When 3.00 g of sodium hydroxide (NaOH) was dissolved in 100.00 g of water a value of 12.00 degree C was obtained for ΔT
Calculate the value (calories) for the heat of solution of 3.00 g of NaOH.
So attempting to solve this question I did the following: (3g NAOH+100g H2O)*(12 degree C)*(4.18 the specific heat of water)= 5166 J I've tried a bunch of other things trying to get the right answer, but out of all of them I thought this made the most sense. If anyone could tell me where I went wrong that would be greatly appreciated.
Calculate the value (calories) for the heat of solution of 3.00 g of NaOH.
So attempting to solve this question I did the following: (3g NAOH+100g H2O)*(12 degree C)*(4.18 the specific heat of water)= 5166 J I've tried a bunch of other things trying to get the right answer, but out of all of them I thought this made the most sense. If anyone could tell me where I went wrong that would be greatly appreciated.
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5166 J / 4.184 J/cal = 1235 calories
Normally 3 g of NaOH is not included:
100 g x 4.18 J/gC x 12 C = 5016 J
5016 J / 4.18 J/calorie = 1200 calories
or simply-
100 g x 1.00 cal/gC x 12C = 1200 calories
Normally 3 g of NaOH is not included:
100 g x 4.18 J/gC x 12 C = 5016 J
5016 J / 4.18 J/calorie = 1200 calories
or simply-
100 g x 1.00 cal/gC x 12C = 1200 calories