Chemistry- IB past paper

Practice Question Solving
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How can i solve such questions?

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Hi
By given values,
mol of HNO3 = 0.05 dm3 * 1 M = 0.05 mol
mol of KOH = 0.05 dm3 * 1 M = 0.05 mol

The balanced reaction for mixture,

KOH + HNO3 —> KNO3 + H2O + heat
1 mol + 1 mol —> 1 mol + 1mol + heat ( 6.4 C)

and also in latter case they didn’t change the Molarity of both solutions and halved the volumes for both.

mol of HNO3 = 0.025 dm3 * 1 M = 0.025 mol
mol of KOH = 0.025 dm3 * 1 M = 0.025 mol

KOH + HNO3 —> KNO3 + H2O + heat
1 mol + 1 mol —> 1 mol + 1mol + heat ( 6.4 C)

Therefore, this would also produce the same temperature value which is 6.4 C

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