Answer:
A) 8.00 mol NH₃
B) 137 g NH₃
C) 2.30 g H₂
D) 1.53 x 10²⁰ molecules NH₃
Explanation:
Let us consider the balanced equation:
N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)
Part A
3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:
Part B:
1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:
Part C:
According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:
Part D:
6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:
Answer:
p3=0.36atm (partial pressure of NOCl)
Explanation:
2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) Kp = 51
lets assume the partial pressure of NO,Cl2 , and NOCl at eequilibrium are P1 , P2,and P3 respectively
![Kp=\frac{[NOCl]^{2} }{[NO]^{2} [Cl_2] }](https://tex.z-dn.net/?f=Kp%3D%5Cfrac%7B%5BNOCl%5D%5E%7B2%7D%20%7D%7B%5BNO%5D%5E%7B2%7D%20%5BCl_2%5D%20%7D)
![Kp=\frac{[p3]^{2} }{[p1]^{2} [p2] }](https://tex.z-dn.net/?f=Kp%3D%5Cfrac%7B%5Bp3%5D%5E%7B2%7D%20%7D%7B%5Bp1%5D%5E%7B2%7D%20%5Bp2%5D%20%7D)
p1=0.125atm;
p2=0.165atm;
p3=?
Kp=51;
On solving;
p3=0.36atm (partial pressure of NOCl)
We use Charles's Law: V1/T1=V2/T2
Standard Temperature: 0 degree Celsius= 273K
333.0 degrees Celsius= 606K
Set up: (1.00L)/ (273K)= V2/ (606.0K)
⇒ V2= (1.00L)/ (273K)* (606.0K)= 2.22L
Hope this would help :))
Answer:
Final temperature: 659.8ºC
Expansion work: 3*75=225 kJ
Internal energy change: 275 kJ
Explanation:
First, considering both initial and final states, write the energy balance:
Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:
The pressure is constant, so: 
(There is a multiplication by 100 due to the conversion of bar to kPa)
So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):
On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:
Subtracting the first from the second:
Isolating 
:
Assuming that it is water steam, n=0.1666 kmol
ºC
Answer:
<h2>2. H2O < CH3OH < CH3CH2OH < C9H20</h2>
Explanation:
Decane is a hydrocarbon so it is a non-polar molecule. Non-polar molecules are soluble in non-polar molecules and polar molecules are soluble in polar molecules. Water is a polar molecule so it is least soluble in decane molecule and 
is a non-polar molecule so it is most soluble in decane molecule. Methanol and ethanol are the derivatives of hydrocarbon so they have more solubility than water but less solubility than nonane molecules.
