Question

Part a the sun produces energy via fusion. one of the fusion reactions that occurs in the sun is 411h→42he+201e how much energy in joules is released by the fusion of 2.58 g of hydrogen-1? express your answer to three significant figures and include the appropriate units. view available hint(s) δe = submit part b this question will be shown after you complete previous question(s). return to assignment provide feedback

Answer 1

The equation for the nuclear fusion reaction is,
4 ¹₁H → ₂⁴He + 2 ₁⁰e
Calculation of mass defect,
Δm = [mass of products - mass of reactants]
      = 4(1.00782) - [4.00260 + 2(0.00054858)]
      = 0.0275828 g/mole
Given that,
Mass of Hydrogen-1 = 2.58 g
The no. of moles of ₁¹H = 2.58 g / 1.00782 = 2.56 moles
Therefore, the mass defect for 2.58 g of ₁¹H is, 
= 2.56 moles * (0.0275828 g / 4) = 0.01765 x 10⁻³ kg
Energy for (0.01765 x 10⁻³ kg) is, 
= (0.01765 x 10⁻³ kg) (3.0 x 10⁸)² = 1.59 x 10¹² J

Answer 2

Answer:

$Energy=-1.59x10^{12}J$

Explanation:

Hello,

In this case, the requested energy is computed via the following equation, whereas the mass defected is the main issue to be calculated:

$Energy=m_{d}*c^2$

Whereas $m_{d}$ is the mass defected and $c$ the speed of light. In this manner, such mass is computed via:

$m_d=m_{products}-m_{reagents}\\m_d=(mass_{He}+m_{electron})-4m_H$

Now, given the mass of the H-1, He, and the electron, one computes such mass as shown below per 4 moles of H as those the involved moles:

$m_d=4.00260g/mol+2*0.00054858g/mol-4*1.00782g/mol\\m_d=-0.027583g/4mol$

Finally, we apply the firstly given formula for the determination of the energy, taking into account that to get Joules we need to convert the mass defected from grams to kilograms as follows:

$Energy=2.58gH*\frac{1molH}{1.00782gH} *-0.027583gH/4molH*\frac{1kgH}{1000gH}*(299 792 458 m / s )^2\\Energy=-1.59x10^{12}J$

The obtained energy turns out negative since is a released type of energy.

Best regards.