Chemistry: Air is a mixture of gases that is about78.0percent n2 by volume. when air is at standard pressure and 25.0 degreec, the n2 component will dissolve in water with a solubility of 4.88 x 10^-4m. what is the value of henry s law constant for n2 under these conditions? express the constant numerically inmoles per liter per atmosphere. =6.26? 10^-4 mol/(l.atm) correct part b as a scuba diver descends under water, thepressure increases. at a total air pressure of 2.73 atm and a temperature of 25.0 degreec,

A substance s molar mass is calculated by multiplying its relative atomic mass by the molar mass constant (1 g/mol). ... By multiplying a given mass by the molar mass, the amount of moles of the substance can be calculated....

Air is a mixture of gases that is about78.0percent

Ответ:

diamondd4

13.01.2022

Chemistry

$(a) 6.256\cdot 10^{-4} M/atm; (b) 1.33\cdot 10^{-3} M$

Explanation:

(a) According to the Henry's law, the solubility is equal to the product between the Henry's law constant and the partial pressure of a gas:

S = k_Hp^o

Air is at standard pressure:

$p = 1.00 atm[/atm]Nitrogen's pecentage is:[tex]\omega = 0.780$

Therefore, its partial pressure is:

$p^o = \omega p$

Solving for the Henry's constant:

$k_H = S/p^o = \frac{S}{\omega p}$

$k_H = \frac{4.88\cdot 10^-4 M}{0.780\cdot 1.00 atm} = 6.256\cdot 10^{-4} M/atm$

(b) Using the constant we've found in the previous part, we know that:

$S = k_H\omega p$

In this case, the percentage is kept the same, however, the total air pressure is:

p = 2.73 atm

Substituting the variables:

$S = 6.256\cdot 10^{-4} M/atm \cdot 0.78\cdot 2.73 atm = 1.33\cdot 10^{-3} M$

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