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:
1st page, number a-ENergy is added or absorbed
I cant see the answers on the 2nd page
Explanation:
Glucose + Oxygen > Carbon Dioxide
C(6)H(12)0(6)+ 0(2) >CO(2)
Once the torch is lit, the acetylene flow must be increased until the flame stops smoking <span>before the oxygen is turned on for adjustment in order to keep the tip of the torch cool.
You should also note that while lighting the torch, you should keep the spark lighter near the tip but not covering it.</span>
