1- Here given when NaOH is added to water, the temperature increases from 24.0oC to 33.7oC. That means there must be a reaction between NaOH and water takes place, in which heat is released and this heat increased the temperature of the solution. So as heat is released during the reaction, this reaction is Exothermic in nature.
2- Now QH2O is the amount of heat absorbed by the water where its temperature increased from 24.0oC to 33.7oC. This amount of heat absorbed by water is calculated by the formula
QH2O = m * C * ΔT
where m = mass of water taken
C = specific heat of water
ΔT = change in temperature = final temp - initial temp
Now putting the given values-
QH2O = m * C * ΔT
= 53 g * 4.186 J/goC * (33.7oC - 24.0oC)
= 53 g * 4.186 J/goC * (9.7oC)
= 2152 J
3- Now this heat that is absorbed by water is actually released from the reaction. So the heat of reaction = ΔH = -2152 J
Here the negative sign indicates the reaction is Exothermic in nature.
4-
In the above question, we get heat of reaction for 2 g od reaction = -2152 J
Now the heat of reaction per 1 g of NaOH = -2152 J/ 2g
= -1076 J/g
5- Now we know mass of 1 mole NaOH = 40 g/mole
Then moles of NaOH in 2 g NaOH = mass / molar mass
= 2 g / 40 g/mole
= 0.05 moles
Then
heat of reaction per 1 mole of NaOH = -2152 J/ 0.05 moles
= -43040 J/ mole
2. When 2.0 g of NaOH were dissolved in 53.0 g water in a calorimeter at...
When 4.102 g of KNO3 is dissolved in 39.825 g of water in a "coffee-cup calorimeter" at 24 degrees Celsius, the temperature of the solution went down to 19 degrees Celsius. a.) Is this solution reaction exothermic or endothermic? Why? b.) Calcualate qH2O using Eq. 1. c.) Find as it occured in the calorimeter (Eq. 4). d.) Find for the solution of one mole KNO3 in water. e.) Find for the solution of 1.00 g KNO3 in water.
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When a 4.00 g sample of KSCN is dissolved in water in a calorimeter that has a total heat capacity of 3.98 kJ.K-1, the temperature decreases by 0.250 K. Calculate the molar heat of solution of KSCN. AH soln = kJ/mol
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When 3.90 g of cesium perchlorate (CsClO4) is dissolved in 112 g of water in a styrofoam calorimeter of negligible heat capacity, the temperature drops from 25.00 to 22.93 °C. Based on this observation, calculate q for the water and ΔH° for the process, assuming that the heat absorbed by the salt is negligible. CsClO4(s) Cs+(aq) + ClO4- (aq) The specific heat of water is 4.184 J °C-1 g-1. Give the answers in kJ. qH2O = kJ ΔH° = kJ