Respuesta :

kenmyna

 The heat required to   vaporize 650 grams  of liquid    with a heat  of vaporization  of 723  joules/grams  is 470,000  J ( answer  E)



calculation


Heat (Q)  =  m Hv   where,

 m(mass)=  650  g

 HV(   heat of vaporization)  = 723 J/g


heat  is therefore =  650 g  x 723  J/g  =469950 J


round off  469950 j into nearest 10,000 =  470,000 J ( answer E)

dexteright02

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:  

[tex]H_v\:(heat\:of\:vaporization) = 723\:\dfrac{J}{g}[/tex]

[tex]q\:(heat) =\:?\:(in\:Joule)[/tex]

[tex]m\:(mass) = 650\:g[/tex]

We apply the data to the formula, see:

[tex]H_v = \dfrac{q}{m}[/tex]

[tex]q = H_v * m[/tex]

[tex]q = 723\:\dfrac{J}{\diagup\!\!\!\!g} * 650\:\diagup\!\!\!\!\!g[/tex]

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

Answer:

E. 470,000 J

_______________________

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

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