Answer:
0.3023 M
Explanation:
Let Picric acid = 
So,
+
⇄
+ 
The ICE table can be given as:
+
⇄
+ 
Initial: 0.52 0 0
Change: - x + x + x
Equilibrium: 0.52 - x + x + x
Given that;
acid dissociation constant (
) = 0.42
![K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cfrac%7B%5BH_3O%5E%2B%5D%5BPicric%5E-%5D%7D%7BH_%7Bpicric%7D%7D)
![0.42 = \frac{[x][x]}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B0.52-x%7D%7D)
![0.42 = \frac{[x]^2}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5E2%7D%7B0.52-x%7D%7D)
0.42(0.52-x) = x²
0.2184 - 0.42x = x²
x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)
Using the quadratic formula;
; ( where +/- represent ± )
= 
= 
=
OR 
=
OR 
=
OR 
= 0.30225 OR - 0.72225
So, we go by the +ve integer that says:
x = 0.30225
x = [
] = [
] = 0.3023 M
∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).
C. Distance from Earth
The observed brightness of a star (the apparent magnitude) to an absolute magnitude, you need to know the distance [d], to the star. Alternatively, if we know the distance and the apparent magnitude [m] of a star, we can calculate its absolute magnitude [d].
*paraphrased from "COSMOS.com"
The answer is volume that is volume limits the precision of data obtained in an experiment .
As while pouring of the liquids, we control the volume, but it can have been off. And this could results in the wrong mole to mole ratio.Thus it limits the precision, So the answer is that volume limits the precision of data obtained in an experiment .
The given question is incomplete. The complete question is:
When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.
Answer: The vant hoff factor for sodium chloride in X is 1.9
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
= freezing point constant
i = vant hoff factor = 1 ( for non electrolyte)
m= molality =

Now Depression in freezing point for sodium chloride is given by:
= Depression in freezing point
= freezing point constant
m= molality =


Thus vant hoff factor for sodium chloride in X is 1.9