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
.
I will have to go with carbon monoxide
Answer:
1. hydrogen - H
2. helium - He
3. sodium - Na
4. magnesium - Mg
5. potassium - K
Explanation:
Hydrogen is the element of group 1 and first period. The atomic number of hydrogen is 1 and the symbol of the element is H.
The electronic configuration of the element hydrogen is:-

Helium is the element of group 18 and first period. The atomic number of helium is 2 and the symbol of the element is He.
The electronic configuration of the element helium is:-

Sodium is the element of group 1 and third period. The atomic number of sodium is 11 and the symbol of the element is Na.
The electronic configuration of the element sodium is:-

Magnesium is the element of group 2 and third period. The atomic number of magnesium is 12 and the symbol of the element is Mg.
The electronic configuration of the element magnesium is:-

Potassium is the element of group 1 and forth period. The atomic number of potassium is 19 and the symbol of the element is K.
The electronic configuration of the element potassium is:-

C16H32O2(aq) --> 16CO2(g) + 16H2O(l) ... said its wrong though?
<span>This is because you haven't added any oxygen needed for the combustion, so your equation does'nt balance. Also a solution in water [aq] doesn't burn! </span>
<span>Try </span><span>C16H32O2(s) + 23O2(g) --> 16CO2(g) + 16H2O(l)
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