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 pressure calculated as the difference between the net hydrostatic pressure and the net colloid osmotic pressure is known as: filtration pressure.
<h3>What is pressure?</h3>
Pressure can be defined as a measure of the force exerted per unit area of an object or body. Thus, it is usually measured in Newton per meter square.
<h3>The types of pressure.</h3>
In Science, there are different types of pressure and these include the following:
Filtration pressure is a pressure that is typically calculated as the difference between the net hydrostatic pressure and the net colloid osmotic pressure. Also, it promotes the filtration of fluid through a membrane.
Read more on pressure here: brainly.com/question/24827501
The answer is six because it is in group six A on the periodic table.
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K = 
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M
Answer:
C. P = nRT
Explanation:
PV = nRT, where n is a number of moles and R is the universal gas constant, R = 8.31 J/mol ⋅ K.
Hope this helps :)