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lana [24]
2 years ago
6

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 55.0 atm oxygen, 90.0 atm nitrogen, and 50.0 atm helium. What is the percent oxygen (by volume) in this trimex blend
Chemistry
1 answer:
gladu [14]2 years ago
8 0

Answer:

The correct answer is 28.2 %.

Explanation:

Based on the given question, the partial pressures of the gases present in the trimix blend is 55 atm oxygen, 50 atm helium, and 90 atm nitrogen. Therefore, the sum of the partial pressure of gases present in the blend is,  

Ptotal = PO2 + PN2 + PHe

= 55 + 90 + 50

= 195 atm

The percent volume of each gas in the trimix blend can be determined by using the Amagat's law of additive volume, that is, %Vx = (Px/Ptot) * 100

Here Px is the partial pressure of the gas, Ptot is the total pressure and % is the volume of the gas. Now,  

%VO2 = (55/195) * 100 = 28.2%

%VN2 = (90/195) * 100 = 46%

%VHe = (50/195) * 100 = 25.64%

Hence, the percent oxygen by volume present in the blend is 28.2 %.  

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A gas that has a volume of 28 liters, a temperature of 45C, And an unknown pressure has its volume increased to 34 liters and it
patriot [66]

Answer:

P1 = 2.5ATM

Explanation:

V1 = 28L

T1 = 45°C = (45 + 273.15)K = 318.15K

V2 = 34L

T2 = 35°C = (35 + 273.15)K = 308.15K

P1 = ?

P2 = 2ATM

applying combined gas equation,

P1V1 / T1 = P2V2 / T2

P1*V1*T2 = P2*V2*T1

Solving for P1

P1 = P2*V2*T1 / V1*T2

P1 = (2.0 * 34 * 318.15) / (28 * 308.15)

P1 = 21634.2 / 8628.2

P1 = 2.5ATM

The initial pressure was 2.5ATM

3 0
2 years ago
When 0.40 mol Al is mixed with 0.40 mol Br2, the following reaction occurs: 2Al(s) + 3Br2(l) → 2AlBr3(s) Identify the limiting r
Lera25 [3.4K]

Answer:

d. The limiting reactant is Br2, and 0.13 mol Al should remain unreacted.

Explanation:

From the reaction, 2Al(s) + 3Br2(l) → 2AlBr3(s)

2 moles of Al combines with 3 moles of Br to form 2 moles of AlBr3

Hence 1 mole of Br combines with 2/3 moles of Al

or 0.4 moles of Br combines with 2/3×0.4 moles of Al or 0.267 Moles of Al leaving

0.4 - 0.267 = 0.133 moles of Al remaining unreacted

7 0
2 years ago
Your task is to measure the amount of energy evolved during the combustion of some hydrocarbon. Which of the following would be
Dahasolnce [82]

Answer:

c. Bomb calorimetry

Explanation:

The hydrocarbons are combustibles, it means that they can react in a combustion reaction to release energy. To measure this amount of energy, it's necessary equipment that the reaction can be placed in a controlled way. The bomb calorimeter is this equipment, which is an adiabatic vessel, with water. The heat is calculated based on the increase in the water temperature.

The coffee-cup calorimetry is used to measure the heat of a dissolution reaction and the bomb manometry is used to measure the pressure.

7 0
2 years ago
If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.50 cm radius? le
ElenaW [278]

The volume of sphere can be calculated using the following formula:


V=\frac{4}{3}\pi r^{3}


Here, r is radius of the sphere which is 3.5 cm. Putting the value,


V=\frac{4}{3}\pi r^{3}=\frac{4}{3}(3.14)(3.5cm)^{3}=180 cm^{3}


This is equal to the volume of lead, density of lead is 11.34 g/cm^{3} thus, mass of lead can be calculated as follows:


m=d×V


Putting the values,


m=11.34 g/cm^{3}\times 180 cm^{3}=2041.2 g


Let the mass of ore be 1 g, 68.6% of galena is obtained by mass, thus, mass of galena obtained will be 0.686 g.


Now, 86.6% of lead is obtained from this gram of galena, thus, mass of lead will be:


m=0.686×0.866=0.5940 g


Therefore, 0.5940 g of lead is obtained from 1 g of ore for 100% efficiency, thus, for 92.5% efficiency

m=\frac{92.5}{100}\times 0.5940=0.54945 g

1 g of lead obtain from\frac{1}{0.54945}=1.82 grams of ore.


Thus, 2041.2 g of lead obtain from:


2041.2\times 1.82=3.715\times 10^{3}g or 3.715 kg


Therefore, mass of ore required to make lead sphere is 3.715 kg.


6 0
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What volume is occupied by 0.34 moles of Helium gas?
guapka [62]

Answer:

0.65882352941

Explanation:

5 0
2 years ago
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