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juin [17]
3 years ago
15

Trimix is a general name for a type of gas blend used by technical divers and contains nitrogen, oxygen and helium. In one Trimi

x blend, the partial pressures of each gas are 55 atm oxygen, 90 atm nitrogen, and 50 atm helium. What is the percent oxygen (by volume) in this trimex blend?
Chemistry
2 answers:
Keith_Richards [23]3 years ago
6 0

Answer:

28.2

Explanation:

Add all of the pressures, 55, 90, and 50, and divide 100 by the answer you get (195). You'll get 0.512820513 and multiply it by .55 (atm of Oxygen) and you'll get 28.2

Korolek [52]3 years ago
3 0

Answer:

The percentage by volume of Oxygen gas in the Trimex blend is 28.2%.

Explanation:

We shall use the formula that connects the mole ratio of a particular gas in a mixture with the partial pressure of that gas and the total pressure of the mixture: which is:

Mole ratio of Oxygen in Trimex = Partial pressure of Oxygen in Trimex divided by the Total pressure of all the gases that made up of Trimex.

Here we go, mole ratio of Oxygen= 55/195. The total pressure is the addition of the partial pressures of all the gases in the Trimex mixture, which is 55+90+50=195.

So the mole ratio of our Oxygen gas will be: 0.282, no unit.

This means we can assume the total moles of all the gas is 1, and oxygen is 0.282.

Since we know that 1 mole of any gas will have 22.4dm cube as volume, then the total volume of the Trimex will be 22.4dm cube, while that of the oxygen gas will be 0.282 X 22.4 dm cube= 6.3179dm cube. Therefore, the percentage by volume of Oxygen gas in the Trimex will be: 6.3179/22.4 X 100 = 28.2%.

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tia_tia [17]

Answer:

3 mol/L

Explanation:

You should know or have the equation to solve for Molarity which is;

M = n/v           (M: Molarity) (n: moles of solute) (v: Liters of solute)

You can start off differently but I would start by converting the mL to L. This is your "v" value.

50.0 mL/ 1000 mL = 0.05 L

Now, you have to convert grams to moles in order to solve for molarity (M).

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Na= 22.99 g/mol         Cl= 35.45g/mol

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22.99 + 35.45= 58.44 g/mol

3.) Now, you're going to use the "picket fence method" or whichever your teacher taught you to convert from grams to moles. This will be your "n" value. (I cannot show it on here without it looking weird, so my sincere apologies.)

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4.)You are now going to plug in your answers into the equation for Molarity.

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5.) I am sure your professor might be a stickler so for sig figs sake when you multiply or divide use the smallest amount of sig figs you see which is 1. Round 3.4222 to 3 mol/L

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If a recycling center collects 3245 aluminum cans and there are 22 aluminum cans in 1 lb what volume in liters
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The volume in liters of 3245 aluminum cans is 24.8 L.

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To convert mass to volume, we need the density of aluminum (5.95 lb/L). The volume corresponding to 147.5 lb of aluminum is:

147.5 lb \times \frac{1L}{5.95 lb} = 24.8 L

The volume in liters of 3245 aluminum cans is 24.8 L.

You  can learn more about density here: brainly.com/question/1841285

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julia-pushkina [17]

Answer: 448 g of O_2 will be required to completely react with 784g moles of CO(g) during this reaction.

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allochka39001 [22]

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<em />

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<em />

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