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never [62]
3 years ago
7

Calculate the heat absorbed by the water in a calorimeter when 175 grams of lead cools from 125.0°C to 22.0°C. The specific heat

capacity of lead is .01295 J/g°C
Chemistry
1 answer:
Iteru [2.4K]3 years ago
5 0

Answer:

Q = 233.42 J

Explanation:

Given data:

Mass of lead = 175 g

Initial temperature = 125.0°C

Final temperature = 22.0°C

Specific heat capacity of lead = 0.01295 J/g.°C

Heat absorbed by water = ?

Solution:

Heat  absorbed by water is actually the heat lost by the metal.

Thus, we will calculate the heat lost by metal.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 22.0°C - 125.0°C

ΔT = -103°C

Q = 175 g × 0.01295 J/g.°C×-103°C

Q = -233.42 J

Heat absorbed by the water is 233.42 J.

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3 years ago
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3 0
3 years ago
Consider the following reaction: 2Mg(s)+O2(g)-->2MgO(s) delta H=-1204kJ
Pachacha [2.7K]

Answer:

a. The reaction is exothermic.

b. -87,9 kJ

c. 9,60g of Mg(s)

d. 602kJ are absorbed

Explanation:

Based on the reaction:

2Mg(s) + O₂(g) → 2MgO(s) ΔH = -1204kJ

a. The reaction is exothermic. Because ΔH<0. That means the reaction produces heat when occurs

b. 3,55g of Mg(s) are:

3,55g Mg × ( 1mol / 24,305g) = 0,146 moles of Mg(s)

As 2 moles of Mg(s) produce -1204 kJ of heat:

0,146 moles of Mg(s) × ( -1204kJ / 2mol Mg) =  <em>-87,9 kJ</em>

c. If -238 kJ of heat were transferred. The moles of Mg(s) that react must be:

-238kJ × ( 2mol Mg / -1204kJ) = 0,395 moles of Mg(s). In grams:

0,395 moles × ( 24,305g / 1mol Mg) = <em>9,60g of Mg(s)</em>

d. The reverse reaction is:

2MgO(s) → 2Mg(s) + O₂(g)  ΔH = +1204kJ

40,5g of MgO(s) are:

40,5g MgO × ( 1mol MgO / 40,3044g) = 1,00 moles of MgO(s)

As 2 moles of MgO absorbe 1204kJ of energy:

1,00 moles of MgO(s) × ( +1204 kJ / 2mol MgO) = <em>602kJ are absorbed</em>

<em></em>

I hope it helps!

7 0
3 years ago
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