1. ms = 4+10 = 14 g c = 1cal/g Δt = 15 - 18 = -3°
Q = m·c·Δt = - 42 cal
n NH4Cl = 4/53,5 = 0,0747moli
0,0747 moli.......... - 42 cal
1 mol......................Qdiz = - 562,248 calorii /mol
2. ms = 13g Δt = 3°
Q = 13·3 = 39 cal.
nKOH = 3/56 = 0,0535 moli ...........................39 cal
1 mol.....................................Qdiz = +728,97cal/ mol