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3 years ago

What are the advantages and disadvantages of fission?

Chemistry

1 answer:

viktelen [127]3 years ago

5 0

<u>Advantages of Nuclear Fission</u>

Nuclear fission provides cheapest energy . Almost 10% of electricity used in the world is obtained from the fission reaction

It offers a low-emission energy solution since there is no carbon dioxide gas emitted during the nuclear fission reaction

A well controlled and maintained nuclear reactor can produce energy for 36 to 40 months so works for .

It is a reliable source of energy as energy is obtained from uranium which is available is plenty.

It provides very concentrations of energy as it can provide large amount of energy from small amount of fuel.

The reaction gives less annual mortality rate of any energy resource with 90 deaths per trillion kilowatt hours

<u>Disadvantages of Nuclear Fission </u>

It is dangerous and also explosive.

It creates harmful and radioactive waste products.

It is not a renewable energy resource like solar and wind energy

It can develop long-term health issues for people exposed to then radioactive waves.

It involves high cost in installation of the reactors.

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GIVING 17 POINTS PLS HELP ME!!!!

Novosadov [1.4K]

Hydropower can affect fish and other marine life which is the true statement and is denoted as option C.

<h3>What is Hydropower?</h3>

This is the process in which water energy is converted into electrical energy through the use of dams and turbines.

When the dams are created, it blocks the path of marine activities and limits them to certain activities such as reproduction.

Read more about Hydropower here brainly.com/question/8934285

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8 0

2 years ago

Helpppppp!!!!! It’s due today <br> URGENT!!!!!

VladimirAG [237]

Answer:

Explanation:

2. a [CO3 2-][H3O+] / [H2O][HCO3-

b. [H2PO4-][H3O+]/[H3PO4][H2O]

5 0

3 years ago

In which situation is no work considered to be done by a force?

Masja [62]

If the angle is either 0 or 180, that means that there is either negative or positive work, so A and D are not correct.

If the angle is 45, then there is still some work involved.

The only option where there is no work done by a force is B. when the angle is between the force and displacement is 90.

7 0

3 years ago

Read 2 more answers

A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane

son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

= 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

= 75 - 67.5

= 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

= 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

= 25 - 21.25

= 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

= 64.125 moles

$CO$ formed is = 0.05 x 67.5

= 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

= 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

= 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21}$

= 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

= 2 x 64.125

= 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

= 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3 = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

= 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b). The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.

6 0

3 years ago

If a pharmacist adds 10 ml of purified water to 30 ml of a solution having a specific gravity of 1.30, calculate the specific gr

xxMikexx [17]

The specific gravity of a sample is the ratio of the density of the sample with respect to one standard sample. The standard sample used in specific gravity calculation is water whose density is 1 g/mL. The solution having specific gravity 1.30 is the density of the sample that is 1.30 g/mL. Thus the weight of the 30 mL sample is (30×1.30) = 39 g.

Now the mass of the 10 mL of water is 10 g as density of water is 10 g/mL. Thus after addition the total mass of the solution is (39 + 10) = 49g and the volume is (30 + 10) = 40 mL. Thus the density of the mixture will be $\frac{49}{40}=1.225$ g/mL. Thus the specific gravity of the mixed sample will be 1.225 g/mL.

6 0

3 years ago