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emmainna [20.7K]
3 years ago
10

Radium decays to form radon. Which equation correctly describes this decay? Superscript 226 Subscript 88 Baseline Upper R a righ

t arrow Superscript 222 Subscript 84 Baseline Upper R n + Superscript 4 Subscript 4 Baseline Upper H e Superscript 226 Subscript 88 Baseline Upper R a right arrow Superscript 222 Subscript 86 Baseline Upper R n + Superscript 0 Subscript negative 1 Baseline e Superscript 226 Subscript 88 Baseline Upper R a right arrow Superscript 222 Subscript 86 Baseline Upper R n + Superscript 0 Subscript + 1 Baseline e Superscript 226 Subscript 88 Baseline Upper R a right arrow Superscript 222 Subscript 86 Baseline Upper R n + Superscript 4 Subscript 2 Baseline Upper H e
Chemistry
1 answer:
KiRa [710]3 years ago
4 0

Answer: 226 Subscript 88 Baseline Upper R a right arrow Superscript 222 Subscript 86 Baseline Upper R n + Superscript 4 Subscript 2 Baseline Upper H e

Explanation:

Alpha decay : When a larger nuclei decays into smaller nuclei by releasing alpha particle. In this process, the mass number and atomic number is reduced by 4 and 2 units respectively.

The general representation of alpha decay reaction is:

^{A}_{Z}\textrm{X}\rightarrow ^{A-4}_{Z-2}\textrm {Rn}+ ^{4}_{2}\textrm{He}

Representation of Radium decays to form Radon

^{226}_{88}\textrm{Ra}\rightarrow ^{222}_{86}\textrm {Rn}+ ^{4}_{2}\textrm{He}

Thus 226 Subscript 88 Baseline Upper R a right arrow Superscript 222 Subscript 86 Baseline Upper R n + Superscript 4 Subscript 2 Baseline Upper H e represents alpha decay.

         

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If we assume m = 1g, we find that:

V_{lithium}=1.88mL

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