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Author Question: 2 LiOH(s) Li2O(s) + H2O(l) rH = 379.1 kJ/mol LiH(s) + H2O(l) LiOH(s) + H2(g) rH = -111.0 kJ/mol 2 ... (Read 194 times)

magmichele12

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Question 1

1674 J of heat are absorbed by 25.0 mL of NaOH (d = 1.10 g/mL, specific heat = 4.10 J/g °C). The temperature of the NaOH goes up by how many degrees?
◦ 14.8 °C
◦ 18.0 °C
◦ 17.2 °C
◦ 14.2 °C
◦ 19.1 °C

Question 2

2 LiOH(s) → Li2O(s) + H2O(l)  Δr = 379.1 kJ/mol
LiH(s) + H2O(l) → LiOH(s) + H2(g)  Δr = -111.0 kJ/mol
2 H2(g) + O2(g) → 2 H2O(l)  Δr = -285.9 kJmol

Compute Δr in kJ/mol for 2 LiH(s) + O2(g) → Li2O(s) + H2O(l)
◦ +125.2 kJ/mol
◦ -17.7 kJ/mol
◦ -128.8 kJ/mol
◦ -303.6 kJ/mol
◦ + 128.8 kJ/mol


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Marked as best answer by magmichele12 on Jul 8, 2021

ricroger

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magmichele12

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Reply 2 on: Jul 8, 2021
Gracias!


Liddy

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Reply 3 on: Yesterday
Thanks for the timely response, appreciate it

 

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