Answer:
The first one.

Explanation:
When comparing two fractions with variables like this, it's important to get to the same denominator in order to compare apples with apples and then be able to do not only comparisons but also perform additions/subtractions.
Question is which denominator to use and how to reach it.
In this case, the question and the answer choices do the work for you. The question asks which one is the LEAST common denominator, and the answers show denominators x² and 4x². The smallest of these is x², however, we can't simplify the first fraction to get to the x² denominator, so we'll go for the 4x².
So, the first fraction has already the correct denominator (4x²), we just have to transform the second one.
We multiply it by 1, expressed in a different way. Since we're multiplying by one, we're not affecting the value, just the way it looks.
Let's do it!, to get the denominator to go from x² to 4x², we need to multiply it by 4... so we'll multiply by 4/4 (which is 1, neutral for the multiplication).

And now you have both fractions on the same denominator, without having changed their value, just their looks
In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
<span>2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l)
</span><span>E = +1.47
</span>
<span>Br(l) + 2e- = 2Br-
</span><span>E = +1.065
</span>
We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V
Answer:

Explanation:
If l = 3, the electrons are in an f subshell.
The number of orbitals with a quantum number l is 2l + 1, so there
are 2×3 + 1 = 7 f orbitals.
Each orbital can hold two electrons, so the f subshell can hold 14 electrons.

Answer:
Explanation:
Reaction given
6 H⁺ + 2 MnO₄⁻ + 5 (COOH)₂ = 10CO₂ +8H₂O + 2 Mn⁺²
Oxidation number of Mn in MnO₄⁻
= x - 4 x 2 = -1
x = 8 -1
+ 7
Oxidation no of Mn in Mn⁺² = +2
So its oxidation no is decreased from + 7 to + 2 . Hence it is reduced.