ok i don't like to ask homework questions but i am somewhat stuck right now. I keep trying the problem but my answers are off. Here is the problem: Consider the dissolution of CaCl2:
CaCl2(s) ---> Ca^(2+) (aq) + 2Cl^(-1) (aq) deltaH = -81.5kJ
An 11.0g sample of CACl2 is dissolved in 125g of water, with both substances at 25.0°C. Calculate the final temperature of the solution assuming no heat lost to the surroundings and assuming the solution has a specific heat capacity of 4.18 J/(°C*g).
If you could help me understand how to do this problem then i think i can figure out the rest. Thanks in advance
2007-11-05 13:14:50 · 1 answers · asked by live in the moment 2 in Science & Mathematics ➔ Chemistry
Atomic weights: Ca=40 Cl=35.5 CaCl2=111
11.0gCaCl2 x 1molCaCl2/40gCaCl2 x -81.5J/1molCaCl2 = -22.4J
The rest is to figure how many degC the temperature isabove 25.0C based on -22.4J and 4.18J/g-C
2007-11-05 13:25:56 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋