How many liters of 0.1107 M KCI contains 15.00 g of KCI (FW= 74.6 g/mol)

The oxidation number is also called as oxidation state. The oxidation number of oxygen in the carbonate ion found in malachite is -2.

What is oxidation number ?

The charge that an atom would have if all of its links to other atoms were fully ionic is known as the oxidation state, also known as the oxidation number. It describes how much an atom in a chemical molecule has been oxidized. The oxidation state might theoretically be positive, negative, or zero.

Antoine Lavoisier used the term "oxidation" to describe the interaction of a material with oxygen. The definition was eventually expanded to cover additional processes in which electrons are lost, regardless of whether oxygen was present, because it was understood that the material loses electrons when it is oxidized.

Thus, The oxidationnumber of oxygen in the carbonate ion found in malachite is -2.

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Answer:

76.3 g of 1 mole of element Z

Explanation:

According to Avogadro’s number the number of particles in 1 mole of a substance is 6.023 x 10²³.  

The mass of 3 x 10²³ particles is 38g  

Therefore, the mass of 6.023 x 10²³ is = 76.3 g

The molarity of the NaOCl solution is 0.10M

The correct answer to the question is Option C. 0.10 M

We'll begin by calculating the number of mole of NaOCl required to produce 0.010 mole of Cl₂.This can be obtained as follow:

NaOCl + 2HCl —> Cl₂ + NaCl + H₂O

From the balanced equation above,

1 mole of NaOCl reacted to produce 1 mole of Cl₂.

Therefore

0.010 mole of NaOCl will also react to produce 0.010 mole of Cl₂.

• Finally,we shall determine the molarity of the NaOCl solution.

Mole of NaOCl = 0.01 mole

Volume = 100 mL = 100 / 1000 = 0.1 L

Molarity of NaOCl =?

Molarity = mole / Volume

Molarity of NaOCl = 0.01 / 0.1

Molarity of NaOCl = 0.1 M

Therefore, the molarity of the NaOCl solution is 0.1 M

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Answer:

Option C 0.10 M

Explanation:

you are not providing the reaction, but luckily we have data to write the equation. We know that we are taking HCl and NaOCl to produce Cl2, the reaction is as follow:

NaOCl + 2HCl ---------> Cl₂ + NaCl + H₂O

From this reaction, we know that we have excess of acid that is added to 100 mL (0.1 L) of NaOCl, and this produces 0.010 moles of Chlorine.

All we have to do here, is apply stechiometry. We already know that the acid is in excess, so the moles produced would be the moles of the limiting reactant, in this case, the NaOCl.

According to the reaction NaOCl and Cl2 has a mole ratio of 1:1, so the moles of Cl2 would be the same moles that NaOCl lost so:

moles Cl2 = moles NaOCl = 0.010 moles

Then the concentration:

M = 0.010 / 0.1

M = 0.10 M

This is the molarity of NaOCl