At the end of the reaction, the catalyst is UNCHANGED.
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Answer: Reducing agent in the given reaction is 
.
Explanation:
A reducing agent is defined as an element which tends to lose electrons to other element leading to an increase in its oxidation number.
In the given reaction, oxidation state of sulfur in is +2 and 
has 0 oxidation state.
In 
oxidation state of S is 2.5 and in 
oxidation state of I is -1.
Since, an increase in oxidation state of S is occurring from +2 to +2.5. Hence, it is acting as a reducing agent.
Thus, we can conclude that reducing agent in the given reaction is .
Answer:
79.04 L
Explanation:
We are given;
Initial Volume; V1 = 6.24L
Initial Pressure; P1 = 760 mm Hg
Final pressure; P2 = 60.0mm Hg
To solve for final volume, we will use Boyles law;
P1•V1 = P2•V2
Let's make V2 which is the final volume the subject;
V2 = (P1•V1)/P2
V2 = (760 × 6.24)/60
V2 = 79.04 L
A low electronegativity
Explanation:
Potassium is a metal that is expected to have a very low electronegativity value.
Electronegativity is the relative tendency by which an atom attracts valence electrons in a chemical bond.
Potassium is an element in the first group on the periodic table.
The common trend is that electronegativity increases from left to right and decreases down a group.
- Potassium as metal will prefer to lose electrons rather than attracting because that will make it achieve the octet configuration that will ensure its stability.
- This is why it will have low electronegativity.
Learn more:
Electronegativity brainly.com/question/11932624
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