Consider the nuclear fusion of hydrogen into helium: \(4{ }_{1}^{1} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+2\) identical subatomic particles (a) What are the two identical subatomic particles? (b) During this transformation, mass is lost and converted into energy, releasing \(2.56 \times 10^{9} \mathrm{~kJ}\) per mole of helium formed. How many gallons of gasoline would you have to burn to release this much energy, assuming gasoline is pure octane? Burning octane releases \(5509.0 \mathrm{~kJ} / \mathrm{mol} ;\) the \(\mathrm{MM}\) of octane is \(114.23 \mathrm{~g} / \mathrm{mol} ;\) the density of octane is \(0.703 \mathrm{~g} / \mathrm{mL} ;\) there are \(3785.408 \mathrm{~mL}\) per gallon. (c) How much mass was lost in grams? Note that a joule is a \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}^{2}\).

In order to identify the two identical subatomic particles, we have to look at the mass and atomic number of the elements involved. The left side of the equation has 4 hydrogen atoms, with the mass of the 4 hydrogen atoms being \(4(1^1)\). The right side of the equation has one helium atom with a mass number of 4, and 2 identical subatomic particles that we need to identify. In nuclear reactions, both the atomic number and mass number must be conserved. Therefore, we need to identify a subatomic particle whose atomic number is 0 (since the atomic number is conserved) and mass number is 1 (since the mass number is also conserved). This subatomic particle is a neutron. Answer (a): The two identical subatomic particles are neutrons.

We are given that during the nuclear fusion of hydrogen into helium, \(2.56 \times 10^9 kJ\) per mole of helium is released. We need to determine how many gallons of gasoline would have to be burned to release this much energy. We are given the energy released by burning octane, which is 5509.0 kJ/mol. First, we need to find the moles of octane needed to release the same amount of energy as the nuclear fusion: \[ \text{moles of octane} = \frac{\text{energy released by nuclear fusion}}{\text{energy released by burning octane}} = \frac{2.56 \times 10^9 kJ}{5509.0 kJ/mol} \] Now, we need to convert the moles of octane to grams using the molar mass of octane (114.23 g/mol): \[ \text{grams of octane} = \text{moles of octane} \times \left(\frac{114.23 g}{mol}\right) \] Next, we need to convert the grams of octane to milliliters using the density of octane (0.703 g/mL): \[ \text{mL of octane} = \frac{\text{grams of octane}}{0.703 g/mL} \] Finally, we need to convert the milliliters of octane to gallons using the conversion factor (3785.408 mL per gallon): \[ \text{gallons of octane} = \frac{\text{mL of octane}}{3785.408 mL/gal} \] Answer (b): Converted energy in the number of gallons of gasoline.

We are given that \(2.56 \times 10^9 kJ\) of energy is released during the nuclear fusion. In order to find the mass lost in grams, we will use Einstein's famous formula \(E = mc^2\), where \(E\) is energy, \(m\) is mass, and \(c\) is the speed of light \(c = 3 \times 10^8 m/s\). First, we need to convert the energy to joules by multiplying by 1000: \[ E = 2.56 \times 10^9 kJ \times 1000 \frac{J}{kJ} = 2.56 \times 10^{12} J \] Now, we will solve for the mass (in kg) lost using Einstein's formula: \[ m = \frac{E}{c^2} = \frac{2.56 \times 10^{12} J}{(3 \times 10^8 m/s)^2} \] Finally, we will convert the mass from kg to grams by multiplying by 1000: \[ m = m_{kg} \times 1000 \frac{g}{kg} \] Answer (c): Mass lost in grams.