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Firdavs [7]
2 years ago
8

A certain radioactive atom has 90 protons and 142 neutrons. If this atom undergoes alpha decay, what is the mass number of the d

aughter atom? The mass number of the daughter atom is .
Chemistry
2 answers:
Hatshy [7]2 years ago
5 0

Answer: the mass number of the daugther atom is 232,


Explanation:


1) Alpha (α) decay is a nuclear reaction in which a nucleus (parent's nucleus) emits an alpha (α) particle and leads to a different atom (daughter atom).


2) The alpha (α) particle is a nucleus of helium atom, i,e, a nucleus with two protons and two neutrons. The symbol used for the α particles is <em>⁴₂He</em>, where the superscript 4 indicates the mass number (2 protons + 2 neutrons = mass number 4), and the subscript 2 indicates the atomic number (number of protons).


3) Then, to determine the mass number of the daughter atom you just need to do a mass number balance:


mass number of the parent atom = mass number of the daugther atom + mass number of the α particle.


The mass number of the radioactive (parent) atom is 90 protons + 142 neutrons = 232.


∴ 232 = x + 4 ⇒ x = 232 - 4 = 228 ← answer.


The full equation may help you to have a wider vision of the problem:


²³²₉₀ X → ⁴₂ He + ²²⁸₈₈ Y

Note this:

  • 232 = 4 + 228 (this is a mass number balance)
  • 90 = 2 + 88 (this is an atomic number balance)
  • X is the parent atom, and Y is the daughter atom
  • You can use a periodic table to determine the identity of the unknown atoms (using the atomic numbers).
Eduardwww [97]2 years ago
3 0
<h2>Answer:</h2>

The mass number of daughter atom is 228.

<h3>Explanation:</h3>
  • As the mass number is the total number of protons and neutrons in an atom, So initial mass number will be 90 + 142 = 232.
  • As the alpha particles are helium nuclei having mass number of 4.
  • So after eliminating alpha particles, the mass number will lower by 4.
  • Mass number of daughter atom = 232 - 4 = 228

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1. Chromium

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Explanation:

Hello,

In this case, the given reaction with the appropriate oxidation states turns out:

2(Cr^{+3}O^{-2}_2)^-(aq) + 2H_2O(l) + 6(Cl^{+1}O^{-2})^-(aq)\longrightarrow 2(Cr^{+6}O^{-2}_4)^{2-}(aq) + 3Cl^0_2(g) + 4OH^-(aq)

In such a way, the oxidation half-reaction is written for chromium as the reducing agent so it is oxidized from +3 to +6, nonetheless, since there are two chromiums undergoing such change, 6 electrons are being transferred as shown below:

2(Cr^{+3}O^{-2}_2)^-(aq) \longrightarrow 2(Cr^{+6}O^{-2}_4)^{2-}(aq)+6e^-

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