Question #6d153

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Answer 1 · 2017-06-16T23:45:01

The answer is C. $_{9}^{19} F$ $""_9^19"F"$ .

The nuclear equation is

$_{7}^{14} N +_{2}^{4} He \to ?$ $""_7^14"N" + ""_2^4"He" → ?$

The sum of the subscripts and of the superscripts must be the same on both sides of the equation.

Thus, the equation becomes

$_{7}^{14} N +_{2}^{4} He \to_{9}^{18} X$ $""_7^14"N" + ""_2^4"He" → ""_9^18"X"$

Element 9 is fluorine, so the complete equation is

$_{7}^{14} N +_{2}^{4} He \to_{9}^{18} F$ $""_7^14"N" + ""_2^4"He" → ""_9^18"F"$