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sleet_krkn [62]
3 years ago
11

The partial pressure of O2 in air at sea level is 0.21atm. The solubility of O2 in water at 20∘C, with 1 atm O2 pressure is 1.38

×10−3 M. Part A Using Henry's law, calculate the molar concentration of O2 in the surface water of a mountain lake saturated with air at 20 ∘C and an atmospheric pressure of 665 torr . Express your answer using two significant figures. nothing
Chemistry
1 answer:
adell [148]3 years ago
7 0

Answer:

1.21x10^{-3} M

Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

0.21 = Hx1.38x10^{-3}

H = \frac{0.21}{1.38x10^{-3} }

H = 152.17 atm/M

For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

P = 0.21*0.875 = 0.1837 atm

Then, the molar concentration [O2], will be:

P = Hx[O2]

0.1837 = 152.17x[O2]

[O2] = 0.1837/15.17

[O2] = 1.21x10^{-3} M

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WITCHER [35]

Answer: Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]

We are given:

\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ

Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

6 0
3 years ago
How many moles is 22.4 liters of oxygen gas at standard temperature and pressure represent
makkiz [27]

Answer:

So 1 mole

Explanation:

PV = nRT

P = Pressure atm

V = Volume L

n = Moles

R = 0.08206 L·atm·mol−1·K−1.

T = Temperature K

standard temperature = 273K

standard pressure = 1 atm

22.4 liters of oxygen

Ok so we have

V = 22.4

P = 1 atm

PV = nRT

n = PV/RT

n = 22.4/(0.08206 x 273)

n  = 22.4/22.40

n = 1 mole

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Explanation:

It is known that a chemical/substance can either be present in a solid, liquid or gaseous state.

So, when the phase of a substance like solid, liquid or gas is present at 1 atmosphere pressure and at a temperature of 25^{o}C then it known as standard state of substance.

Thus, we can conclude that standard state refers to 1 atm and 25^{o}C.

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