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Complete the following nuclear reactions:

(a) $?+\frac{14}{7} \mathrm{N} \rightarrow_{1}^{1} \mathrm{H}+_{8}^{17} \mathrm{O} \quad(\mathrm{b})_{3}^{7} \mathrm{L} \mathrm{i}+_{1}^{1} \mathrm{H} \rightarrow_{2}^{4} \mathrm{He}+?$

a. _{2}^{4} H e+\frac{14}{7} N \rightarrow \frac{1}{1} H+_{8}^{17} O

b. _{3}^{7} L i+_{1}^{1} H \rightarrow_{2}^{4} H e+_{2}^{4} H e

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for number 42. We need to complete both of these equations. Um, so I'm just gonna go across the top with the mass numbers on this side of the equation. I've got 18. So on this side of the equation, I need 18 and already 14. So you can tell you need four more, and then with the bottom, basically the number of protons on this side of the equation, I start with nine. So in this side equation, I need nine. I already have seven, so I need two more. So, um, periodic table Ellen with two protons is healing. So this isn't out for particle on down here. I'm this little equations complete. So on the top, I start with a mass number of eight over here already have four. So I need four more. I start with a total number of protons of four. So here I have to I need two more. So this one is also another powerful particle

University of Virginia