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guapka [62]
3 years ago
7

Why does a catalyst work for a long time before it needs replacing?

Chemistry
1 answer:
DedPeter [7]3 years ago
7 0
Because the catalyst is not really part of the reaction. it is something that speed up a reaction by lowering the energy need for the reaction to take place. however, in the end the catalyst is brought back to its initial state. that's why it is long lasting
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PLEASE ANSWER THIS QUESTION ASAP 50 POINTS
quester [9]

Answer:

A very large amount of energy is produced from a series of chemical reactions.

Explanation:

Nuclear fission is the process of splitting apart nuclei (usually large nuclei). When large nuclei, such as uranium-235, fissions, energy is released. So much energy is released that there is a measurable decrease in mass, from the mass-energy equivalence. This means that some of the mass is converted to energy.

4 0
3 years ago
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Use a sheet of paper to answer the following question. Take a picture of your answers and attach to this assignment. Draw the st
pentagon [3]

Answer:

Explanation:

Structure of the 2,2,4,4-tetramethyl-3-pentanone is give in the attachment

In 2,2,4,4-tetramethyl-3-pentanone, no alpha hydrogen is present, therefore, enol form is not possible and hence, exist only in keto form.

Explanation for existence of cyclohexa-2,4-diene-1-one only in enol form:

keto form of cyclohexa-2,4-diene-1-one not aromatic and hence less stable.

Whereas enol form it is aromatic which makes it highly stable. that's why cyclohexa-2,4-diene-1-one exists only in enol form.

8 0
3 years ago
Fog is an example of a
Crank

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colloidal system liquid dispersed in gas.

6 0
3 years ago
What season starts in the northern hemisphere as the suns direct rays strike the equator moving from the southern hemisphere to
laila [671]

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3 0
3 years ago
The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.
madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

H_3BO_3\rightarrow H^+H_2BO_3^-

at t=0  cM              0             0

at eqm c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 4.4 M and \alpha = ?

K_a=5.8\times 10^{-10}

Putting in the values we get:

5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}

(\alpha)=0.000011

[H^+]=c\times \alpha

[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M

Also pH=-log[H^+]

pH=-log[4.8\times 10^{-5}]=4.3

Thus pH of a 4.4 M H_3BO_3 solution is 4.3

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3 years ago
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