What is the heat required to vaporize 650 grams of a liquid with a heat of vaporization of 723 joules/gram? a. 0.90 j b. 1.1 j c. 73 j d. 1,370 j e. 470,000 j

 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)

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:  

We apply the data to the formula, see:

E. 470,000 J

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

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."