Question

Table 4.3: Enthalpy of Dissolution for a Salt Caci, Na,CO, 104,0 Volume of water 90. 1 mL Mass of water (assume a density of 1.00 g/mL) Initial temperature of water 90, 1mL 90.19 22.10 13.050g 16.5°C -5.6°C 90.1g 21.9°C 13.0489 39.6°0 17.7°C Mass of salt Final temperature of solution Change in temperature (47) qwater qcal Arxn AHS per mole of salt in kJ/mol (AH)

Answer 1

C chy) Caleulation you call Sellh . STA 17.7°C mass of water = goolg Ccal = 340 25/°C Sh2O = 4.186 J/gie Solution © Q weiter = mundur no AT weitu = goolgx 4.18 6 J/g • c X 17.7°C mater = 6675.7 J quatur a 6.7 kJ Teal - Cool XAT = 34. 20/8 x 1707 q cal = 605.345 Kala 0.6 kJ Von = a water + Ucal Ason a 6675. I +60 5.345 Arona 7281.04 J = 7.3 KJ

ma was calde = mass molar mass = 13.048 g = 0.11757 20.989/mat mal - Als xmales = -% xn ans = -9 ron moles Ons =- 7.3 kJ e 01175 mal = -62.09 kumor S enthalpy of dissociation you calle Cithalso and the ans of Que 3) = 2) Caleulation for Na₂CO₃.10H20 - ST= -5.6°C : muratur = goolg Cal = 28.6575/¢ , swater = 401865 / poc Solution al amater = mSAT Fuatu = 90.19 X4 1861/9•c X(-5.6) Iwetuz -2112. 10 ou -2.1 KO

с се & Wear & Ceal AT Vcal ? 28. 567 Foc X (+506°c) Real = -159.985 ore - 0.16 kJ Aron = quatu + acal Ogun = - 2112.13 + (-159.98) Yon=-22720 1 Jou - 2. 27 kJ та4а : © И. Он imass melis of Na, o, 101,0 = malademous e 13.059 286.14g/mol & 4056810² mol Duis = -loxn - - (-20 27 KT) males 4056 Xlornol Als = 49.76 kmal Thank you