I’ve been having problems too honestly lol
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)
You want to achieve the lowest number possible
Because <span>3-aminopropan-2-ol is bigger than </span><span>1-aminopropan-2-ol number wise, so you should name it as </span>1-aminopropan-2-ol
Have a wonderful day :) thanks for the points