Answer:
Explanation:
So, just find 55 on the solubility side and with your finger just move to the right untill you touch the line you should get in between 46-48. I would go more for 48.
Hope you have a wonderful day!
The number of moles of Nitrogen : == 7.975
<h3>Further explanation
</h3>
Given
9.60x10²⁴ molecules of nitrogen dioxide (NO2)
Required
The number of moles
Solution
Reaction
N2 + 2O2 → 2NO2
moles of NO2 :
= 9.6 x 10²⁴ : 6.02 x 10²³
= 15.95
From the equation, mol N2 :
= 1/2 x mol NO2
= 1/2 x 15.95
= 7.975
The number of molecules :
= 8 x 6.02 x 10²³
= 4.816 x 10²⁴
Answer : The value of acid dissociation constant is, 
Solution : Given,
Concentration pyridinecarboxylic acid = 0.78 M
pH = 2.53
First we have to calculate the hydrogen ion concentration.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
![2.53=-\log [H^+]](https://tex.z-dn.net/?f=2.53%3D-%5Clog%20%5BH%5E%2B%5D)
![[H^+]=2.95\times 106{-3}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D2.95%5Ctimes%20106%7B-3%7DM)
Now we have to calculate the acid dissociation constant.
The equilibrium reaction for dissociation of (weak acid) is,

initially conc. 0.78 0 0
At eqm. (0.78-x) x x
The expression of acid dissociation constant for acid is:
![k_a=\frac{[C_6H_4NO_2^-][H^+]}{[C_6H_4NO_2]}](https://tex.z-dn.net/?f=k_a%3D%5Cfrac%7B%5BC_6H_4NO_2%5E-%5D%5BH%5E%2B%5D%7D%7B%5BC_6H_4NO_2%5D%7D)
As, ![[H^+]=[C_6H_4NO_2^-]=x](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BC_6H_4NO_2%5E-%5D%3Dx)
So, 
Now put all the given values in this formula ,we get:



Therefore, the value of acid dissociation constant is, 
Answer:
C6H6
Explanation:
We can obtain the molecular formula from the empirical formula.
What we need do here is:
(CH)n = 78
The n shows the multiples of both element present in the actual compound. It can be seen that carbon and hydrogen have the same element ratio here. We then use the atomic masses of both elements to get the value of n. The atomic mass of carbon is 12 a.m.u while the atomic mass of hydrogen is 1 a.m.u
(1 + 12)n = 78
13n = 78
n = 78/13 = 6
The molecular formula is
(CH)n = (CH)6 = C6H6