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3 years ago

At 1 atm, how much energy is required to heat 75.0 g H 2 O ( s ) at − 20.0 ∘ C to H 2 O ( g ) at 119.0 ∘ C?

Chemistry

1 answer:

Hitman42 [59]3 years ago

3 0

Answer:

238,485 Joules

Explanation:

The amount of energy required is a summation of heat of fusion, capacity and vaporization.

Q = mLf + mC∆T + mLv = m(Lf + C∆T + Lv)

m (mass of water) = 75 g

Lf (specific latent heat of fusion of water) = 336 J/g

C (specific heat capacity of water) = 4.2 J/g°C

∆T = T2 - T1 = 119 - (-20) = 119+20 = 139°C

Lv (specific latent heat of vaporization of water) = 2,260 J/g

Q = 75(336 + 4.2×139 + 2260) = 75(336 + 583.8 + 2260) = 75(3179.8) = 238,485 J

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<h3>Answer:</h3>

128 g HCl

<h3>General Formulas and Concepts:</h3>

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<u>Step 1: Define</u>

[RxN - Unbalanced] Mg (s) + HCl (aq) → MgCl (aq) + H₂ (g)

↓

[RxN - Balanced] 2Mg (s) + 2HCl (aq) → 2MgCl (aq) + H₂ (g)

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<u>Step 2: Identify Conversions</u>

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The answer to your question is given after the questions so I just explain how to get it.

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To learn more about heat check the link below:

brainly.com/question/13439286

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