J.R. S. answered 12/02/21
Ph.D. University Professor with 10+ years Tutoring Experience
11.0 g ice
6.25 kJ heat energy
assuming the initial temperature of the ice is at 0ºC, and using the following:
∆Hfusion = 334 J/g
C liquid water = 4.184 J/gº
∆Hvap = 2260 J/g
To melt the 11.0 g of ice:
11.0 g x 334 J/g = 3674 J = 0.367 kJ
To raise temp of 11 g of water from 0 to 100º :
11.0 g x 4.184 J/gº x 100º = 4602 J = 4.60 kJ
So, 4.60 kJ + 0.37 kJ = 4.97 kJ and the water has just reached boiling, So there are still 1.28 kJ of heat energy that haven't been used. So, the answer would be ...
A. The ice will melt.
B. The liquid water will vaporize.
E. The final state of sample is gas.
2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g) ... balanced equation
1 g Al x 1 mol Al / 26.982 g x 3 mols H2 / 2 mols Al = 0.05556 mols H2
At STP 1 mol = 22.4 L
0.05556 mols x 22.4 L / mol = 1.25 L (option A)
Previously answered this question. Propanol is the gas.
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