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kvasek [131]
3 years ago
8

Use the ideal gas law to calculate the concentrations of nitrogen and oxygen present in air at a pressure of 1.0 atm and a tempe

rature of 298 K. Assume that nitrogen comprises 78% of air by volume and that oxygen comprises 21%. Express your answers using two significant figures separated by a comma.
Chemistry
1 answer:
notka56 [123]3 years ago
4 0

Answer:

[N₂] = 0.032 M

[O₂] = 0.0086 M

Explanation:

Ideal Gas Law → P . V = n .  R . T

We assume that the mixture of air occupies a volume of 1 L

78% N₂ → Mole fraction of N₂ = 0.78

21% O₂  → Mole fraction of O₂ = 0.21

1% another gases  → Mole fraction of another gases = 0.01

In a mixture, the total pressure of the system refers to total moles of the mixture

1 atm . 1L = n . 0.082L.atm/mol.K . 298K

n = 1 L.atm / 0.082L.atm/mol.K . 298K → 0.0409 moles

We apply the mole fraction to determine the moles

N₂ moles / Total moles = 0.78 → 0.78 . 0.0409 mol = 0.032 moles N₂

O₂ moles / Total moles = 0.21 → 0.21 . 0.0409 mol = 0.0086 moles O₂

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The answer from this questions is the letter

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5 0
3 years ago
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A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate
inna [77]

Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams  and dry-weight of 85.2 grams

a) wet density of the CL sample

wet density can be expressed as  p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

p = 0.0453 lbs/in³

so the wet density of the CL sample is 0.0453 lb/in³

b)

water content of sample is taken as;

w =  (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

c)

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 /  1.6537

Sd = 0.0274 lb/in³

therefore the dry density of the CL sample is 0.0274 lb/in³

8 0
3 years ago
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cupoosta [38]

Answer : The value of reaction quotient, Q is 0.0625.

Solution : Given,

Concentration of N_2 = 2.00 M

Concentration of H_2 = 2.00 M

Concentration of NH_3 = 1.00 M

Reaction quotient : It is defined as a concentration of a chemical species involved in the chemical reaction.

The balanced equilibrium reaction is,

N_2+3H_2\rightleftharpoons 2NH_3

The expression of reaction quotient for this reaction is,

Q=\frac{[Product]^p}{[Reactant]^r}\\Q=\frac{[NH_3]^2}{[N_2]^1[H_2]^3}

Now put all the given values in this expression, we get

Q=\frac{(1.00)^2}{(2.00)^1(2.00)^3}=0.0625

Therefore, the value of reaction quotient, Q is 0.0625.

3 0
3 years ago
___ Particles of which of the following have the greatest thermal energy
abruzzese [7]
Salutations!

<span>___ Particles of which of the following have the greatest thermal energy

Gas is the particle that has the greatest thermal energy. Gas has the greatest thermal energy because the movement between particles is clump. This thermal energy is then converted into kinetic energy.

Thus, your answer is option C.

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Is acetone soluble in water or hexane​
GaryK [48]

Answer:

a

Explanation:

6 0
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