When 50.0 mL of 0.400 M Ca(NO3)2 is added to 50.0 mL of 0.800 M ...

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Question 1

Sodium metal reacts with water to produce hydrogen gas and sodium hydroxide according to the chemical equation shown below. When 0.015 mol of Na is added to 100.00 g of water, the temperature of the resulting solution rises from 25.00°C to 31.45°C. If the specific heat of the solution is 4.18 J/(g ∙ °C), calculate ΔH for the reaction, as written.

2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)

ΔH= ?

◦ -90.0 kJ
◦ -180 kJ
◦ -360 kJ
◦ -5.41 kJ

Question 2

When 50.0 mL of 0.400 M Ca(NO3)2 is added to 50.0 mL of 0.800 M NaF, CaF2 precipitates, as shown in the net ionic equation below. The initial temperature of both solutions is 25.00°C. Assuming that the reaction goes to completion, and that the resulting solution has a mass of 100.00 g and a specific heat of 4.18 J/(g ∙ °C), calculate the final temperature of the solution.

Ca2+(aq) + 2 F-(aq) → CaF2(s)

ΔH° = -11.5 kJ

◦ 26.10°C
◦ 26.65°C
◦ 25.55°C
◦ 24.45°C

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[asan beg]

Thanks for the timely response, appreciate it

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