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Neporo4naja [7]
3 years ago
11

There is a 30g of be-11 it has a half-life of about 14 seconds how much will be left in 28 seconds

Chemistry
1 answer:
Lelu [443]3 years ago
6 0

There will be 7.5 g of Be-11 remaining after 28 s.

If 14 s = 1 half-life, 28 s = 2 half-lives.

After the first half-life, ½ of the Be-11 (15 g) will disappear, and 15 g will remain.

After the second half-life, ½ of the 15 g (7.5 g) will disappear, and 7.5 g will remain.

In symbols,

<em>N</em> = <em>N</em>₀(½)^<em>n</em>

where

<em>n</em> = the number of half-lives

<em>N</em>₀ = the original amount

<em>N</em> = the amount remaining after <em>n</em> half-lives

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Combustion is a type of exothermic chemical reaction that generates light and heat in most cases, and occurs fairly quickly.

<h3>What is an exothermic reaction?</h3>

An exothermic reaction is one that when it occurs releases energy in the form of heat or light to the environment.

<h3>Characteristics of an exothermic reaction</h3>

  • When this type of reaction occurs, the products obtained have lower energy than the initial reactants.

  • Most of the exothermic reactions are oxidation, and when they are very violent they can generate fire, as in combustion.

  • Combustion is a very fast oxidation reaction that occurs between materials called fuels and oxygen.

Therefore, we can conclude that combustion is an example of an exothermic reaction that releases large amounts of heat, which can lead to fire.

Learn more about an exothermic reaction here: brainly.com/question/14969584

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2 years ago
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Which of the following statements is true?
EastWind [94]
The first statement is False... as
For exothermic reaction :
A+B》 C+D + HEAT..(heat is considered as a product)... as for endo.. heat is a reactant.
So tjey can't be of the same energy...

2nd one...based on the
A+B》 C+D+HEAT...For exo reaction... the product have more Heat energy than potential...so its false
Recall...energy can nither be created nor destroyed but converted from one form to another....

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3 years ago
9. Given the following reaction:CO (g) + 2 H2(g) CH3OH (g)In an experiment, 0.45 mol of CO and 0.57 mol of H2 were placed in a 1
WITCHER [35]

Answer:

Keq=11.5

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)

We can write the law of mass action as:

Keq=\frac{[CH_3OH]}{[CO][H_2]^2}

That in terms of the change x due to the reaction extent we can write:

Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}

Nevertheless, for the carbon monoxide, we can directly compute x as shown below:

[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\

[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\

[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\

x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M

Finally, we can compute the equilibrium constant:

Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5

Best regards.

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