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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 : The value of K_{goal} for the final reaction is, 1.238\times 10^{-7}

Explanation :

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CO_2(g)\rightleftharpoons C(s)+O_2(g) K_{goal}=?

Now we have to calculate the value of K_{goal} for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

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K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}

K_{goal}=1.238\times 10^{-7}

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