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Alina [70]
2 years ago
12

A "shielding gas" mixture of argon and carbon dioxide is sometimes used in welding to improve the strength of the weld. if a gas

cylinder of this two-part mixture was at 4.0 atm pressure and this mixture was 90.% argon, what would be the pressure due to the carbon dioxide gas component?
Chemistry
2 answers:
kodGreya [7K]2 years ago
6 0
Dalton's Law of Partial Pressures, commonly applied to ideal gases, explains that the partial pressures of individual, non-reacting gases are equal to the total pressure exerted by the gas mixture. The given gas mixture composed of 90% argon and 10% carbon dioxide has the following partial pressures: 3.6 atm for argon and 0.4 atm for carbon dioxide (answer).
Radda [10]2 years ago
3 0

Answer is: the pressure due of carbon dioxide gas is 0.4 atm.

p(mixture) = 4.0 atm; mixture of argon and carbon dioxide gas.

ω(Ar) = 90% ÷ 100%.

ω(Ar) = 0.9; percentage of argon gas in mixture.

p(Ar) = 0.9 · 4.0 atm.

p(Ar) = 3.6 atm; pressure of argon gas.

The total pressure of an ideal gas mixture is the sum of the partial pressures of the gases in the mixture.

p(mixture) = p(Ar) + p(CO₂).

p(CO₂) = 4 atm - 3.6 atm.

p(CO₂) = 0.4 atm; pressure of carbon dioxide gas.

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Answer:

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In a container containing co, h2, and o2, what is the mole fraction of co if the h2 mole fraction is 0.22 and the o2 mole fracti
almond37 [142]
Thank you for posting your question here at brainly. The <span>mole fraction of co if the h2 mole fraction is 0.22 and the o2 mole fraction is 0.58 is 0.20, below is the solution:
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mole fraction CO + mole fraction H2 + mole fraction O2 = 1 

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6 0
2 years ago
A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?
Tanzania [10]
<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume =  0.50 cm × 10.0 cm × 2.0 cm

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Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

                                      = 19.3 g/cm³

Thus, the density of the gold bar is 19.3 g/cm³

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2 years ago
Explain how manipulation of light waves can cause reflection, refraction, diffusion, and absorption; and describe how different
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Answer:is A

Explanation:

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The gas in a sealed container has an absolute pressure of 125.4 kilopascals. If the air around the container is at a pressure of
AlexFokin [52]

Answer: C. 25.6 kPa

Explanation:

The Gauge pressure is defined as the amount of pressure in a fluid that exceeds the amount of pressure in the atmosphere.

As such, the formula will be,

PG = PT – PA

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PG is Gauge Pressure

PT is Absolute Pressure

PA is Atmospheric Pressure

Inputted in the formula,

PG = 125.4 - 99.8

PG = 25.6 kPa

The gauge pressure inside the container is 25.6kPa which is option C.

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2 years ago
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