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Natalija [7]
3 years ago
13

How many calories of heat were added to 347.9 g of water to raise its temperature from 25oC to 55oC?

Chemistry
1 answer:
saw5 [17]3 years ago
5 0

Answer:

10437calories

Explanation:

The following data were obtained from the question given:

M = 347.9g

C = 4.2J/g°C

T1 = 25°C

T2 = 55°C

ΔT = 55 — 25 = 30°C

Q =?

Q = MCΔT

Q = 347.9 x 4.2 x 30

Q = 43835.4J

Converting this to calories, we obtained the following:

4.2J = 1 calorie

43835.4J = 43835.4/ 4.2 = 10437calories

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Exactly 1 mol Hg2(NO3)2 contains how many moles of Hg , N , and O ?
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A sample of 10.6 g of KNO3 was dissolved in 251.0 g of water at 25 oC in a calorimeter. The final temperature of the solution wa
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Answer:

36.55kJ/mol

Explanation:

The heat of solution is the change in heat when the KNO3 dissolves in water:

KNO3(aq) → K+(aq) + NO3-(aq)

As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.

To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:

<em>Moles KNO3 -Molar mass: 101.1032g/mol-</em>

10.6g * (1mol/101.1032g) = 0.1048 moles KNO3

<em>Change in heat:</em>

q = m*S*ΔT

<em>Where q is heat in J,</em>

<em>m is the mass of the solution: 10.6g + 251.0g = 261.6g</em>

S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-

And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C

q = 261.6g*4.184J/g°C*3.5°C

q = 3830.87J

<em>Molar heat of solution:</em>

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36554J/mol =

<h3>36.55kJ/mol</h3>

<em />

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