The complete question is as follows: The equilibrium constant, Kp, for the following reaction is 10.5 at 350 K: 2CH2Cl2(g) CH4(g) + CCl4(g) Calculate the equilibrium partial pressures of all species when CH2Cl2(g) is introduced into an evacuated flask at a pressure of 0.968 atm at 350 K. PCH2Cl2 = atm PCH4 = atm PCCl4 = atm
Answer: The equilibrium partial pressures of all species, that is, 
, 
and 
is 0.420 atm, 0.420 atm and 0.128 atm.
Explanation:
For the given reaction equation, the initial and equilibrium concentration of involved species is as follows.
Initial: 0.968 atm 0 0
Equilibrium: (0.968 - 2x) x x
Now, 
for this reaction is as follows.
Thus, we can conclude that the equilibrium partial pressures of all species, that is, , and is 0.420 atm, 0.420 atm and 0.128 atm.
Explanation:
hope it is helpful to you ☺️
I'm not sure if this is your question but with NaCl (salt) and water it is homogeneous because when mixed, all the salt particles are even throughout and diffuse throughout the water.
ATP is formed
NADH releases electrons
Oxygen Accepts electrons
Answer:
0.13 M ( 2 s.f)
Explanation:
2Cl2O5 (g)-->2Cl2(g) +5O2 (g)
rate= (17.4 M -1 .s -1 ) [Cl2O5]2
From the rte above, we can tell that our rate constant (k) = 17.4 M -1 .s -1
The units of k tells us this is a second order reaction.
Initial Concentration [A]o = 1.46M
Final Concentration [A] = ?
Time = 0.400s
The integrated rate law for second order reactions is given as;
1 / [A] = (1 / [A]o) + kt
1 / [A] = [ (1/ 1.46) + (17.4 * 0.4) ]
1 / [A] = 0.6849 + 6.96
1 / [A] = 7.6496
[A] = 1 / 7.6496
[A] = 0.13073 M ≈ 0.13 M ( 2 s.f)
