When 7.59 grams of sodium hydroxide (NaOH) are dissolved in 80.0 grams of water at 25.0 °C in an insulated container, the temperature of the water increases to 48.0 °C. Assuming that the specific heat of the solution is 4.184 J/(g°C) and that no heat is gained or lost by the container, what is the AH of solution of NaOH in kJ/mol? Determine the enthalpy of reaction for PO, (s) + 6PCI,(g) → 10POCI, (g) P (s) + 6Cl₂(g) →→ 4PCI,(g) AH = -1225.6 kJ/mol P(s) + 50,(g) →→→ PO, (s) AH = -2967.3 kJ/mol PCI (g) + Cl₂(g) →→ PCI (g) AH = -84.2 kJ/mol PCI (g) + 1/2O₂(g) →→ POCI (g) AH = -285.72 kJ/mol Determine the enthalpy of reaction for HCI(g) + NaNO₂ (s) → HNO₂(1) + NaCl(s)
When 7.59 grams of sodium hydroxide (NaOH) are dissolved in 80.0 grams of water at 25.0 °C in an insulated container, the temperature of the water increases to 48.0 °C. Assuming that the specific heat of the solution is 4.184 J/(g°C) and that no heat is gained or lost by the container, what is the AH of solution of NaOH in kJ/mol? Determine the enthalpy of reaction for PO, (s) + 6PCI,(g) → 10POCI, (g) P (s) + 6Cl₂(g) →→ 4PCI,(g) AH = -1225.6 kJ/mol P(s) + 50,(g) →→→ PO, (s) AH = -2967.3 kJ/mol PCI (g) + Cl₂(g) →→ PCI (g) AH = -84.2 kJ/mol PCI (g) + 1/2O₂(g) →→ POCI (g) AH = -285.72 kJ/mol Determine the enthalpy of reaction for HCI(g) + NaNO₂ (s) → HNO₂(1) + NaCl(s)
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Transcribed Image Text:When 7.59 grams of sodium hydroxide (NaOH) are dissolved in 80.0
grams of water at 25.0 °C in an insulated container, the temperature of
the water increases to 48.0 °C. Assuming that the specific heat of the
solution is 4.184 J/(g°C) and that no heat is gained or lost by the
container, what is the AH of solution of NaOH in kJ/mol?
Determine the enthalpy of reaction for PO,(s) + 6PCI(g) —
10POCI,(g)
P (s) + 6Cl₂(g) →→ 4PCI,(g) AH = -1225.6 kJ/mol
P (s) +50₂(g) → PO, (s) AH = -2967.3 kJ/mol
PCI₂(g) + Cl₂(g) →→ PCI (g) AH = -84.2 kJ/mol
PCI (g) + 1/2O₂(g) → POCI,(g) AH = -285.72 kJ/mol
10
Determine the enthalpy of reaction for HCI(g) + NaNO,(s) → HNO₂(I) +
NaCl(s)
2NaCl(s) + H₂O(1)→ 2HCI(g) + Na₂O(s) AH = -507.1 kJ/mol
NO(g) + NO₂(g) + Na₂O(s) → 2NaÑO,(s) AH° = -427.0 kJ/mol
NO(g) + NO₂(g) →→ NO(g) + O₂(g) AH = -43.01 kJ/mol
2HNO₂(1)→ N₂O(g) + O₂(g) + H₂O(1) AH = +34.02 kJ/mol
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