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Arlecino [84]
3 years ago
12

During nuclear fission, great amounts of energy are produced from____. A. particle accelerators B. tremendous amounts of mass C.

a series of chemical reactions D. very small amounts of mass.
Chemistry
2 answers:
LekaFEV [45]3 years ago
6 0

Answer: Option (C) is the correct answer.

Explanation:

Nuclear fission is a process in which a large nucleus splits into two or more small nuclei.

For example, ^{239}_{94}Pu \rightarrow ^{134}_{54}Xe + ^{103}_{40}Zr + 3^{1}_{0}n

The neutrons produced in the reaction will again react with plutonium and thus it will form a series of chain reaction. As a result, there will be continuous reaction of neutrons with the plutonium atom thus, it will create a great amount of energy.

Thus, we can conclude that during nuclear fission, great amounts of energy are produced from a series of chemical reactions.

Phoenix [80]3 years ago
3 0
During nuclear fission, great amounts of energy are produced from____.

Answer: Out of all the options presented above, the one that completes the statement and makes it true, is answer choice D) very small amounts of mass. Nuclear fission power does not produce carbon and also produces 12% of the world's power.

I hope it helps, Regards.
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The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A
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Solution :

Given :

The steady state flow = 8000 $ m^3 /d $

                                    $= 80 \times 10^5 \ I/d $

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L $

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d $

If 60 mg of alum $ [Al_2(SO_4)_3.14 H_2O] $ required for one litre of the water treatment.

So Alum required for  $ 80 \times 10^5 \ I/d $

$= 80 \times 15^5 \ I/d  \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d $

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $ = 60 \times 0.234 \text{ of alum ppt. per litre} $

      $= 14.04 $ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

                                        = 112.32 kg/d ppt of alum

Daily total solid load is  $= 144 \ kg/d + 112.32 \ kg/d$

                                       = 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234 $

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

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After addition, the aluminium hydroxide pH of water will increase due to increase in $ OH^- $ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm $

And the pH is reduced into the range of 5.9 to 6.4

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Matter that is the same throughout is called what
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