A.) During an experiment, a student adds 2.90 g of CaO to 400.0 mL of 1.500 M HCl. The student observes a temperature increase of 6.00 °C. Assuming the solution/'s final volume is 400.0 mL, the density if 1.00 g/mL, and the heat capacity is 4.184 J / (g·°C), calculate the heat of the reaction, ΔHrxn.

B.) During an experiment, a student adds 1.05 g of calcium metal to 200.0 mL of 0.75 M HCl. The student observes a temperature increase of 17.0 °C for the solution. Assuming the solution/'s final volume is 200.0 mL, the density is 1.00 g/mL, and the specific heat is 4.184 J / (g·°C), calculate the heat of the reaction, ΔHrxn.

C.) Write the balanced chemical equation for the standard enthalpy of formation of magnesium oxide. Include the phase labels. The arrow has been provided for you. Do not include charges.

Question

Chemistry

Duke

Today, 02:56

A.) During an experiment, a student adds 2.90 g of CaO to 400.0 mL of 1.500 M HCl. The student observes a temperature increase of 6.00 °C. Assuming the solution/'s final volume is 400.0 mL, the density if 1.00 g/mL, and the heat capacity is 4.184 J / (g·°C), calculate the heat of the reaction, ΔHrxn.

B.) During an experiment, a student adds 1.05 g of calcium metal to 200.0 mL of 0.75 M HCl. The student observes a temperature increase of 17.0 °C for the solution. Assuming the solution/'s final volume is 200.0 mL, the density is 1.00 g/mL, and the specific heat is 4.184 J / (g·°C), calculate the heat of the reaction, ΔHrxn.

C.) Write the balanced chemical equation for the standard enthalpy of formation of magnesium oxide. Include the phase labels. The arrow has been provided for you. Do not include charges.

+2

Answers (1)

Know the Answer?

Skeeter · Answer 1

a. 1.00 x 10⁴ J = 10.0 kJ

b. 1.42 x 10⁴ J = 14.2 kJ

Explanation:

Given the change in temperature during the reaction and assuming the volume of water and density remains constant, the change in enthalpy for the reaction will be given by

ΔHxn = Q = mCΔT where,

m = mass of water

C = specific heat of water, and

ΔT = change in temperature

a. mH₂O = 400.0 mL x 1.00g/mL = 400.00 g

Q = ΔHrxn = 400.00g x 4.184 J/gºC x 6.00 ºC = 1.00 x 10⁴ J = 10.0 kJ

b. mH₂O = 200.0 mL x 1.00 g/mL = 200.0 g

Q = ΔHrxn = 200.00 g x 4.184 J/gºC x 17.0ºC = 1.42 x 10⁴ J = 14.2 kJ