Solution:

1. Write the balanced chemical equation:

   MgCO₃(s) + 2HCl(aq) → MgCl₂(aq) + CO₂(g) + H₂O(l)

2. Calculate the moles of CO₂ produced:

   Using the ideal gas law (PV = nRT), where P = 1 atm, V = 1.00 L, R = 0.0821 L atm/mol K, and T = 298 K, we can calculate the number of moles of CO₂.

   n(CO₂) = PV / RT = (1 atm * 1.00 L) / (0.0821 L atm/mol K * 298 K) = 0.0414 mol

3. Determine the moles of MgCO₃ reacted:

   From the balanced equation, 1 mole of MgCO₃ produces 1 mole of CO₂. Therefore, n(MgCO₃) = 0.0414 mol.

4. Calculate the moles of MgO produced:

   From the balanced equation, 1 mole of MgCO₃ produces 1 mole of MgO. Therefore, n(MgO) = 0.0414 mol.

5. Calculate the mass of MgO produced:

   Molar mass of MgO = 24.31 g/mol + 16.00 g/mol = 40.31 g/mol

   Mass of MgO = n(MgO) * Molar mass of MgO = 0.0414 mol * 40.31 g/mol = 1.67 g

Solution:

1. Write the balanced chemical equation for the combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). The enthalpy change for this reaction is -890 kJ/mol.
2. We want to find the enthalpy change for: C₂H₄(g) + H₂(g) → C₂H₆(g).
3. Manipulate the given equation to obtain the desired reaction. We need to find a way to get C₂H₆(g) on both sides of the equation. We can do this by adding the given equation to the reverse of the desired reaction:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

C₂H₆(g) → C₂H₄(g) + H₂(g)

Adding these two equations gives: CH₄(g) + 2O₂(g) + C₂H₆(g) → CO₂(g) + 2H₂O(g) + C₂H₄(g) + H₂(g)

4. Now, we can rearrange the equation to isolate the desired reaction: C₂H₄(g) + H₂(g) → C₂H₆(g). The enthalpy change for this reaction is the same as the enthalpy change for the combustion of methane, which is -890 kJ/mol.

Answer: -890 kJ/mol

Solution:

1. Calculate the molar masses:

   • Molar mass of C = 12.01 g/mol
   • Molar mass of H = 1.01 g/mol
   • Molar mass of O = 16.00 g/mol

2. Calculate the mass percentages of each element in glucose:

   • Mass of C in glucose = (6 x 12.01) = 72.06 g/mol
   • Mass of H in glucose = (12 x 1.01) = 12.12 g/mol
   • Mass of O in glucose = (6 x 16.00) = 96.00 g/mol

3. Convert mass percentages to moles of each element:

   • Moles of C = 72.06 g/mol / 12.01 g/mol = 6.00 mol
   • Moles of H = 12.12 g/mol / 1.01 g/mol = 12.00 mol
   • Moles of O = 96.00 g/mol / 16.00 g/mol = 6.00 mol

4. Divide the moles of each element by the smallest number of moles:

   • C: 6.00 mol / 6.00 mol = 1
   • H: 12.00 mol / 6.00 mol = 2
   • O: 6.00 mol / 6.00 mol = 1

5. The empirical formula is CH₂O.