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)
Answer:
c i guess, decrease temperature
Just look it up you’ll get the answers faster
Multiply the volume by the density. 1 ml = 1 cm^3
1.11 g/cm^3 * 1 cm^3/mL * 386 mL = 428 g
All your units cancel out except for g which is what you would expect since you want units of mass.