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Hey guys, Here are few questions from final review that I know the answers to bu

ID: 1000873 • Letter: H

Question

Hey guys,

Here are few questions from final review that I know the answers to but I simply dont know the reasoning behind:

1) A mixture of 1.0 mol He and 1.0 mol Ne are at STP in a rigid container. Which of the following statements is TRUE?

A) Both gases have the same average kinetic energy.

B) Both gases contribute equally to the density of the mixture under these conditions.

C) Both gases have the same molecular speed.

D) The mixture has a volume of 22.4 L

E) All of the above are TRUE.

2) How many valence electrons does an atom of S have?

A) 3

B) 1

C) 2

D) 4

E) 6

3)Give the ground state electron configuration for Pb.

A) [Xe]6s26p2

B) [Xe]6s25d106p2

C) [Xe]6s25f146d106p2

D) [Xe]6s24f145d106p2

E) [Xe]6s24f145d106s26p2

4) Which Brønsted-Lowry acid is not considered to be a strong acid in water?

A) HBr

B) HCl

C) HNO2

D) HClO4

E) H2SO4

112) A solution with a hydroxide ion concentration of 4.15 × 10-4 M is _____________.

A) acidic

B) basic

C) neutral

D) 2.41 × 10-11 M

E) not enough information

113) What is the pH of a 0.040 M HClO4 solution?

A) 0.040

B) 0.080

C) 1.40

D) 12.60

E) 7

114) What is the pH of a 0.040 M Ca(OH)2 solution?

A) 1.10

B) 1.40

C) 12.60

D) 12.90

E) 7.00

Explanation / Answer

1) (A) Both gases have the same average kinetic energy.

2) (E) 6 16 total and 6 in the outermost shell

3) D) [Xe]6s24f145d106p2

Elements Pb.Lead: properties of free atoms. Lead atoms have 82 electrons and the shell structure is 2.8.18.32.18.4. The ground state electron configuration of ground state gaseous neutral lead is [Xe].4f14.5d10.6s2.6p2 and the term symbol is 3P0.

4) E) H2SO4

112) D) 2.41 × 10-11 M

The equation to use is: [H+] [OH-] = 10^-14
You have [OH-] = (4.15*10^-6)
Substitute into the equation and solve for [H+]
[H+] * (4.15*10^-6) = 10^-14
[H+] = 10^-14 / (4.15*10^-6)
[H+] = 2.41*10^-9M

113) (A) 0.040

HClO4 is a strong acid => pH = - log 0.040 = 1.40
[H+] =10^-1.40 = 0.0398
CCl3CO2H <------> CCl3CO2- + H+ at equilibrium

114) (C) 12.60

Ca(OH)2 is a strong base and completely dissolves into: Ca(OH)2 -> Ca+2 + 2 OH-. For every 1 Ca(OH)2 you will get 2 OH-. That means that 0.020 M Ca(OH)2 produces 0.040 M OH-.

At 25ºC, Kw = [H+][OH-], so 1.0 x 10^-14 = [H+] x 0.040. [H+] = 1.0 x 10^-14/0.040 = 2.5 x 10^-13

pH = -log[H+]. Putting in our answer from above into this equation gives us pH = -log (2.5 x 10^-13). This gives us a pH = 12.6