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Ch.21 - Radioactivity & Nuclear Chemistry

Chapter 21, Problem 77

Complete each nuclear equation and calculate the energy change (in J/mol of reactant) associated with each (Be-9 = 9.012182 amu, Bi-209 = 208.980384 amu, He-4 = 4.002603 amu, Li-6 = 6.015122 amu, Ni-64 = 63.927969 amu, Rg-272 = 272.1535 amu, Ta-179 = 178.94593 amu, and W-179 = 178.94707 amu). a. _____ + 94Be → 63Li + 42He

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Welcome back everyone. In this example, we need to write the complete nuclear equation for the below reaction. We need to also give the associated energy change in jewels per mole of our reactant for our reaction. And we're given the atomic masses of our isotopes. So what we should recognize is that we're missing an entire isotope reactant here. So we would say that it's X, that we're missing an unknown element with the unknown atomic number Z and the unknown mass number, which you recall is represented by the symbol A. And so we would say that based on our given equation above, we can say therefore, a plus the mass number for magnesium 24 is equal to on our product side, the mass number for aluminum 27 plus our second product helium which has a mass number of four. And we would say that A is now equal to a value of seven. So so far, we can fill in seven for our unknown element as its mass number. And then looking at our atomic number, Z, we can say that Z plus the atomic number for magnesium being 12 is equal to on our product side, our atomic number for aluminum being 13 plus our atomic number for our second product helium being two. And so therefore, we can say that Z is equal to a value of three. And so filling in our value for three, we would say that because we know our atomic number is three, we would refer to our periodic tables and see that our atom is lithium. And so we would fill in for our symbol the atom lithium. And so we can say that our equation should read as lithium seven, reacting with magnesium 24 produces as products, aluminum 27 and helium four. And so this would be our first answer as our complete nuclear equation for part one of the prompt. And so now we can move on to part two, which is calculating associated energy change in jewels per mole. However, before we can do that, we need to make note of our mass for lithium seven. And when we look this up in our textbooks or online, we would see that lithium seven has an atomic mass of 7.01600 A. And so we want to calculate our mass effect first before calculating associated energy change and recall that mass effect is found from taking our sum of our masses of our reactants subtracted from the sum of our masses of our products. And so we're going to use the info given in the prompt. So to calculate mass effect. We would say that we have for the sum of our masses of reactants, we have our mass of lithium seven which has a mass of 7.01600 AM use added to the mass of our second reactant for magnesium which according to the prompt has a mass of 23.9 and sorry, that should be nine 850 4 a.m. use which is then added to the masses of our products where we have our first product being aluminum 27 which according to the prompt has an atomic mass of 26.98 1:54 a.m. U which is then added to our atomic mask for helium four which according to the prompt is 4.00260 AM U. And so this would complete our calculation for mass effect and we want to actually express this in kilograms. So we call that mast effect is in units of kilograms. So what we would get here is that our mast effect is equal to a value of 0.0169 g. We can say which we will convert to kilograms by recalling that our prefix kilo tells us we have 10 to the third power grams for 1 kg. So we can get rid of grams and we're left with kilograms. And this gives us a mass effect value equal to 1.69 times 10 to the negative fifth power kilograms. And now we can use this to calculate our associated energy change. So we're going to recall our following formula for the energy mass relation formula where we say that energy is equal to the mass times the speed of light squared. And so we would say that our mass which we just calculated as 1.69 times 10 to the negative fifth power multiplied by the speed of light, which we should recall is 3.00 times 10 to the eighth power. And then squared is equal to a value of 1.5 to 1 times 10 to the 12th power. And so to show our units for this, we would recall that these units here would be kilograms per mole since we have grams per mole, technically, when we converted from a US to grams. And so to write that down here, we would have kilograms per mole recall that the speed of light has units of meters per second. They're going to be squared. And so this would now give us kilograms times meters squared divided by seconds squared chimes moles. But we don't really actually have to write in the moles here because it's just understood. So we can just say we have units of kilograms times meter squared divided by second squared. And we can recall that our units, kilograms times meter squared divided by second squared is equivalent to one Joel. And so we would express our answer in terms of jewels. So we could say that this is equal to 1.5 to 1 times 10 to the 12 Power Jews. And so this would be our associated energy change as our final answer to complete this example. So everything highlighted in box and yellow represents our two final answers. So I hope that everything that I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.