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Fynjy0 [20]
4 years ago
7

The resulting atomic number of an element that emits 1 alpha particle, 1 positron, and 3 beta particles is

Chemistry
1 answer:
Natalka [10]4 years ago
4 0

Answer: -

Thus the resulting atomic number of an element that emits 1 alpha particle, 1 positron, and 3 beta particles is the same as original.

Explanation: -

Let the mass number be A and the atomic number be Z.

When an alpha particle is emitted, the mass number decreases by 4 and the atomic number decreases by 2.

Thus after alpha particle the new mass number = A - 4

Thus after alpha particle the new atomic number = Z - 2

When a positron is emitted, the atomic number only decreases by 1.

Thus after positron emission the new mass number = A - 4

Thus after positron emission the new atomic number = Z - 2 - 1

= Z - 3

When a beta particle is emitted, the atomic number increases by 1.

For 3 beta particles, the atomic number will increase by 3.

Thus after beta particle emission the new mass number = A - 4

Thus after beta particle emission the new atomic number = Z - 3+3

= Z

Thus the resulting atomic number of an element that emits 1 alpha particle, 1 positron, and 3 beta particles is the same as original.

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Answer:

\rm _{38}^{91}Sr \longrightarrow \,  _{-1}^{0}e + \,  _{39}^{91}Y

Explanation:

The unbalanced nuclear equation is

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Let's write the question mark as a nuclear symbol.

\rm _{38}^{91}Se} \longrightarrow \,  _{-1}^{0}e \, + \,  _{Z}^{A}X

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.  

Then

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Then, your nuclear equation becomes

\rm _{38}^{91}Sr \longrightarrow \,  _{-1}^{0}e + \,  _{39}^{91}X

Element 39 is yttrium, so the balanced nuclear equation is

\rm _{38}^{91}Sr \longrightarrow \,  _{-1}^{0}e + \,  _{39}^{91}Y

7 0
3 years ago
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When compounds possess a plane of symmetry between the chiral centers they are called achiral or meso compounds. Among the given compounds (A) compound 1 have a plane of symmetry. So we can say compound one is a meso or achiral compound. Compounds two, three, and four have no plane of symmetry, as you can see in the structures attached. So all other compounds (compound 2, compound 3, and compound 4) except compound one are not meso or achiral.

You can also learn about meso compounds from the following question:

brainly.com/question/29022658

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