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)
<u>Answer:</u> The entropy change of the ethyl acetate is 133. J/K
<u>Explanation:</u>
To calculate the number of moles, we use the equation:

Given mass of ethyl acetate = 398 g
Molar mass of ethyl acetate = 88.11 g/mol
Putting values in above equation, we get:

To calculate the entropy change for different phase at same temperature, we use the equation:

where,
= Entropy change = ?
n = moles of ethyl acetate = 4.52 moles
= enthalpy of fusion = 10.5 kJ/mol = 10500 J/mol (Conversion factor: 1 kJ = 1000 J)
T = temperature of the system = ![84.0^oC=[84+273]K=357K](https://tex.z-dn.net/?f=84.0%5EoC%3D%5B84%2B273%5DK%3D357K)
Putting values in above equation, we get:

Hence, the entropy change of the ethyl acetate is 133. J/K
Answer:When a strontium atom loses two electrons, it becomes a(n) cation with a charge of 2+. when any neutral atom loses an electron it becomes cation that is positively charged ions and an ion gets the charge according to the number of electrons loses.
you are very welcome but I may be inaccurate.