Asked by Temmick
ZN(s) + 2H+(aq) produces ZN2+(aq) +H2(g)
A sample of Zinc is placed in the ice calorimeter. if 0.0657g of Zinc causes a decrease of 0.109ML in the ice/water volume of the calorimeter, what is the enthalpy change, per mole of Zinc, for the above reaction per mole of zinc.
PLEASE HELP! DUE TOMORROW. I DON'T KNOW WHAT TO DO. It looks like there is not enough information given to find enthalpy change.
A sample of Zinc is placed in the ice calorimeter. if 0.0657g of Zinc causes a decrease of 0.109ML in the ice/water volume of the calorimeter, what is the enthalpy change, per mole of Zinc, for the above reaction per mole of zinc.
PLEASE HELP! DUE TOMORROW. I DON'T KNOW WHAT TO DO. It looks like there is not enough information given to find enthalpy change.
Answers
Answered by
DrBob222
H2O(s) ==> H2O(l)
The density of liquid H2O = 1.00 g/mL.
The density of ice is 0.917 g/mL.
Suppose we start with 1 mol (18 g) H2O(s) and it melts to the liquid.
As a liquid it will occupy 18g/1.00 = 18 mL.
As a solid it will occupy 18/0.917 = 19.63 mL.
19.63-18.0 = 1.63 mL diffeence.
The heat fusion of ice is 6.01 kJ/mol; therefore, 6.01/1.63 = about 3.69 kJ for each mL difference.
In this experiment we have a difference of 0.109 mL; then 3.69 kJ/mL x 0.109 mL = ? kJ for 0.0657 g Zn. You want per mol Zn and you have 0.0657/65.39 = about 0.001 mol Zn so
?kJ/0.001 = ? kJ/mol Zn.
The density of liquid H2O = 1.00 g/mL.
The density of ice is 0.917 g/mL.
Suppose we start with 1 mol (18 g) H2O(s) and it melts to the liquid.
As a liquid it will occupy 18g/1.00 = 18 mL.
As a solid it will occupy 18/0.917 = 19.63 mL.
19.63-18.0 = 1.63 mL diffeence.
The heat fusion of ice is 6.01 kJ/mol; therefore, 6.01/1.63 = about 3.69 kJ for each mL difference.
In this experiment we have a difference of 0.109 mL; then 3.69 kJ/mL x 0.109 mL = ? kJ for 0.0657 g Zn. You want per mol Zn and you have 0.0657/65.39 = about 0.001 mol Zn so
?kJ/0.001 = ? kJ/mol Zn.
Answered by
Chemx10^-313
402 kJ/mol Zn
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