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Molodets [167]
3 years ago
9

What is the boiling point of propanone if the pressure on its surface is 48 kilopascals?

Chemistry
1 answer:
Scorpion4ik [409]3 years ago
8 0
<span>The boiling point would be 35 degrees Celsius. </span>
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When 55.0 grams of metal at 75.0°C is added to 100. grams of water at 15.0°C, the temperature of the water rises to 18.7°C. Assu
olga2289 [7]

Answer:

The specific heat of the metal is 0,50 J/gºC

Explanation:

Assume that no heat is lost to the surroundings

(Q = m . C . ΔT)metal + (Q = m . C . ΔT)water = 0

Let's replace our values.

55g . C . (18,7ºC - 75ºC) + 100g . 4,184 J/g·°C . (18,7ºC - 15ºC) = 0

55g . C . -56,3 ºC + 418,4J/·°C . 3,7ºC = 0

-3096,5 gºC . C + 1548,08 J = 0

1548,08 J = 3096,5 gºC . C

1548,08 J / 3096,5 gºC  = C = 0,50 J/gºC

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3 years ago
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Answer:

Explanation:

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8 0
2 years ago
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Answer:

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