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beks73 [17]
3 years ago
13

Write a balanced half-reaction describing the reduction of gaseous dioxygen to aqueous oxide anions.

Chemistry
1 answer:
maksim [4K]3 years ago
8 0

<u>Answer:</u> The balanced half reaction is written below.

<u>Explanation:</u>

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

X^{n+}+ne^-\rightarrow X

When oxygen gas is reduced to oxide ions, the number of electron transferred are 2

The chemical equation for the reduction of oxygen gas to oxide ions follows:

O_2+2e^-\rightarrow 2O^{2-}

Hence, the balanced half reaction is written above.

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Answer:

Complex  System

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8 0
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What is the Mr of a substance where 1 mole has a mass of 11 g?
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Answer:

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Explanation:

6 0
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When cations and anions join, they form what
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ionic

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6 0
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The diagram below shows two natural processes. Which of these statements is correct about Process 1 and Process 2?
dem82 [27]

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3 0
3 years ago
The half-life of radioactive substance is 2.5 minutes. what fraction of the origional radioactive remains after 10 mins
saul85 [17]
The answer is 1/16.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. (1/2)^{n} = x,
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample

2. t_{1/2} = \frac{t}{n}
where:
<span>t_{1/2} - half-life
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
So, we know:
t = 10 min
<span>t_{1/2} = 2.5 min

We need:
n = ?
x = ?
</span>
We could first use the second equation to calculate n:
<span>If:
t_{1/2} = \frac{t}{n},
</span>Then: 
n = \frac{t}{ t_{1/2} }
⇒ n = \frac{10 min}{2.5 min}
⇒ n=4<span>
</span>
Now we can use the first equation to calculate the remained fraction of the sample.
<span>(1/2)^{n} = x
</span>⇒ x=(1/2)^4
<span>⇒x= \frac{1}{16}</span>
3 0
3 years ago
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