Question

#4

4) Given the data in the table below. Δ1Prn for the reaction Ca(OH)2 + 2H3AsO4 → Ca(H2AsO4)2 + 2H2O is kJ. Substance AH't(k/mol) Ca(OF)2 H3As04 Ca(H2As04)2 -23460 H20 986.6 -900.4 -285.9 A)-744.9 B)-4519 C)-4219 D) -130.4 E) -76.4 5) The combustion of titanium with oxygen produces titanium dioxide: Ti (s) +02 (g) → TiO2 (s) When 0.610 g of titanium is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00 °C to 50.50 °C. In a separate experiment, the heat capacity of the calorimeter is measured to be 9.84 kJ/K. The heat of reaction for the combustion of a mole of Ti in this calorimeter is A) 2.09 B) 4.14 C)-311 D)-0.154 k.J/mol. E)-1.98 x 104 6) A 4.50-g sample of liquid water at 25.0 °C is heated by the addition of 133 J of energy. The final temperature of the water is 。 J/g-K. A) 149 B) 25.1 C)-17.9 D) 32.1 C. The specific heat capacity of liquid water is 4.18 E) 7.07

Answer 1

So, 250.92 kJ is evolved on combustion of .013 moles of Ti Heat evolved for 1 mole combustion = 250.92/.013 almostequalto 1.98 times 10^4 kJ Therefore The heat of combustion of Ti is -1.98 times 10^4 kJ/mol(-ve is for evolution of heat) E is the correct answer Let, the final temperature of the sample be t degree c so, according to conservation of energy law, the heat provided results in increase in temp 133 = 4.50 times 4.18 times (t - 25) 7.1 = t - 25 t = 32.1 Therefore D is the correct answer \r\n Given reaction: Ca(OH) _2 + 2 H_3AsO_4 rightarrow Ca(H_2AsO_4) _2 + 2H_2O we know, Delta H degree_rxn = sigma n H degree _f(products) - sigma m H degree_f(reactants) (i) where n and m are the respective stoichiometric coefficients. so, Delta H degree_f(Ca(OH) _2) = -986.6 kJ/mol Delta H_f(Ca(H_2O_4)) _2 = -2346.0 kJ/mol Delta H degree_f(H_3AsO_4) = -900.4 kJ/mol Delta H_f(H_2O) = -285.9 kJ/mol Putting these values in the equation (i)