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KatRina [158]
2 years ago
11

What type of radioactive decay will the isotopes 13B and 188Au most likely undergo?

Chemistry
1 answer:
scoundrel [369]2 years ago
8 0

Answer:

b. Beta emission, beta emission

Explanation:

A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).

Now let us look at the N/P ratio of each atom;

For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6

For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4

For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.

For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.

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7 0
3 years ago
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LekaFEV [45]
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5 0
3 years ago
Write balance complete molecular equation, ionic equation, and net ionic equations for the reactions that occur when each of the
AveGali [126]

<u>Answer:</u> The complete molecular, ionic, and net ionic equations are given below. The spectator ions are sodium and nitrate ions.

<u>Explanation:</u>

The ionic equation is defined as the equation in which all the substances that are strong electrolytes present in an aqueous state and are represented in the form of ions.

The net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

The balanced molecular equation for the reaction of lead (II) nitrate and sodium sulfide follows:

Pb(NO_3)_2(aq)+Na_2S(aq)\rightarrow PbS(s)+2NaNO_3(s)

The ionic equation follows:

Pb^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+S^{2-}(aq)\rightarrow PbS(s)+2Na^+(aq)+2NO_3^-(aq)

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

Pb^{2+}(aq)+S^{2-}(aq)\rightarrow PbS(s)

3 0
2 years ago
Of the following: H2(g); He(g); CO2(g); which would behave least like an ideal gas? Why?
Margarita [4]
Answer:

CO2(g)

Because CO2 is the larges molecule with specific geometric, therefore it is not likely to behave as an ideal gas.
4 0
2 years ago
I need help asap. which are the products?
beks73 [17]

Answer:

A

Explanation:

The products of a chemical equation is what is being experimented with. However, the results in which you get are known as your solution.

4 0
2 years ago
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