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3 years ago

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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Assuming complete distribution, what is the molarity of 10 milligrams of lisinopril in 100 liters?

SVEN [57.7K]

Answer:

$2.47x10^{-7} M$

Explanation:

Lisoprisil's molecular mass is 405.488g/mol, we'll use this fact to calculate molarity, which units are mol/L, and we proceed to the calculus:

First, we'll unify unities, the 10 milligrams of lisinopril we'll transform into grams.

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3 years ago

A 0.500-g sample containing Na2CO3 plus inert matter is analyzed by adding 50.0 mL of 0.100 M HCl, a slight excess, boiling to r

Yakvenalex [24]

The percentage by mass of Na2CO3 in the sample is 48%.

The equation of the reaction of Na2CO3 with HCl;

Na2CO3(aq) + 2HCl(aq) ------> 2NaCl(aq) + H2O(l) + CO2(g)

Since the HCl is in excess, the excess is back titrated with NaOH as follows;

NaOH (aq) + HCl(aq) ---->NaCl(aq) + H2O(l)

Number of moles of HCl added = 0.100 M × 50/1000 L = 0.005 moles

Number of moles of NaOH added = 5.6/1000 × 0.100 M = 0.00056 moles

Since the reaction of NaOH and NaOH is 1:1, 0.00056 moles of HCl reacted with excess HCl.

Amount of HCl that reacted with Na2CO3 = 0.005 moles - 0.00056 moles = 0.0044 moles

Now;

1 mole of Na2CO3 reacts with 2 moles of HCl

x moles of Na2CO3 reacts with 0.0044 moles of HCl

x = 1 mole × 0.0044 moles / 2 moles

x = 0.0022 moles

Mass of Na2CO3 reacted = 0.0022 moles × 106 g/mol = 0.24 g

Percentage of Na2CO3 in the sample = 0.24 g/ 0.500-g × 100/1 = 48%

Learn more about back titration: brainly.com/question/25485091

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