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Sergeeva-Olga [200]
3 years ago
6

Sometimes a particle of radiation penetrates a substance and continues on its way, and sometimes it doesn't. For the particles t

hat don't make it through a substance, what happened to them? At the atomic level what part of an atom did they hit? As for the particles that weren't shielded or blocked by the material, how did they manage to get through? Also, explain why denser substances shield better than lighter substances.
Chemistry
1 answer:
Amanda [17]3 years ago
6 0

Explanation:

According to the atomic model of Rutherford, an atom has most of empty space inside the region where electrons reside.

And, nucleus is the smaller part that is, present at the center of an atom. It is solid and inside the nucleus there are  protons and neutrons are present.

  • When radiation hits a solid particle then they hit the positively charged nucleus due to the positive charge of protons. So, they tend to face a strong repulsion and then they retrace the path. As a result, they make it through the substance.
  • Now, at the atomic level they hit the nucleus of the atom that s, positively charged.
  • Particles that are not shielded or blocked by material tend to pass directly through the large empty space as in this empty space there is no repulsion because of nucleus. As a result, they tend to pass directly through the material without any disturbance.
  • Denser is a substance, larger will be its nucleus. This means that then there will be more number of protons and neutrons inside the nucleus.  Hence, there will be high positive charge on the nucleus and very less free space that will allow very less number of radiations to pass through without any disturbance.

Thus, denser substance will shield better as compared to the lighter substances.

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Answer:

ΔH of dissociation is 38,0 kJ/mol

Explanation:

The dissociation reaction of KBrO₃ is:

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This dissolution consume heat that is evidenced with the decrease in water temperature.

The heat consumed is:

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Where C is specific heat of water (4,186 J/mol°C)

ΔT is the temperature changing (18,0°C - 13,0°C = 5,0°C)

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Replacing, heat consumed is:

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13,8 g of KBrO₃ are:

13,8 g×(1mol/167g) = 0,0826 moles

Thus, ΔH of dissociation is:

3,14kJ / 0,0826mol = <em>38,0 kJ/mol</em>

<em></em>

I hope it helps!

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