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Vinil7 [7]
3 years ago
5

Calculate the change in heat (Q) that occurs when 7.58grams of steam at 101°C cools to liquid water at 40°C. (cp = 4.184 J/g°C)

Chemistry
1 answer:
musickatia [10]3 years ago
3 0

Answer:

Q = 19065.4 J

Explanation:

Q₁ = the required energy for phase change of H₂O from vapor to liquid

Q₁ = mass of H₂O * heat of vaporization

Q₁ = 7.58 g * 2260 J/g

Q₁ = 17130.8 J

Q₂ = energy required to cool the water fro 101°C to 40°C

Q₂ = mass of H₂O * specific heat of H₂O * ΔT

Q₂ = 7.58 × (4.184 J/gC) × (101 - 40)

Q₂ = 1934.6 J

Q = Q₁ + Q₂

Q = (17130.8 + 1934.6) J

Q = 19065.4 J

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A. 1.0 liter of a 1.0 M mercury (II) chloride (HgCl2) solution<br> IMO)
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What mass of the following chemicals is needed to make the solutions indicated?

Answer:

271.6g

Explanation:

The mass of the chemicals need to make the needed solution can be derived by obtaining the number of moles first.

Given parameters:

Volume of solution  = 1L

Molarity of HgCl₂  = 1M

    number of moles of HgCl₂  = molarity of solution x volume

                                                   =   1 x 1

                                                    = 1 mole

From;

           Mass of a substance  = number of moles x molar mass;

  we can find mass;

          Molar mass of HgCl₂  = 200.6 + 2(35.5)  = 271.6g/mol

       Mass of the substance  = 1 x 271.6  = 271.6g

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DD

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