<u>Answer:</u> The mass of iron in the ore is 10.9 g
<u>Explanation:</u>
We are given:
Mass of iron (III) oxide = 15.6 g
We know that:
Molar mass of Iron (III) oxide = 159.69 g/mol
Molar mass of iron atom = 55.85 g/mol
As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.
To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:
In 159.69 g of iron (III) oxide, mass of iron present is 
So, in 15.6 g of iron (III) oxide, mass of iron present will be = 
Hence, the mass of iron in the ore is 10.9 g
By measuring them, by monitoring them, or by separating them.
Fertilizer use the run on can go into fresh water
1. A
2.C
3.A
4.C
5.B
I really hope this isn't cheating and that you can actually do this for yourself and you just want to double check. To study just read the Scientific Method and memorize it
Answer:The 1st and 2nd reactions are the example of oxidation -reduction.
Explanation:
Oxidation is basically when a species loses electrons and reduction is basically when the species gains electrons.
A reaction is known as an oxidation -reduction reaction only if oxidation and reduction simultaneously occur in the reaction. It basically means if a species is getting oxidized in the reaction then the other species present in the system must be reduced in the reaction.
Oxidation-reduction reactions are also known as redox reactions.
In the 1st reaction the oxidation state of Na in reactant is 0 and in products is +1 hence Na is oxidized and the oxidation state of chlorine is 0 in reactants and in products is -1 so chlorine is reduced. Hence Na is oxidized and Cl is reduced so the reaction is a example of oxidation-reduction.
2Na(s)+Cl₂(g)→2NaCl(s)
In the second reaction the oxidation state of Na in reactant is 0 and in products is +1 hence Na is oxidized and the oxidation state of Cu is +1 in reactant and 0 in products so Cu is reduced. Hence Na is oxidized and Cu is reduced so the reaction is an example of oxidation-reduction.
Na(s)+CuCl(aq)→NaCl(aq)+Cu(s)
In the third reaction the oxidation state of Na changes from +1 to +1 and that of Cu also changes from +1 to +1. So there is no change in oxidation state of the species present in reactants and products. Hence this reaction is not an example of oxidation and reduction.
