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4 years ago

Which of the following is not an oxidation-reduction reaction?

Chemistry

2 answers:

fredd [130]4 years ago

6 0

Answer : The option (C) is not an oxidation-reduction reaction.

Explanation :

A : $XeF_6(s)\righatrrow XeF_4(s)+F_2(g)$

This is an oxidation-reduction reaction. In this reaction, the oxidation state of 'Xe' changes from (+6) to (+4) and the oxidation state of 'F ' changes from (-1) to (0). That means 'Xe' shows reduction and 'F' shows oxidation in the reaction.

B : $2Cs(s)+I_2(g)\righatrrow 2Csl(s)$

This is an oxidation-reduction reaction. In this reaction, the oxidation state of 'Cs' changes from (0) to (+1) and the oxidation state of 'I' changes from (0) to (-1). That means 'I' shows reduction and 'Cs' shows oxidation in the reaction.

D : $Zn(s)+2AgNo_3(aq)\rightarrow Zn(NO_3)_2(aq)+Ag(s)$

This is an oxidation-reduction reaction. In this reaction, the oxidation state of 'Zn' changes from (0) to (+2) and the oxidation state of 'Ag' changes from (+1) to (0). That means 'Ag' shows reduction and 'Zn' shows oxidation in the reaction.

C : $2H_2SO_4(aq)+2Ba(OH)_2(aq)\rightarrow 2BaSO_4(s)+4H_2O(I)$

This is not an oxidation-reduction reaction. In this reaction, the oxidation states are not changes.

Hence, the option (C) is not an oxidation-reduction reaction.

saveliy_v [14]4 years ago

3 0

An oxidation-reduction reaction happens when hydrogen, oxygen or halogen atoms are transferred, so in this case, your answer would be B.

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Determine the molecular formula for the compound with a molar mass of 46.08 g/mol and the following percent compostion:

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Answer:

$\large\boxed{\large\boxed{CH_6N_2}}$

Explanation:

1. First determine the empirical formula.

a) Base: 100 g of compound

mass atomic mass number of moles

g g/mol mol

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H 13.13 1.008 13.13/1.008 = 13.03

N 60.81 14.007 60.81/14.007 = 4.34

b) Divide every number of moles by the smallest number: 2.17

mass number of moles proportion

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H 13.03/2.17 6

N 4.34/2.17 2

c) Empirical formula

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d) Mass of the empirical formula

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2. Molecular formula

Since the mass of one unit of the empirical formula is equal to the molar mass of the compound, the molecular formula is the same as the empirical formula:

$CH_6N_2$

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