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3 years ago

Study this chemical reaction:

Chemistry

1 answer:

photoshop1234 [79]3 years ago

4 0

Answer:

The oxidation reaction: Ni → Ni⁴⁺ + 4e.

The reduction reaction: 2Cl₂ + 4e → 4Cl⁻.

Explanation:

The mentioned reaction: Ni + 2Cl₂ → NiCl₄.

It is an oxidation-reduction reaction comes from two half reactions.

The oxidation reaction: in it Ni losses 4 electrons and is oxidized to Ni⁴⁺ according to the reaction:

Ni → Ni⁴⁺ + 4e.

The reduction reaction: every molecule of Cl₂ will gain 2 electrons and is reduced to 2Cl⁻ according to the reaction:

2Cl₂ + 4e → 4Cl⁻.

The summation of the two-half reactions will give the reaction:

Ni + 2Cl₂ → NiCl₄.

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The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate

Tems11 [23]

Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

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One problem with elimination reactions is that mixtures of products are often formed. For example, treatment of 2-bromo-2-methyl

NikAS [45]

Answer:

Structures are given below.

Explanation:

Treatment of 2-bromo-2-methylbutane with KOH in ethanol will give elimination of HBr through E2 mechanism.
H atoms adjacent to Br will be eliminated.
2-bromo-2-methylbutane has two possible adjacent H atoms that can be eliminated giving mixture of products.
Product of this elimination reaction is alkene. Here saytzeff fule is followed during elimination. So most substituted alkene will be major product.
Structure of alkenes are given below.

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What information would you find on a WHIMIS label?

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Answer/explanation:

Product identifier – the brand name, chemical name, common name, generic name or trade name of the hazardous product.
Initial supplier identifier – the name, address and telephone number of either the Canadian manufacturer or the Canadian importer.

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Li2O is the formula for lithium oxide

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