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vova2212 [387]
4 years ago
6

Large area lakes and rivers are an important natural resource. These Major bodies of water help create ________________ for huma

ns by providing a source of food and water.
A) space
B) shelter
C) a habitat
D) recreation
Chemistry
1 answer:
PilotLPTM [1.2K]4 years ago
4 0
I think it would be habitat
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At room temperature, the solubility of which solute in water would be most affected by a change in pressure?
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<h3><u>Answer;</u></h3>

At room temperature, the solubility of which solute in water would be most affected by a change in pressure?

<h3><u>Explanation;</u></h3>
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over a 12.3 minuete period 5.13 E-3 moles of F2 gas effuses from a contaier. How many moles of CH4 gas could effuse from from th
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Answer : The moles of methane gas could be, 7.90\times 10^{-3}mol

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}

[\frac{(\frac{n_1}{t_1})}{(\frac{n_2}{t_2})}]=\sqrt{\frac{M_2}{M_1}}

where,

R_1 = rate of effusion of fluorine gas

R_2 = rate of effusion of methane gas

n_1 = moles of fluorine gas = 5.13\times 10^{-3}mol

n_2 = moles of methane gas = ?

t_1=t_2 = time = 12.3 min  (as per question)

M_1 = molar mass of fluorine gas  = 38 g/mole

M_2 = molar mass of methane gas = 16 g/mole

Now put all the given values in the above formula 1, we get:

[\frac{(\frac{5.13\times 10^{-3}mol}{12.3min})}{(\frac{n_2}{12.3min})}]=\sqrt{\frac{16g/mole}{38g/mole}}

n_2=7.90\times 10^{-3}mol

Therefore, the moles of methane gas could be, 7.90\times 10^{-3}mol

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