Answer : The pressure in the flask after reaction complete is, 2.4 atm
Explanation :
To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.
where,
P = final pressure in the flask = ?
R = gas constant = 0.0821 L.atm/mol.K
T = temperature = 
V = volume = 4.0 L
= moles of 
= 0.20 mol
= moles of 
= 0.20 mol
Now put all the given values in the above expression, we get:
Thus, the pressure in the flask after reaction complete is, 2.4 atm
The equilibrium constant Kc for this reaction is calculated as follows
from the equation N2 + 3H2 =2 NH3
qc = (NH3)2/{(N2)(H2)^3}
Qc is therefore = ( 0.001)2 /{(0.1) (0.05)^3} = 0.08
Answer:
1/3
Explanation:
Pyruvate is produced by the glycolysis in cytoplasm. The oxidation of pyruvate takes place in mitochondrial matrix.
Pyruvate is converted to acetyl-CoA in the reaction given below:
Pyruvate + NAD⁺ + CoA-SH ⇒ acetyl-CoA + NADH + CO₂
1 molecule of carbon dioxide is eliminated from 1 molecule of pyruvate.
Also,
2 molecules of carbon dioxide is eliminated from 2 molecules of pyruvate (as glucose on glycolysis yields 2 molecules of pyruvate).
Also, acetyl-CoA further goes into the citric acid cycle and produces 2 molecules of carbon dioxide.
Thus pyruvate produces total 3 molecules of CO₂ and hence glucose produces 6 molecules of CO₂ (as glucose on glycolysis yields 2 molecules of pyruvate)
Thus,
<u>Fraction = 2/6 = 1/3</u>
Answer:
A 3s orbital is at a greater average distance from the nucleus than a 2s orbital
Explanation:
As the principal quantum number n increases, the distance of the orbital from the nucleus increases. Hence if we consider the 2s and 3s orbitals, it is easy to see that the 3s orbital is at a greater distance from the nucleus than the 2s orbitals.
This is clearly seen when we plot the radial distribution against the distance from the nucleus. This enables us to visualize the region in space in which an electron may be found.
