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yawa3891 [41]
3 years ago
10

For each reaction, write the chemical formulae of the oxidized reactants in the space provided. Write the chemical formulae of t

he reduced reactants in the space provided.
Chemistry
1 answer:
liberstina [14]3 years ago
5 0

Explanation:

First Reaction;

Ca + ZnCl2 --> CaCl2 + Zn

Oxidized Reactant: Ca. There is increase in oxidation number from 0 to +2

Reduced Reactant: Zn. There is decrease in  oxidation number form +2 to 0

Second Reaction:

FeI2 + Mg --> Fe + MgI2

Oxidized Reactant: Mg. There is increase in oxidation number from 0 to +2

Reduced Reactant: Fe. There is decrease in  oxidation number form +2 to 0

Third Reaction;

Mg + 2AgNO3 --> Mg(NO3)2 + Ag

Oxidized Reactant: Mg. There is increase in oxidation number from 0 to +2

Reduced Reactant: Ag. There is decrease in  oxidation number form +1 to 0

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How many atoms of oxygen are present in 7.51 grams of<br> glycine with formula C₂H5O2N?
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1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.

How to calculate number of atoms?

The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

However, the number of moles of oxygen in glycine can be calculated using the following expression:

Molar mass of C₂H5O2N = 75.07g/mol

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Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles

Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms

Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.

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Why is air pumped through the chamber instead of just letting the fuel use the air that is
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