Homework Answers
1)
NH4Cl(aq) + NaOH(aq) -----> NH4OH(aq) + NaCl(aq)
∆H = -3.6kJ
Given moles of NH4Cl = ( 2.0mol/1000ml)× 50ml = 0.1mol
given moles of NaOH = (2.0mol/1000ml) × 50ml = 0.1mol
0.1mole of NH4Cl react with 0.1mole of NaOH
q = ∆H
q = ( -3.6kJ/0.10mol) × 1mol
q = - 36kJ
36 kJ of heat is produced
2)
q = m × ∆T ×Cs
m = mass, 105g
∆T = Temperature difference
Cs = specific heat capacity ,0.216cal/g℃
542 cal = 105g × ∆T × 0.216cal/g℃
∆T = 542cal /( 105g × 0.216cal/g℃)
∆T = 23.90℃
New temperature = 25℃ + 23.90℃
New temperature = 48.90℃
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