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Bond [772]
3 years ago
10

Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volu

me was 322 milliliters, but its pressure was the same. If the final temperature of the balloon is the same as the freezer’s, what is the temperature of the freezer?
Chemistry
1 answer:
katrin2010 [14]3 years ago
8 0

Answer:

T2=276K

Explanation:

Given:

Initial volume of the balloon V1 = 348 mL

Initial temperature of the balloon T1 = 255C

Final volume of the balloon V2 = 322 mL

Final temperature of the balloon T2 =

To calculate T1 in kelvin

T1= 25+273=298K

Based on Charles law, which states that the volume of a given mass of a ideal gas is directly proportional to the temperature provided that the pressure is constant. It can be applied using the below formula

(V1/T1)=(V2/T2)

T2=( V2*T1)/V1

T2=(322*298)/348

T2=276K

Hence, the temperature of the freezer is 276 K

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We are aware that concentration and volume are related to each other by the formula -

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Hence, the final concentration is 1.4%.

Learn more about concentration -

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The complete question is -

A 53.5 mL sample of an 5.4 % (m/v) KBr solution is diluted with water so that the final volume is 205.0 mL.

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