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hodyreva [135]
3 years ago
15

What is responsible for moving the water to the ground?

Chemistry
2 answers:
Mandarinka [93]3 years ago
6 0

Answer:

Water applied to the surface of a relatively dry soil infiltrates quickly due to the affinity of the soil particles for water. As time passes and the soil becomes wet, the force of gravity becomes the dominant force causing water to move.

Explanation:

DIA [1.3K]3 years ago
6 0

Answer:

Gravity

Explanation:

Water applied to the surface of a relatively dry soil infiltrates quickly due to the affinity of the soil particles for water. As time passes and the soil becomes wet, the force of gravity becomes the dominant force causing water to move.

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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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How much of a 3.75 M KI solution would you need to prepare 252.5 mL of a 0.780 M KI solution?
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