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erastovalidia [21]
3 years ago
14

PLEASE HELP 30 POINTS

Chemistry
2 answers:
Annette [7]3 years ago
5 0

Answer:

1) Only Organism Z is alive, and Organism X has been dead longer than Organism Y.

2) S (sulfur) was oxidized and N (nitrogen) was reduced.

3) Chlorine is reduced in reaction 1 and iron is oxidized in reaction 2.

Explanation:

1) After death of an organism there is radioactive disintegration of C-14 allotrope of carbon, which increases the ratio of C-12 to C-14 in a dead organism as compared to a living organism.

Longer the time of death, more the ratio of the two isotopes.

So as the ratio of isotopes is same in Z, it is alive.

The ratio of isotopes if higher in X than Y,X has been dead longer than Organism Y.

2) oxidation is increase in  oxidation state due to loss of electrons and reduction is decrease in oxidation sate due to gain of electrons.

Here oxidation state of hydrogen is same on both the sides of equation.

The oxidation state of sulfur increases and that of nitrogen decreases so sulfur has undergone oxidation while nitrogen reduction.

3) As mentioned above, the oxidation state of iron increases (0 to +2) so it is undergoing oxidation while that of chlorine decreases (0  to -1) so it is undergoing reduction.

posledela3 years ago
4 0

The answears are in the attached photo.

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The concentration of solution in M or mol/L can be calculated using the following formula:

C=\frac{n}{V} .... (1)

Here, n is number of moles and V is volume of solution in L.

The molecular formula of potassium sulfate is K_{2}SO_{4} thus, there are 2 moles of potassium in 1 mol of potassium sulfate.

1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass

Putting the values,

n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol

Thus, number of moles of  K_{2}SO_{4} will be 0.1061\times 0.5=0.053 mol.

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1 mL=10^{-3}L

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225 mL=0.225 L

Putting the values in equation (1),

C=\frac{(0.053 mol}{0.225 L}=0.236 M

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