Electron affinity for fluorine is than chlorine most likely , due to the electron repulsion that occur between the electron where n= 2 . the elements in the second period have such small electron clouds that electron repulsion is greater than that of the rest of the family.
The answer to the question is a.
Answer:
a) Kb = 10^-9
b) pH = 3.02
Explanation:
a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:
[NaA] and [A-] = 0.05/0.6 = 0.083 M
Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9
b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:
[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M
pKb = 10^-9
Ka = 10^-5
HA = H+ + A-
Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])
[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0
Clearing [H+]:
[H+] = 0.00095 M
pH = -log([H+]) = -log(0.00095) = 3.02
Answer:
B - (C , Al, P, Cl)
Explanation:
How I got this answer was by looking at my periodic table it shows you how much it contains by the Atomic number.
Atomic number on C (Carbon) is- 6
Atomic number on Al (Aluminum) is - 13
Atomic number on P (Phosphorus) is - 15
Atomic number on Cl (Chlorine) is - 17
Now it says least to greatest and the other options are wrong I did the work for you hope this helps :)) I also had this project you didnt ask but the answer for the The Lesson are {B E M S} which as the code numbers are gonna be -7494- Im glad to help if you need more help I will give you the other answers as well :) !
