You will need to know the volume and the concentration of the stock dilution in order to find the concentrations of the rest of the solutions you create.
The equation given in question is as follow,
ZnS + NO₃⁻ → Zn²⁺ + S + NO
Oxidation state of elements in reactant side are as follow,
Zn = +2
S = -2
N = +5
O = -2
Oxidation state of elements in product side are as follow,
Zn = +2
S = 0
N = +2
O = -2
So, In this reaction Nitrogen is Reduced while Sulfur is Oxidized.
Now Split the reaction into two half cell reactions,
Reduction Reaction,
ZnS → S + 2e⁻
Oxidation Reaction,
NO₃⁻ + 3e⁻ → NO
As the oxygen atoms are not balance, So, in acidic medium add H⁺ on the side having greater number of Oxygen atom and H₂O on the side having less number of Oxygen atoms,
Hence,
H⁺ + NO₃⁻ + 3e⁻ → NO + H₂O
Now Balance the reaction,
4H⁺ + NO₃⁻ + 3e⁻ → NO + 2H₂O
So, write both half cell equations as,
ZnS → S + 2e⁻ --------- (1)
4H⁺ + NO₃⁻ + 3e⁻ → NO + 2H₂O -------(2)
Multiply eq. 1 with 3 and eq. 2 with 2 to equalize the electron.
So,
3ZnS → 3S + 6e⁻
8H⁺ + 2NO₃⁻ + 6e⁻ → 2NO + 4H₂O
_____________________________________(e⁻ cancelled)
3ZnS + 2NO₃⁻ + 8H⁺ → 3Zn + 3S + 2NO + 4H₂O
0.409 g!
Pls mark brainlest.!:)
Answer:
1. Mixture
2. Compound
3. Compound
4. Element
Explanation:
Number 1 shows two completely different types of compounds in the same space, so it is a mixture.
Numbers 2 and 3 are compounds because there aren't different types compounds as in Number 1, but the same compound repeated. 2 is a compound because while all of the circles are the same element, they are connected. O2 for example, oxygen, is a compound.
Number 4, finally, is all of the same substance, non-connected, so it shows an element.
Brainliest, please :) (Trying to become a genius so that I can continue helping more people)
