The heat of solution of ammonium nitrate is 26.2 kJ/mol. If a 5.368 g sample of NH4NO3 is added to 40.0 mL of water in a calorimeter at 23.5°C, what is the minimum temperature reached by the solution? (The specific heat of water = 4.18 J/g·°C; the heat capacity of the calorimeter = 650. J/°C.) (A) 14.3°C (B) 20.8°C (C) –7.7°C (D) 25.6°C (E) 21.4°C

The heat of solution of ammonium nitrate is 26.2 kJ/mol. If a 5.368 g

1.	The heat of solution of ammonium nitrate is 26.2 kJ/mol. If a 5.368 g sample of NH4NO3 is added to 40.0 mL of water in a calorimeter at 23.5°C, what is the minimum temperature reached by the solution? (The specific heat of water = 4.18 J/g·°C; the heat capacity of the calorimeter = 650. J/°C.) (A) 14.3°C (B) 20.8°C (C) –7.7°C (D) 25.6°C (E) 21.4°C
Answer» The Correct option is (E) 21.4°C

Discussion

No Comment Found

Related InterviewSolutions

Calculate the standard enthalpy change for the reaction2C8H18(l) + 21O2(g) → 8CO(g) + 8CO2(g) + 18H2O(l). Given: 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l) ΔH° = –11,020 kJ/mol2CO(g) + O2(g) → 2CO2(g) ΔH° = –566.0 kJ/mol(A) 1.0454 x 104 kJ/mol(B) –8,756 kJ/mol(C) 1.1586 x 104 kJ/mol(D) –6,492 kJ/mol (E) –1.0454 ´ 104 kJ/mol
The combustion of butane produces heat according to the equation 2C4H10(g) + 13O2(g) →8CO2(g) + 10H2O(l) ΔH°rxn= –5,314 kJ/mol How many grams of butane must be burned to release 1.00 x 104 kJ of heat?(A) 30.9 g (B) 61.8 g (C) 109 g (D) 153 g (E) 219 g
The combustion of butane produces heat according to the equation 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l) ΔH°rxn = –5,314 kJ/mol What is the heat of combustion per gram of butane?(A) –32.5 kJ/g(B) –45.7 kJ/g(C) –91.5 kJ/g(D) –2,656 kJ/g(E) –15,440 kJ/g
For which of these reactions will the difference between ΔH° and ΔE° be the smallest? (A) N2(g) + 3H2(g) → 2NH3(g) (B) 4PH3(g) → P4(g) + 6H2(g) (C) H2(g) + Cl2(g) → 2HCl(g) (D) CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) (E) P4(s) + 10Cl2(g) → 4PCl5(s)
Methanol (CH3OH) burns according to the equation2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(l), ΔH°rxn = –1454 kJ/mol.How much heat, in kilojoules, is given off when 75.0 g of methanol is burned?(A) 727 kJ(B) 3.22 x 103 kJ(C) 1.45 x 103 kJ(D) 1.70 x 10–3 kJ(E) 3.41 x 103 kJ
For the reactionC(graphite) + O2(g) → CO2(g) ΔH° = –393 kJ/molHow many grams of C(graphite) must be burned to release 275 kJ of heat?(A) 22.3 g (B) 0.70 g (C) 12.0 g (D) 17.1 g (E) 8.40 g
At 25°C, the following heats of reaction are known: 2ClF(g) + O2(g) → Cl2O(g) + F2O(g ΔH°rxn = 167.4 kJ/mol2ClF3(g) + 2O2(g) → Cl2O(g) + 3F2O(g) ΔH°rxn = 341.4 kJ/mol2F2(g) + O2(g) → 2F2O(g) ΔH°rxn = –43.4 kJ/molAt the same temperature, use Hess's law to calculate ΔH°rxn for the reaction:ClF(g) + F2(g) → ClF3(g)(A) –217.5 kJ/mol(B) –130.2 kJ/mol(C) 217.5 kJ/mol(D) –108.7 kJ/mol
Pentaborane B5H9(s) burns vigorously in O2 to give B2O3(s) and H2O(l). Calculate ΔH°rxn for the combustion of 1 mol of B5H9. ΔH°f[B2O3(s)] = –1,273.5 kJ/mol ΔH°f[B5H9(s)] = 73.2 kJ/mol ΔH°f[H2O(l)] = –285.8 kJ/mol(A) –1,2735 kJ/mol(B) –4,543 kJ/mol(C) –18,170 kJ/mol(D) –9,086 kJ/mol(E) –8,448 kJ/mol
Given the thermochemical equation 2SO2(g) + O2(g) → 2SO3(g), ΔH°rxn= –198 kJ/mol, how much heat is evolved when 600. g of SO2 is burned?(A) 5.46 x 10–2 kJ(B) 928 kJ(C) 1.85 x 103 kJ(D) 59,400 kJ(E) 3.71 x 103 kJ
Given that CaO(s) + H2O(l) →Ca(OH)2(s), ΔH°rxn = –64.8 kJ/mol, how many grams of CaO must react in order to liberate 525 kJ of heat? (A) 6.92 g (B) 56.1 g (C) 455 g (D) 606 g (E) 3.40 x 104 g

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment