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3 years ago

The ability for an pond to freeze over in the winter

Chemistry

2 answers:

Otrada [13]3 years ago

8 0

Answer:

The ability for an pond to freeze over in the winter

Explanation:

The ability for an pond to freeze over in the winter

Brums [2.3K]3 years ago

3 0

Answer:

The ability for a pond to freeze over in the winter is a physical property.

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4) What volume will the gas in the balloon at right occupy at 250k?<br><br> balloon: 4.3L 350K

swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

$Pv =nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 4.3 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

4.3 liter = $4.3\times10^{-3}=4.3\times10^{-3} m^{3}$

And initial temperature of balloon, $T_{1}$ = 350 K

Let the final volume of balloon is $V_{2}$

And as given, final temperature of balloon, $T_{2}$ is 250 K

Now, $V_{1} = KT_{1}$

$4.3\times10^{-3}=K\times350\ (equation\ 1 )$

$V_{2} = KT_{2}$

$=K\times250\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}$

K cancelled by K.

By cross multiplication:

$350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\ = \frac{1075\times10^{-3}}{350} \\ =2.87\times10^{-3}m^{3}$

Now, convert it into liter with the help of calculation done above,

$2.87\times10^{-3} \times1000\\2.87\times10^{-3} \times10^{3} \\2.87\ liter$

Therefore, volume of the gas in the balloon at 250 K will be 2.87 liter.

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