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CaHeK987 [17]
2 years ago
11

What substances dissociate into ions when dissolved in water

Chemistry
1 answer:
dusya [7]2 years ago
5 0

Polar or ionic substances dissociate into ions when dissolved in water.

<h3>What substances dissociate into ions when dissolved in water?</h3>

Ionic substances dissociate into ions when dissolved in water because water is also a polar by nature. We know that like dissolve like so polar substances dissolve in water and dissociate into ions i.e. positive an negative ions.

So we can conclude that polar or ionic substances dissociate into ions when dissolved in water.

Learn more about ions here: brainly.com/question/14511468

#SPJ1

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First find the number of moles of sulfur using dimensional analysis with avogadro’s number as the conversion factor. 4.2*10^24 atoms * (1 mol/6.022*10^23 atoms) = 7.0 mol sulfur. The molar mass of sulfur is 32.06 g/mol, which is found on the periodic table as sulfur’s (S) atomic weight. Use dimensional analysis again with the molar mass of sulfur as the conversion factor. 7.0 mol * 32.06 g/mol = 224.42 g sulfur. Since the problems gives us two significant figures, round the mass of sulfur to 220 grams, or 2.2 * 10^2 g.
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Neptunium-237 undergoes a series of α-particle and β-particle productions to end up as thallium-205. How many α particles and β
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<h3>Answer:</h3>

8 alpha particles

4 beta particles

<h3>Explanation:</h3>

<u>We are given;</u>

  • Neptunium-237
  • Thallium-205
  • Neptunium-237 undergoes beta and alpha decay to form Thallium-205.

We are required to determine the number of beta and alpha particles produced to complete the decay series.

  • We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
  • When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.

In this case;

Neptunium-237 has an atomic number 93, while,

Thallium-205 has an atomic number 81.

Therefore;

²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti

We can get x and y

237 = 4x + y(0) + 205

237-205 = 4x

4x = 32

 x = 8

On the other hand;

93 = 2x + (-y) + 81

but x = 8

93 = 16 -y + 81

y = 4

Therefore, the complete decay equation is;

²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti

Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.

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