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January 30, 2015

January 30, 2015

Posted by **Anonymous** on Monday, November 17, 2008 at 7:29pm.

assume solution has same physical properties as water.

deltaH of solution = -82.8 kJ/mol

specific heat of water = 4.184J/g degrees celsius

density of water = 1.00 g/cm^3

DOES ANYONE HAVE ANY CLUE AS 2 HOW 2 START THIS?

any help will b appreciated

- chemistry -
**DrBob222**, Monday, November 17, 2008 at 7:50pmHow much heat must be added to the water to move the T from Iinitial = 10 C to Tfinal = 16.4 C. That will be

q = massH2O x specific heat water x (Tfinal-Tinitial) = ??

Knowing that 1 mol CaCl2 dissolving in water will produce 82.8 kJ, how many mols CaCl2 are required. Then convert that to grams. Post your work if you get stuck.

- chemistry -
**Anonymous**, Monday, November 17, 2008 at 8:02pmdelta T = 6.4 degrees celsius

q = 26.4g x (4.184J/g*degrees celsius) x 6.4 deg celsius = 706 J = 0.706 kJ

82.8 kJ/mol/ 0.706kJ = 117 mol

(117mol)(111g/mol) = 1.3 x 10^4 g

this doesn't ssem right...did i do something wrong?

and would i use 2 significant digits in the final answer because of the temperature?

- chemistry -
**Anonymous**, Monday, November 17, 2008 at 8:10pmoh never mind...i was supposed to take 0.706kJ and divide by 82.8kJ and the answer would be 0.95g, which makes a lot more sense.

- chemistry -
- chemistry -
**DrBob222**, Monday, November 17, 2008 at 8:16pmRight. My answer was 0.944 grams. As to the number of s.f., I assume the 10 C is really 10.0 C. If so, then there are 3 s.f. figures in delta T and the other values; therefore, 3 s.f. are justified. By the way, the mass of water is 26.3 instead of 26.4 and the molar mass of CaCl2 is closer to 111 than it is to 117. You might want to calculate those parts again and see if you get my value.

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