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pentagon [3]
3 years ago
5

Nuclear fusion is the process used to generate electricity in nuclear power plants. is the process in which a large nucleus spon

taneously splits into two or more smaller nuclei. is the process in which two smaller nuclei combine to form a larger nucleus. is the process in which a large nucleus is bombarded with a neutron and then splits into two or more smaller nucleir.
Chemistry
2 answers:
Dovator [93]3 years ago
7 0

Answer:

is the process in which two smaller nuclei combine to form a larger nucleus.

Explanation:

Nuclear fusion -

It is the type of reaction , where two or more lighter nuclei combines to form a larger nuclei , with large amount of energy released in the form of heat and light is referred to as a nuclear fusion .

The process of fusion is exactly opposite of the nuclear fission , where a single nucleus is broken down into many smaller nuclei , and is used in the nuclear power plants to generate energy.

In the nuclear fusion process , lighter nucleus like nickel - 62 and iron - 56 are used .

Arte-miy333 [17]3 years ago
3 0

Answer:

Nuclear fusion can be defined asva reaction in which two or more atomic nuclei are combined to form one or more different atomic nuclei and subatomic particles ( neutrons or protons). The difference in mass between the reactants and products is manifested as either the release or absorption of energy . This difference in mass arises due to the difference in atomic " binding energy " between the atomic nuclei before and after the reaction. Fusion is the process that powers active or " main sequence " stars , or other high magnitude stars.

A fusion process that produces nuclei lighter than iron-56 or nickel-62 will generally release energy. These elements have relatively small mass per nucleon and large binding energy per

nucleon . Fusion nuclei lighter than these releases energy (an exothermic process), while fusion of heavier nuclei results in energy retained by the product nucleons, and the resulting reaction is endothermic .

Nuclear fusion generates electricity (as a proposed form of power) by using heat from nuclear fusion reactions. In a fusion process, two lighter atomic nuclei combine to form a heavier nucleus, while releasing energy. The devices designed to harness this energy are known as fusion reactors.

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In this case, in order to substantiate whether the cabin meet the federal standards, we need to convert the 500 mmHg to atm and compare the result with 0.72 atm by knowing that 1 atm equals 760 mmHg:

500mmHg*\frac{1atm}{760mmHg} \\\\=0.66atm

Thus, since 0.66 atm is 0.06 atm away from the federal standard we can infer that it may meet the federal standard, however, it would not be recommended to let the pressure decrease more than that.

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3 years ago
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3 years ago
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"What volume of" a 0.300 M BaF2 "solution is needed to prepare" 240.0 mL of a 0.100 M F- solution
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4 years ago
Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and
skelet666 [1.2K]

Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

Explanation :

The given balanced chemical reaction is,

C_6H_6(l)\rightarrow C_6H_6(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0_{(C_6H_6(g))} = standard enthalpy of formation  of gaseous benzene = 82.93 kJ/mol

\Delta H_f^0_{(C_6H_6(l))} = standard enthalpy of formation  of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]

\Delta H^o=33.89kJ/mol=33890J/mol

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0_{(C_6H_6(g))} = standard entropy of formation  of gaseous benzene = 269.2 J/K.mol

\Delta S^0_{(C_6H_6(l))} = standard entropy of formation  of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]

\Delta S^o=95.94J/K.mol

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 25^oC\text{ or }298K.

\Delta G^o=(33890J)-(298K\times 95.94J/K)

\Delta G^o=5299.88J/mol=5.299kJ/mol

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

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3 years ago
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