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NeTakaya
2 years ago
4

What is the heat required to vaporize 650 grams of a liquid with a heat of vaporization of 723 joules/gram?

Chemistry
2 answers:
babymother [125]2 years ago
4 0

<u>Given:</u>

Mass of liquid = 650 g

Heat of vaporization = 723 J/g

<u>To determine:</u>

Heat required to vaporize 650 g of liquid

<u>Explanation:</u>

The heat required to vaporize 1 g of the liquid is 723 J

Therefore, the heat required to vaporize 650 g is-

= 650 g * 723 J/ 1 g = 469,950 J

Ans: Heat required is 469.95 kJ

11Alexandr11 [23.1K]2 years ago
3 0

Hello!

What is the heat required to vaporize 650 grams of a liquid with a heat of vaporization of 723 joules/gram?

We have the following data:  

H_v\:(heat\:of\:vaporization) = 723\:\dfrac{J}{g}

q\:(heat) =\:?\:(in\:Joule)

m\:(mass) = 650\:g

We apply the data to the formula, see:

H_v = \dfrac{q}{m}

q = H_v * m

q = 723\:\dfrac{J}{\diagup\!\!\!\!g} * 650\:\diagup\!\!\!\!\!g

q = 469,950 \to \boxed{\boxed{q \approx 470,000\:J}}\end{array}}\qquad\checkmark

Answer:

E. 470,000 J

_______________________

I Hope this helps, greetings ... Dexteright02! =)

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Final temperature in degree celsius = (1633 - 273)^{o}C

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It is known that for NH_{3} the value of C_{v} = 0.028 kJ/mol.

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