+ AH;= -91.3 kJ (1) NO(g) +ÎN28) +2028) (2) N28) + O2(8)+NO28) NO(g) + O2(g) +NO2(8) 102 ΔΗ? = ? AHx= -58.1 kJ Determi
Posted: Wed May 18, 2022 1:45 pm
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Determine the enthalpy change of equation number 4 + 10 CHO 218) (D 1 4C + 5H 24C + 4H264 → CH 3 H) + 120218) AH = -1320K; AH = -430K; AH = -732Kj → HO + 4CHO CHI HO
Determine the enthalpy change of equation number 5. CH (OH)zimp) CHO210) + H 1218) How + 0. H,02104 28) ΔΗ =- 453 kJ ΔΗ =- 352kJ AH =- 152 kJ AH = - 250 kJ Hie + 20 → HO HO HO CH (OH)200) + H, 02100) → CH2(g) + 2H,0 AH = ? O360
Determine the change of enthalpy of the last equation. Ca → Calw + 20 200 Н. → HOT + HW → Ca(OH) → 2 Ca(OH)2 + Ca(OH)2 (a) + 73036 028 0) AH = - 70.4 kJ AH = - 34.7 kJ AH = - 64.8 kJ AH = -30.9 kJ Ca + H₂ 200 Cao + HO → Ca(OH), 2(aq) AH = ? (C)