Answer:
2NO(g) + O2(g) --> 2NO2(g)
now 400 ml of NO × 2 mol of NO2/2 mol of NO
= 400 ml of NO2
now 500 ml of O2 × 2 mol of NO2/1 mol of O2
= 1000 ml of NO2
now 400 ml of NO2 × 1 mol of O2/2 mol of NO
= 200 ml
subtract that from 500 ml of total i.e. 500-200 =300 ml
The total volume of the reaction mixture is 1000 ml -300ml = 700 ml
Answer:
I believe it's D
Explanation:
they both are worked a lot and it makes the most sense.
Answer : The correct option is, (C) 1.1
Solution : Given,
Initial moles of 
= 1.0 mole
Initial volume of solution = 1.0 L
First we have to calculate the concentration .
The given equilibrium reaction is,
Initially c 0
At equilibrium 
The expression of 
will be,
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
where,
= degree of dissociation = 40 % = 0.4
Now put all the given values in the above expression, we get:
Therefore, the value of equilibrium constant for this reaction is, 1.1
Answer:
c. Forms different hydrates which have different colors
Explanation:
So half life is the time taken for a sample to decay to half its original mass, its a constant and applies to any original mass, it could be 5g or 1kg, it will take the same amount of time for the original mass to half. In this case the half life is 3 days.
After 3 days the sample will be at half its original mass, now 50g.
Now we can treat the 50g as if its a new sample. After another 3 days (6 days in total) there will be half of 50g left, = 25g.
