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Anestetic [448]
4 years ago
13

Calculate the mass percentage of oxygen in dry air

Chemistry
2 answers:
grigory [225]4 years ago
5 0
              <span> (0.20948 x 31.998) / (
(0.78084 x 28.013) +
(0.20948 x 31.998) +
(0.00934 x 39.948) +
(0.000375 x 44.0099) +
(0.00001818 x 20.183) +
(0.00000524 x 4.003) +
(0.000002 x 16.043) +
(0.00000114 x 83.80) +
(0.0000005 x 2.0159) +
(0.0000005 x 44.0128) ) = 0.23140 = 23.140% O by mass </span>
Oxana [17]4 years ago
4 0

Answer:

0.23 = 23% O by mass

Explanation:

Hi! Let's solve this!

We have the percentage data of each component in the air and its masses.

Nitrogen 0.78084 28.013

Oxygen 0.20948 31.998

Argon 0.00934 39.948

Carbon dioxide 0.000375 44.0099

Neon 0.00001818 20.183

Helium 0.00000524 4.003

Methane 0.000002 16.043

Krypton 0.00000114 83.80

Hydrogen 0.0000005 2.0159

Nitrous oxide 0.0000005 44.0128

Now we have to calculate a molar fraction.

(0.20948 * 31.998) = (0.78084 * 28.013)

+ (0.20948 * 31.998) + (0.00934 * 39.948) + (0.000375 * 44.0099) + (0.00001818 * 20.183) + (0.00000524 * 4.003) + (0.000002 * 16.043) + (0.00000114 * 83.80) + (0.0000005 * 2.0159) + ( 0.0000005 * 44.0128) = 0.23 = 23% O by mass

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Explanation:

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3 years ago
If an atom gains an electron it is known as what ? And has a negative charge
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Account for the change when NO2Cl is added using the reaction quotient Qc. Match the words in the left column to the appropriate
ryzh [129]

Answer:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases.

Explanation:

Hello,

In this case, for the equilibrium reaction:

NO_2Cl(g)+NO(g)\rightleftharpoons NOCl(g)+NO_2(g)

Whose equilibrium expression is:

Kc=\frac{[NO_2][NOCl]}{[NO_2Cl][NO]}

The proper matching is:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc, since the denominator becomes greater, therefore, Qc decreases.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases, since the lower the denominator, the higher Qc as it has the concentration of reactants.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases, since the reactants must be consumed in order to reestablish equilibrium by shifting the reaction towards the products.

Best regards.

8 0
4 years ago
How is mass calculated given density and volume? (1 point) Question 8 options: 1) Volume divided by volume 2) Sum of volume and
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Answer:

3) Density multiplied by volume

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

So by rearranging this formula mass can be determine.

 d = m/v

m = d×/v

Unit of mass when volume in mL and ddensity is g/mL

m = g/mL×mL

m = g

Example:

density = 5g/mL

volume = 3 mL

mass = ?

m = d×v

m = 5g/mL ×3 mL

m = 15 g

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