Unlike chemical reactions, for which mass is conserved, nuclear reactions result

Question

Unlike chemical reactions, for which mass is conserved, nuclear reactions result in slight changes in mass. When mass is lost, it becomes energy according to the equation

E=mc2

where E is the energy in joules, m is the mass defect in kilograms, and c is the speed of light (c=3.00×108 m/s). The mass defect is the difference between the total mass of the products and the total mass of reactants. The following values can be used to calculate the mass defect.

Part A

The sun produces energy via fusion. One of the fusion reactions that occurs in the sun is

411H42He+201e

How much energy in joules is released by the fusion of 1.80 g of hydrogen-1?

the correct answer I got is ( -1.11x10^12 ).

Part B

Nuclear power plants produce energy using fission. One common fuel, uranium-235, produces energy through the fission reaction

235 92U+10nfission fragments+neutrons+3.20×1011 J/atom

What mass of uranium-235 is needed to produce the same amount of energy as the fusion reaction in Part A?

Express your answer to three significant figures and include the appropriate units

????????

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The following table provides standard enthalpies of formation for methane (CH4), carbon dioxide (CO2), and water (H2O).

By definition, the enthalpy of formation of any element in its standard state is 0 kJ/mol.

Part C

Another way to produce energy is by burning natural gas, which is mostly methane (CH4). The reaction

CH4(g)+2O2(g)CO2(g)+2H2O(g)

shows the combustion of methane. Assuming complete combustion, what mass of methane is needed to produce the same amount of energy as the fusion reaction in Part A?

Express your answer to three significant figures and include the appropriate units.

??????????

Please help....

Explanation / Answer

0.00054858+1.00782+4.00260=5.01096858

1.80/-0.359=5.01(in negative because its fusion)

5.01096858-5.01=9.6858x10^-4

9.6858x10^4 * (3x10^8) ^2 = -1.11x10^12

*****

CH4(g)+2O2(g)?CO2(g)+2H2O(g)

Let us calculate Hrxn

Hrxn = Hprod - Hreact = (-393.5 + 2*-285.83) - (-74.8 + 0 ) = -890.36 kJ/rxn

So, from part A

E = -7.82*10^11 J or -7.82*10^8 kJ

Divide E/Hrxn = (-7.82*10^8) / (-890.36) = 878296.42 times the original reaction

original reaction has 1 mol of CH4, therefore w eneed 878296.42 moels of CH4

1 mol = 16 g

878296.42 = 1.4*10^7 g of CH4

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