Question

4HE + 241 AM ___> ______ + 2 1n
2 95 0

Answer 1

Answer on Question #73494 – Chemistry – Organic Chemistry

Task:

$${ } _ { 2 } ^ { 4 } H e + { } _ { 9 5 } ^ { 2 4 1 } A m = \underline { { - } } + 2 { } _ { 0 } ^ { 1 } n$$

Solution:

$$\frac {A _ {x}}{z _ {x}} \boldsymbol {X}$$ $$A _ {x} + 2 * A (n) = A (A m) + A (H e);$$ $$A _ {x} + 2 * 1 = 2 4 1 + 4;$$ $$A _ {x} = 2 4 5 - 2 = 2 4 3.$$ $$z _ {x} + 2 * z (n) = z (A m) + z (H e);$$ $$z _ {x} + 2 * 0 = 9 5 + 2;$$ $$z _ {x} = 9 7 - 0 = 9 7.$$

Then,

$$\frac {A _ {x}}{z _ {x}} \boldsymbol {X} = \begin{array}{c} 2 4 3 \\ 9 7 \end{array} B k$$ $${ } _ { 9 7 } ^ { 2 4 3 } B k - \text { B e r k e l i u m }$$ $${ } _ { 2 } ^ { 4 } H e + { } _ { 9 5 } ^ { 2 4 1 } A m = { } _ { 9 7 } ^ { 2 4 3 } B k + 2 { } _ { 0 } ^ { 1 } n$$

Answer: ${}_{2}^{243}Bk - \text{Berkelium}$ ;

$${ } _ { 2 } ^ { 4 } H e + { } _ { 9 5 } ^ { 2 4 1 } A m = { } _ { 9 7 } ^ { 2 4 3 } B k + 2 { } _ { 0 } ^ { 1 } n$$

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