1) ΔrH = 2mol·ΔfH(NO) - (ΔfH(O₂) + ΔfH(N₂)).
ΔrH = 2 mol · 90.3 kJ/mol - (0 kJ/mol + 0 kJ/mol).
ΔrH = 180.6 kJ.
2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).
ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).
ΔS = 24.8 J/K.
3) ΔG = ΔH - TΔS.
55°C: ΔG = 180.6 kJ - 328.15 K · 24.8 J/K = 172.46 kJ.
2570°C: ΔG = 180.6 kJ - 2843.15 K · 24.8 J/K = 110.09 kJ.
3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.
3) CH₃-COOH + NH₃ → CH₃-COO⁻NH₄⁺
4) 2 FeCl₃ + 3 Ag₂SO₃ → Fe₂(SO₃)₃ + 6 AgCl
5) 2 Al + 3 NiCl₂ → 2 AlCl₃ + 3 Ni
6) 4 LiCl + Pb(NO₂)₄ → 4 LiNO₂ + PbCl₄
7) 3 H₂SO₄ + 2 Al(OH)₃ → Al₂(SO₄)₃ + 6 H₂O
8) Cd(NO₃)₂ + Na₂S → CdS + 2 NaNO₃
9) Cr₂(SO₄)₃ + 3 (NH₄)₂CO₃ → Cr₂(CO₃)₃ + 3 (NH₄)₂SO₄
Reduced stroke volume and cardiac index results in a drop in blood pressure often seen when a patient is turned from supine to prone. Respiratory changes include a 30% to 35% decrease in respiratory compliance and increase in peak airway pressure. This in turn decreases venous return and cardiac output.
IT IS EQUAL TO 5 MOLE OF HYDROGEN ATOMS
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