Question

1. When a 4.00 g sample of RbBr is dissolved in water in a calorimeter that has a total heat capacity of 1.26 kJ⋅K−1, the temperature decreases by 0.420 K. Calculate the molar heat of solution of RbBr.

ΔHsoln=

2. In a constant‑pressure calorimeter, 60.0 mL of 0.340 M Ba(OH)2 was added to 60.0 mL of 0.680 M HCl.

The reaction caused the temperature of the solution to rise from 22.24 ∘C to 26.87 ∘C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184J/g⋅°C,) respectively), what is ΔH for this reaction (per mole H2O produced)? Assume that the total volume is the sum of the individual volumes.

ΔH=

Answer 1

2 & mass of the RbBr is n=4.09 410 ²3 kg. Total heat capacity of the calorimetoris c= 12605/ AT=0.420K RbBr molar mass = 165.4 g/mol. molas = spot 481165.vg) mola 0.0242 mot. Q = CAT = 12 to 80.4205 Time - 126080.420 1/mol 010242 = 21.8 kJ/mol Theis we have the molar heat of solution of Rbby equal to 21.8 kJ/mol. B BALOH)2 + 2 Hd Back, + 2H2O The two treactants are present in their Stokhiometic amounts, so neither is in exces and either could be considered limiting: (0.06001) x10.660 md. L. Hellx (210 Hp / 2mal hely = 0.0408 mol HLO (60-omi + booml) x191ml) XC4.1845 18°c) 1 (26:87(-22.24) = 2324.63.5 absorbed by the solution | (2324.639) | Loro yo x mol Mg) = 58946518 = 57.0 KT) mol H20