrogen}}{\text{Rate of gas}}\right)^2\)

Step 3: Substitute the known values (\(M_{\text{hydrogen}} = 2.00\,\text{g/mol}\)):

\(M_{\text{gas}} = 2.00 \times (3.0)^2 = 2.00 \times 9 = 18.00\,\text{g/mol}\)

Final Answer: The molar mass of the gas is \(18.00\,\text{g/mol}\).

Step 1: Write the balanced chemical equation:

\(\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\)

Step 2: Calculate moles of \(\text{CH}_4\):

\(\text{Molar mass of CH}_4 = 12 + (4 \times 1) = 16\,\text{g/mol}\)

\(\text{Moles of CH}_4 = \frac{15.0}{16} = 0.9375\,\text{mol}\)

Step 3: Use the mole ratio to find moles of \(\text{CO}_2\):

\(\text{Moles of CO}_2 = 0.9375 \times 1 = 0.9375\,\text{mol}\)

Step 4: Convert moles of \(\text{CO}_2\) to volume at STP:

\(\text{Volume} = \text{Moles} \times 22.4\,\text{L/mol}\)

\(\text{Volume} = 0.9375 \times 22.4 = 21.00\,\text{L}\)

Final Answer: \(21.00\,\text{L}\) of \(\text{CO}_2\) are produced.

Step 1: Write Coulomb's Law equation:

\(F = k \frac{|q_1 q_2|}{r^2}\)

Step 2: Substitute the known values:

\(F = (8.99 \times 10^9) \frac{(1.0 \times 10^{-6})(2.0 \times 10^{-6})}{(0.50)^2}\)

\(F = (8.99 \times 10^9) \frac{2.0 \times 10^{-12}}{0.25}\)

\(F = (8.99 \times 10^9) \times 8.0 \times 10^{-12}\)

\(F = 0.0719\,\text{N}\)

Final Answer: The magnitude of the force is \(0.0719\,\text{N}\).

Step 1: Write the Gibbs free energy equation:

\(\Delta G = \Delta H - T \Delta S\)

Step 2: Convert \(\Delta S\) to \(\text{kJ/mol}\,\text{K}\):

\(\Delta S = \frac{-300}{1000} = -0.300\,\text{kJ/mol}\,\text{K}\)

Step 3: Substitute the known values:

\(\Delta G = -250 - (298 \times -0.300)\)

\(\Delta G = -250 + 89.4\)

\(\Delta G = -160.6\,\text{kJ/mol}\)

Final Answer: \(\Delta G = -160.6\,\text{kJ/mol}\)