ANSWER:
4 a) Specific elements have more than one oxidation state, demonstrating variable valency.
For example, the following transition metals demonstrate varied valence states:
,
,
, etc.
Normal metals such as
also show variable valencies. Certain non-metals are also found to show more than one valence state 
4 b) Isotopes are members of a family of an element that all have the same number of protons but different numbers of neutrons.
For example, Carbon-14 is a naturally occurring radioactive isotope of carbon, having six protons and eight neutrons in the nucleus. However, C-14 does not last forever and there will come a time when it loses its extra neutrons and becomes Carbon-12.
5 a)
→
5 b)
→ 
5 c)
→
(already balanced so don't need to change)
5 d)
→
5 e)
→ 
EXPLANATION (IF NEEDED):
1. Write out how many atoms of each element is on the left (reactant side) and right (product side) of the arrow.
2. Start multiplying each side accordingly to try to get atoms of the elements on both sides equal.
EXAMPLE OF BALANCING:
<u>Given information:</u>
Concentration of HCl = 0.035 M
<u>To determine:</u>
pH of the solution
<u>Explanation:</u>
Hydrochloric acid, HCl is a strong acid. It will completely dissociate to give H+ and Cl- ions
HCl → H+ + Cl-
Hence the concentration of H+ = Cl- = 0.035M
Now, pH measures the strength of H+ ions in a given solution. It is expressed as:
pH = -log[H+]
pH (HCl) = -log(0.035) = 1.46
Ans: pH of 0.035M HCl is 1.46
Answer:
C. Lithium
Explanation:
I Goo gled it and I think that's right.
Answer:
0.1035 M
Explanation:
Considering:
Sodium chloride will furnish Sodium ions as:
Given :
For Sodium chloride :
Molarity = 0.288 M
Volume = 3.58 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 3.58×10⁻³ L
Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:
Moles of sodium ions by sodium chloride = 0.00103104 moles
Sodium sulfate will furnish Sodium ions as:
Given :
For Sodium sulfate :
Molarity = 0.001 M
Volume = 6.51 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 6.51 ×10⁻³ L
Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:
Moles of sodium ions by Sodium sulfate = 0.00001302 moles
Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles
Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L
Concentration of sodium ions is:
<u>
The final concentration of sodium anion = 0.1035 M</u>