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S_A_V [24]
3 years ago
9

Balance each of the following redox reactions occurring in basic solution.MnO−4(aq)+Br−(aq)→MnO2(s)+BrO−3(aq)Express your answer

as a chemical equation.Identify all of the phases in your answer.
Chemistry
1 answer:
Ahat [919]3 years ago
8 0

Answer : The balanced chemical equation is,

2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)

Explanation :

Rules for the balanced chemical equation in basic solution are :

  • First we have to write into the two half-reactions.
  • Now balance the main atoms in the reaction.
  • Now balance the hydrogen and oxygen atoms on both the sides of the reaction.
  • If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.
  • If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion (OH^-) at that side where the less number of hydrogen are present.
  • Now balance the charge.

The half reactions in the basic solution are :

Reduction : MnO_4^-(aq)+2H_2O(l)+3e^-\rightarrow MnO_2(s)+4OH^-(aq) ......(1)

Oxidation : Br^-(aq)+6OH^-(aq)\rightarrow BrO_3^-(aq)+3H_2O(l)+6e^-  .......(2)

Now multiply the equation (1) by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)

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Then proceed with the following equations.


100g Na_3N*(\frac{1molNa_3N}{82.98gNa_3N})*(\frac {6mol Na}{2molNa_3N})*(\frac {22.99gNa}{1molNa})=83.12gNa

The answer is 83.12gNa.
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Which numbers are used? Consider the balanced chemical equation that follows. You are asked to determine how many moles of water
velikii [3]
I am assuming that the balanced equation is 
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Since the hydrogen is the limiting reagent (meaning if the hydrogen runs out the reaction stops), we need to compare the mole ratio between hydrogen and water.

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7 0
4 years ago
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The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.13 M, it takes ________ min for
Zigmanuir [339]

Answer:

Therefore it takes 8.0 mins for it to decrease to 0.085 M

Explanation:

First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.

A→ product

Let the concentration of A = [A]

\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]

k=\frac{2.303}{t} log\frac{[A_0]}{[A]}

[A₀] = initial concentration

[A]= final concentration

t= time

k= rate constant

Half life: Half life is time to reduce the concentration of reactant of its half.

t_{\frac{1}{2} }=\frac{0.693}{k}

Here t_{\frac{1}{2} }=0.13 min

k=\frac{0.693}{t_{\frac{1}{2}} }

\Rightarrow k=\frac{0.693}{13 }

To find the time takes for it to decrease to 0.085 we use the below equation

k=\frac{2.303}{t} log\frac{[A_0]}{[A]}

\Rightarrow t=\frac{2.303}{k} log\frac{[A_0]}{[A]}

Here ,   k=\frac{0.693}{13 },  [A₀] = 0.13 m and [ A] = 0.085 M

t=\frac{2.303}{\frac{0.693}{13} } log(\frac{0.13}{0.085})

\Rightarrow t= 7.97\approx 8.0

Therefore it takes 8.0 mins for it to decrease to 0.085 M

7 0
3 years ago
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