only your final answer in a box whenever appropriate. Correctness of units and number of significant figures expressed in your final answers is ghly anticipated. 1. Calculate the heat evolved when 13.56g of oxygen gas (O2) is used to burn white phosphorus at standard temperature and pressure according to the equation: P4(s) + 502(g) → P4O10(5), AH = -3013 kJ/mol 2. Calculate the AH for the reaction: B2Hela) + 6Cl2lg) → 2BCI3(a) + 6HCI(a) %3D

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Chapter6: Thermochemistry
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Problem 105AE: Combustion of table sugar produces CO2(g) and H2O( l). When 1.46 g table sugar is combusted in a...
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Calculate the ΔH for the reaction: B2H6(g) + 6Cl2(g) → 2BCl3(g) + 6HCl(g)
The following are given:
a. BCl3(g) + 3H2O(l) → H3BO3(g) + 3HCl(g) ΔH = -112.5 kJ
b. B2H6(g) + 6H2O(l) → 2H3BO3(g) + 6H2(g) ΔH = -493.4 kJ
c. 12H2(g) + 12Cl2(g) → HCl(g) ΔH = -92.3 kJ

Solve the following problems to the best of your ability. Enclose
only your final answer in a box whenever appropriate. Correctness of
units and number of significant figures expressed in your final answers is
ghly anticipated.
1. Calculate the heat evolved when 13.56g of oxygen gas (O2) is used to burn
white phosphorus at standard temperature and pressure according to the
equation: P4(s) + 502(g) → P.O10(5), AH = -3013 kJ/mol
2. Calculate the AH for the reaction: B2H6(g) + 6CI2(g) → 2BCI3(9) + 6HCI(g)
The following are given:
a. BCI3(g) + 3H2O1) → H3BO3(g) + 3HCI9) AH = -112.5 kJ
b. B2H69) + 6H2O) →2H3BO3(g) + 6H2(g) AH = -493.4 kJ
c. H2(g) + Cl2(9) →HCI(g)
C.:
AH = -92.3 kJ
2
Transcribed Image Text:Solve the following problems to the best of your ability. Enclose only your final answer in a box whenever appropriate. Correctness of units and number of significant figures expressed in your final answers is ghly anticipated. 1. Calculate the heat evolved when 13.56g of oxygen gas (O2) is used to burn white phosphorus at standard temperature and pressure according to the equation: P4(s) + 502(g) → P.O10(5), AH = -3013 kJ/mol 2. Calculate the AH for the reaction: B2H6(g) + 6CI2(g) → 2BCI3(9) + 6HCI(g) The following are given: a. BCI3(g) + 3H2O1) → H3BO3(g) + 3HCI9) AH = -112.5 kJ b. B2H69) + 6H2O) →2H3BO3(g) + 6H2(g) AH = -493.4 kJ c. H2(g) + Cl2(9) →HCI(g) C.: AH = -92.3 kJ 2
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