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trapecia [35]
3 years ago
11

How many moles of gas are contained in a 60.0L cylinder at a pressure of 200.0 atm and a temperature of 25.0? A.(490.48 mol) B.(

590.84 mol) C.(340.56 mol) D.(540.67 mol)​
Chemistry
1 answer:
ExtremeBDS [4]3 years ago
5 0

Answer: A. 490.48mol

Explanation:

Using the ideal gas equation; PV =nRT

where P = Presssure = 200 atm

          V = Volume = 60L

          n = Number of moles

          R = Gas constant = 0.08206 L.atm/mol/K

          T = Temperature = 25 °C  + 273 = 298K

Make "n" the subject of the formular;

n =  PV /  RT

   =  200   x    60   /    0.08206 x  298

   =  490.73  mol, which is closest to  option A. 490.48mol

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Answer:

Option (2) 2

Explanation:

NO3- + 4H+ + Pb → Pb2+ + NO2 + 2H2O

The equation above can be balance as follow:

There are 3 atoms of the left side and a total of 4 atoms on the right side. It can be balance by putting 2 in front NO3- and 2 in front of NO2 as shown below:

2NO3- + 4H+ + Pb → Pb2+ + 2NO2 + 2H2O

Now the equation is balanced.

The coefficient of NO2 is 2

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Answer:

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Explanation:

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Explanation:

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What is the mole fraction of each component if 3.9 g of benzene (C6H6) is dissolved in 4.6 g of toluene (C7H8)
Savatey [412]

Answer:

Step 1 of 6

(a)

The mass of benzene is  , so calculate the moles of benzene as follows:



The mass of toluene is, so calculate the moles of toluene as follows:



Now, calculate the mole fraction as follows:





Therefore, the mole fraction of benzene and toluene is  and  respectively.

Step 2 of 6

(b)

The formula to calculate the partial pressure is as follows:



Here,  is the partial pressure of benzene,  is the vapour pressure of pure benzene and  is the mole fraction of benzene.

Vapour pressure of pure benzene at  is.

Substitute the values in the equation as follows:



Therefore, the partial pressure is  .

Step 3 of 6

(c)

Vapor pressure of the solution at 1 atm is  .

When the total pressure of the vapour pressure of the mixture is  at a temperature, then, the solution boils. It corresponds to the boiling point of the solution.

Calculate the total pressure of the solution at  as follows:



Since, the total pressure is less than the atmospheric pressure, the solution will not boil at  .

Calculate the total pressure of the solution at  as follows:



Since, the total pressure is greater than the atmospheric pressure, the solution will boil at  .

Therefore, the boiling point of the solution is  .

Step 4 of 6

(d)

Mole fraction of benzene at  is calculated as follows:



Mole fraction of toluene at  is calculated as follows:



Therefore, the mole fractions of benzene and toluene are  and  respectively.

Step 5 of 6

(e)

Vapor pressure of benzene at  is  .

Partial pressure of benzene is calculated as follows:



Vapor pressure of toluene at  is  .

Partial pressure of toluene is calculated as follows:



Step 6 of 6

Weight composition of the vapour that is in equilibrium with the solution is calculated as follows:



Weight composition of the vapour that is in equilibrium with the solution is calculated as follows:



Explanation:

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