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3 years ago

13

A flexible container is put in a deep freeze. Its original volume is 3.00 m3 at 25.0°C. After the container cools, it has shrunk

to 2.00 m3. Its new temperature in degrees Celsius is _____°C

2 answers:

cestrela7 [59]3 years ago

6 0

The answer would be 37.5

makvit [3.9K]3 years ago

5 0

The correct answer is 37.5

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Answer: 0.028 grams

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

$\Delta T_f=k_f\times m$

or,

$\Delta T_f=k_f\times \frac{\text{ Mass of solute in g}\times 1000}{\text {Molar mass of solute}\times \text{ Mass of solvent in g}}$

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m = molality

Putting in the values we get:

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