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iragen [17]
3 years ago
11

Calculate the amount of heat needed to convert 50.0 g of liquid water at to steam at 100ºC.

Chemistry
1 answer:
Kipish [7]3 years ago
6 0

Answer:

We need 113000 J of heat (option 2 is correct)

Explanation:

Step 1: Data given

Mass of liquid water = 50.0 grams

ΔHVap = 2260 J/g

Temperature = 100 °C

ΔHVap = The amount of heat released to change phase of a liquid water to steam = 2260 J/g

Step 2: Calculate the heat needed

Q =m* ΔHVap

⇒with Q = the amount of heat needed = TO BE DETERMINED

⇒with m = the mass of water = 50.0 grams

⇒with ΔHVap = 2260 J/g

Q = 50.0 grams * 2260 J

Q = 113000 J

We need 113000 J of heat (option 2 is correct)

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Solution here,

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Now,

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OLga [1]

Answer:

0.962 atm.

97.4 kPa.

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Explanation:

Hello.

In this case, given the available factors equaling 1 atm of pressure, each required pressure turns out:

- Atmospheres: 1 atm = 760 mmHg:

p=731mmHg*\frac{1atm}{760mmHg} =0.962atm

- Kilopascals:: 101.3 kPa = 760 mmHg:

p=731mmHg*\frac{101.3kPa}{760mmHg} =97.4kPa

- Torrs: 760 torr = 760 mmHg:

p=731mmHg*\frac{760 torr}{760mmHg} =731 torr

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p=731mmHg*\frac{101300Pa}{760mmHg} \\\\p=97,434.6Pa

Best regards.

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