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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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Step2247 [10]

Answer:

See explanation

Explanation:

The equation of the reaction is;

C3H8 + 5O2 ----> 3CO2 + 4H2O

Number of moles of C3H8 = 132.33g/44g/mol = 3 moles

1 mole of C3H8 yields 3 moles of CO2

3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2

Number of moles of oxygen = 384.00 g/32 g/mol = 12 moles

5 moles of oxygen yields 3 moles of CO2

12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2

Hence C3H8 is the limiting reactant.

Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2

1 moles of C3H8 yields 4 moles of water

3 moles of C3H8 yields 3 × 4/1 = 12 moles of water

Mass of water = 12 moles of water × 18 g/mol = 216 g of water

b) Actual yield = 269.34 g

Theoretical yield = 396 g

% yield = actual yield/theoretical yield × 100/1

% yield = 269.34 g /396 g × 100

% yield = 68%

4 0
3 years ago
What is the electron configuration of a ground-state Nb2+ ion? Select one: a. 1s22s22p63s23p64s23d104p65s24d1 b. 1s22s22p63s23p6
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Consider the neutralization reaction 2 HNO 3 ( aq ) + Ba ( OH ) 2 ( aq ) ⟶ 2 H 2 O ( l ) + Ba ( NO 3 ) 2 ( aq ) A 0.120 L sample
k0ka [10]

Answer:

The concentration of the HNO3 solution is 0.150 M

Explanation:

<u>Step 1:</u> Data given

Volume of the unknown HNO3 sample = 0.120 L

Volume of the 0.200 M Ba(OH)2 = 45.1 mL

<u>Step 2:</u> The balanced equation

2HNO3 + Ba(OH)2 ⟶ Ba(NO3)2 + 2H2O

<u>Step 3:</u> Calculate moles Ba(OH)2

moles Ba(OH)2 = molarity * volume

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For 0.00902 moles of Ba(OH)2 we need 2*0.00902 = 0.01804 moles

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Answer:

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