3 years ago

If 600.0 ml of air is heated from 293 k to 333 k what volume will it occupy

Chemistry

1 answer:

mestny [16]3 years ago

6 0

Answer:

Explanation:

V1= 600.0 mL

T1= 293 K

V2 unknown

T2= 333K

Charles law V1*T2=V2*T1

where V1 and T1 are the initial volume and temperature. V2 and T2 are the final temperature and volume.

V2=(V1*T2)/T1 = ( 600.0*333)/293= 682 mL (with significant figures)(681.911 mL before signifcant figures.)

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