Answer:
pH = 11.216.
Explanation:
Hello there!
In this case, according to the ionization of ammonia in aqueous solution:

We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:
![Kb=\frac{[NH_4^+][OH^-]}{[NH_3]} \\\\1.80x10^{-5}=\frac{x*x}{0.150-x}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D%20%5C%5C%5C%5C1.80x10%5E%7B-5%7D%3D%5Cfrac%7Bx%2Ax%7D%7B0.150-x%7D)
However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:

Which is also:
![[OH^-]=1.643x10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.643x10%5E%7B-3%7DM)
Thereafter we can compute the pOH first:

Finally, the pH turns out:

Regards!
Answer:
The correct option is: stable environment
Explanation:
According to the Darwin's theory, natural selection is the concept by which all the small useful variations of traits are preserved.
According to Darwin, there are three <u>necessary and sufficient conditions</u> for the occurrence of natural selection:
1. struggle for existence
2. variation
3. inheritance
These conditions are said to be necessary because if these conditions are not satisfied then natural selection does not occur.
These conditions are said to be sufficient because if these conditions are satisfied, then natural selection will most definitely occur.
I don’t! Know bit I guess it’s C
The rate constant of the reaction K we can get it from this formula:
K=㏑2/ t1/2 and when we have this given (missing in question):
that we have one jar is labeled t = 0 S and has 16 yellow spheres inside and the jar beside it labeled t= 10 and has 8 yellow spheres and 8 blue spheres and the yellow spheres represent the reactants A and the blue represent the products B
So when after 10 s and we were having 16 yellow spheres as reactants and becomes 8 yellow and 8 blue spheres as products so it decays to the half amount so we can consider T1/2 = 10 s
a) by substitution in K formula:
∴ K = ㏑2 / 10 = 0.069
The amount of A (the reactants) after N half lives = Ao / 2^n
b) so no.of yellow spheres after 20 s (2 half-lives) = 16/2^2 = 4
and the blue spheres = Ao - no.of yellow spheres left = 16 - 4 = 12
c) The no.of yellow spheres after 30 s (3 half-lives) = 16/2^3 = 2
and the blue spheres = 16 - 2 = 14