Answer:
(a).
![order \: of \: A = 2 \\ order \: of \:B = 0 \\ rate = k[A] {}^{2} [B]](https://tex.z-dn.net/?f=order%20%5C%3A%20of%20%5C%3A%20A%20%3D%202%20%5C%5C%20order%20%5C%3A%20of%20%5C%3AB%20%20%3D%200%20%5C%5C%20rate%20%3D%20k%5BA%5D%20%7B%7D%5E%7B2%7D%20%5BB%5D)
(b)

(c). The rate increases because kinetic energy of reactant molecules increases.
Answer:
3.24
Explanation:
The dissociation equation for the carboxylic acid can be represented as follows:
RCOOH —-> RCOO- + H+
We can use an ICE table to get the value of the concentration of the hydrogen ion. ICE stands for initial, change and equilibrium.
RCOOH RCOO- H+
Initial 0.2 0.0. 0.0
Change -x +x. +x
Equilibrium 0.2-x. x. x
We can now find the value of x as follows:
Ka = [RCOO-][H+]/[RCOOH]
(1.66* 10^-6) = (x * x)/(0.2-x)
(1.66 * 10^-6) (0.2-x) = x^2
x^2 = (3.32* 10^-7) - (1.66*10^-6)x
x^2 + (1.66 * 10^-6)x - (3.32* 10^-7) = 0
Solving the quadratic equation to get x:
x = 0.0005753650094369094 or - 0.0005753650094369094
As concentration cannot be negative, we discard the negative answer
Hence [H+] = 0.0005753650094369094
By definition, pH = -log[H+]
pH = -log(0.0005753650094369094)
pH = 3.24
Answer:
He traveled at a speed of 4 m/s with a velocity of 4m/s north, and his acceleration changed when stopping for lunch.
Explanation:
Trust me took test got it right!
Sir cn you tell me that can we use this formula
Given that solubility product of AgCl = 1.8 X 10^-10
Dissociation of AgCl can be represented as follows,
AgCl(s) ↔ Ag+(ag) + Cl-(aq)
Let, [Ag+] = [Cl-] = S
∴Ksp = [Ag+][Cl-] = S^2
∴ S = √Ksp = √(1.8 X 10^-10) = 1.34 x 10^-5 mol/dm3
Now, Molarity of solution =

∴ 1.34 x 10^-5 =

∴ Weight of AgCl present in solution = 1.92 X 10^-3 g
Thus,
mass of AgCl that will dissolve in 1l water = 1.92 x 10^-3 g