Zoe....
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When 50 cm3 of hydrochloric acid of concentration 2.0 mol dm−3 is added to 50cm3 of sodium*hydroxide solution of concentration 2.0 mol dm−3, the temperature increase is 13.0 C

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
The experiment is repeated using 25 cm3
of the same hydrochloric acid and 50 cm3 of
the same sodium*hydroxide solution.

Q what is the temperature change?
the answer is 8.7C but how???
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charco
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(Original post by Zoe....)
When 50 cm3 of hydrochloric acid of concentration 2.0 mol dm−3 is added to 50cm3 of sodium*hydroxide solution of concentration 2.0 mol dm−3, the temperature increase is 13.0 C

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

The experiment is repeated using 25 cm3 of the same hydrochloric acid and 50 cm3 of the same sodium*hydroxide solution.


Q what is the temperature change? the answer is 8.7C but how???
Both parts require the use of q = mcΔT to calculate the energy of the particular reaction.

Use the text in red to calculate the enthalpy of neutralisation, i.e. the energy released per mol of water formed.

Then apply this value to the text in blue.
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Zoe....
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I do not get the answer!!!!!!
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