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:
The nervous system sends signals to the muscles to shiver when our body temperature begins to drop to a lower than normal temperature. the slight movement of the muscles will work to bring temperature back to homeostasis
Answer:
Second element(Titanium); [Ar] 3d2 4s2
Third element(Vanadium):Ar 3d3 4s2
Explanation:
Given that there are only three d orbitals in universe L instead of five, the electronic configuration of the second and third elements in the first transition series will now look thus;
Second element(Titanium); [Ar] 3d2 4s2
Third transition element(Vanadium):Ar 3d3 4s2
Hence, the electronic configuration of Titanium and Vanadium in universe L is just the same as what it is on earth.
Well you should study how different chemicals work, and make flashcards to carry around or make notes during class to study for later
Answer:
Number of molecules = 1.8267×10^20
Explanation:
From the question, we can deuced that the gases behave ideally, the we can make use of the ideal gas equation, which is expressed below;
PV = nRT
where
P =pressure
V =volume
n = the number of moles
R is the gas constant equal to 0.0821 L·atm/mol·K
T is the absolute temperature
Given:
P = 6.75 atm;
T = 290.0 k,
; V = 1.07 cm³ = 0.001 L
( 6.75 atm)(0.00107 L) = n(0.0821 L·atm/mol·K)(290K)
n = 3.0335167*10^-4 moles
But there are 6.022×10²³ molecules in 1 mole,
Number of molecules = 1.8267×10^20
