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saveliy_v [14]
2 years ago
9

If the air from the problem above contains 22% oxygen, 75% nitrogen and 3% argon, what is the partial pressure of oxygen near a

blast furnace in mmHg?
Chemistry
1 answer:
melomori [17]2 years ago
7 0

Assuming the total pressure is 1 atm, the partial pressure of oxygen is 0.22 atm.

<h3>What is the partial pressure of oxygen if air contains 22% oxygen, 75% nitrogen and 3% argon?</h3>

The partial pressure of a gas in a mixture of gases is calculated using the formula below:

  • Partial pressure of a gas = mole fraction/total pressure

Assuming:

  • the total pressure is 1 atm.
  • mole fraction of oxygen = 22% = 0.22

Partial pressure of oxygen = 0.22/1

Partial pressure of oxygen = 0.22 atm.

Therefore, assuming the total pressure is 1 atm, the partial pressure of oxygen is 0.22 atm.

Learn more about partial pressure at: brainly.com/question/14119417

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B.90.5 g

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For an alloy that consists of 33.5 wt% Pb and 66.5 wt% Sn, what is the composition (a) of Pb (in at%), and (b) of Sn (in at%)? T
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Answer:

Pb: 22.4 at%

Sn: 77.6 at%

Explanation:

It is possible to find at% of Pb and Sn converting mass in moles using molar mass assuming a basis of 100g, thus:

Pb: 33.5g × (1mol / 207.2g) = <em>0.1617mol</em>

Sn: 66.5g × (1mol / 118.7g) = <em>0.5602mol</em>

<em />

Total moles: 0.1617mol + 0.5602mol = 0.7219mol

Composition in at%:

Pb: 0.1617mol / 0.7219mol × 100 = <em>22.4 at%</em>

Sn: 0.5602mol / 0.7219mol × 100 = <em>77.6 at%</em>

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I hope it helps!

5 0
3 years ago
Which nonmetals have similar chemical properties? Check all that apply.
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An 80.0g sample of an unknown metal is at an initial temperature of 55.5oC. Afer 540 J of energy is absorbed by the metal, the t
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Answer:

Specific heat of metal = 0.26 j/g.°C

Explanation:

Given data:

Mass of sample = 80.0 g

Initial temperature = 55.5 °C

Final temperature = 81.75 °C

Amount of heat absorbed = 540 j

Specific heat of metal = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT =  81.75 °C - 55.5 °C

ΔT =  26.25 °C

540 j = 80 g × c × 26.25 °C

540 j = 2100 g.°C× c

540 j / 2100 g.°C = c

c = 0.26 j/g.°C

7 0
3 years ago
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