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When 10 cm3
of 1.0 mol dm−3 hydrochloric acid is added to 5.0 cm3
of
0.50 mol dm−3 sodium hydroxide in a polystyrene cup, the temperature rises by 2.3 °C.
HCl(aq) + NaOH(aq) o NaCl(aq) + H2O(l)
The temperature rise when 10 cm3
of 1.0 mol dm−3 hydrochloric acid is added to
20 cm3
of 0.50 mol dm−3 sodium hydroxide is
A 18.4 °C.
B 9.2 °C.
C 4.6 °C.
D 2.3 °C.
how is this c
of 1.0 mol dm−3 hydrochloric acid is added to 5.0 cm3
of
0.50 mol dm−3 sodium hydroxide in a polystyrene cup, the temperature rises by 2.3 °C.
HCl(aq) + NaOH(aq) o NaCl(aq) + H2O(l)
The temperature rise when 10 cm3
of 1.0 mol dm−3 hydrochloric acid is added to
20 cm3
of 0.50 mol dm−3 sodium hydroxide is
A 18.4 °C.
B 9.2 °C.
C 4.6 °C.
D 2.3 °C.
how is this c
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(Original post by staerryynighht)
When 10 cm3
of 1.0 mol dm−3 hydrochloric acid is added to 5.0 cm3
of
0.50 mol dm−3 sodium hydroxide in a polystyrene cup, the temperature rises by 2.3 °C.
HCl(aq) + NaOH(aq) o NaCl(aq) + H2O(l)
The temperature rise when 10 cm3
of 1.0 mol dm−3 hydrochloric acid is added to
20 cm3
of 0.50 mol dm−3 sodium hydroxide is
A 18.4 °C.
B 9.2 °C.
C 4.6 °C.
D 2.3 °C.
how is this c
When 10 cm3
of 1.0 mol dm−3 hydrochloric acid is added to 5.0 cm3
of
0.50 mol dm−3 sodium hydroxide in a polystyrene cup, the temperature rises by 2.3 °C.
HCl(aq) + NaOH(aq) o NaCl(aq) + H2O(l)
The temperature rise when 10 cm3
of 1.0 mol dm−3 hydrochloric acid is added to
20 cm3
of 0.50 mol dm−3 sodium hydroxide is
A 18.4 °C.
B 9.2 °C.
C 4.6 °C.
D 2.3 °C.
how is this c
hence
So you can see that since Q is quadrupled and m is doubled, the change in temperature is doubled
So you can see that since Q is quadrupled and m is doubled, the change in temperature is doubled
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