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The specific heat of lead is 0.129 J/gC. Find the amount of heat released when 2.4 moles of lead are cooled from 37.2C to 22.5C.

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DrippQueenMo

Answer:

-943J or 943J released

Explanation:

m=2.4 mol x 207.g/ mol=497.3g

c=.129 J/gC

T= T f-Ti=(22.5-37.2)=-14.7C

Q=?

Q=(49.3g)(.129 J/g C) (-14.7 C)

= -943 J or 943 J released

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