Answer:
By a factor of 12
Explanation:
For the reaction;
A + 2B → products
The rate law is;
rate = k[A]²[B]
As you can see, the rate is proportional to the square of the concentration of A and the of the concentration of B
.
Let's say initially, [A] = x, [B] = y
The rate law in this case is equal to;
rate1 = k. x².y
Now you double the concentration of A and triple the concentration of B.
[A] = 2x, [B] = 3y
The new rate law is given as;
rate2 = k . (2x)². (3y)
rate2 = k . 4x² . 3y
rate2 = 12 k . x² . y
Comparing rate 2 and rate 1, the ratio is given as; rate 2/ rate 1 = 12
Therefore the rate has increased by a factor of 12.
Answer:
CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
Explanation:
To answer this question we must know Kb of CH3CH2NH2 is 5.6x10⁻⁴, and for C6H5NH2 is 4.0x10⁻¹⁰. And the CH3CH2NH3+ and C6H5NH3+ are related with these substances because are their conjugate base. That means:
pKa of CH3CH2NH3+ = CH3CH2NH2; C6H5NH3+ = C6H5NH2
Also, Kw / Kb = Ka
Thus:
pKa of CH3CH2NH3+/CH3CH2NH2 is:
Kw / kb = Ka = 1.79x10⁻¹¹
-log Ka = pKa
pKa = 10.75
pKa of C6H5NH3+/ C6H5NH2 is:
Kw / kb = Ka = 2.5x10⁻⁵
-log Ka = pKa
pKa = 4.6
That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
We know that, M1V1 = M2V2
(Initial) (Final)
where, M1 and M2 are initial and final concentration of soution respectively.
V1 and V2 = initial and final volume of solution respectively
Given: M1 = 12 m, V1 = 35 ml and V2 = 1.2 l = 1200 ml
∴ M2 = M1V1/V2 = (12 × 35)/ 1200 = 0.35 m
Final concentration of solution is 0.35 m
100% find the gfm of both sides then divide
