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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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Ne4ueva [31]
I think it'd be gravity. 
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2 years ago
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Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
2 years ago
What is the total probability of finding a particle in a one-dimensional box in level n = 4 between x = 0 and x = L/8?
Lubov Fominskaja [6]

Answer:

P = 1/8

Explanation:

The wave function of a particle in a one-dimensional box is given by:

\psi = \sqrt \frac{2}{L} sin(\frac{n \pi x}{L})

Hence, the probability of finding the particle in the  one-dimensional box is:

P = \int_{x_{1}}^{x_{2}} \psi^{2} dx

P = \int_{x_{1}}^{x_{2}} (\sqrt \frac{2}{L} sin(\frac{n \pi x}{L}))^{2} dx

P = \frac{2}{L} \int_{x_{1}}^{x_{2}} (sin^{2}(\frac{n \pi x}{L}) dx

Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:

P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]

P = \frac{1}{8} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})    

Solving for n=4:

P = \frac{1}{8} (1 - 4\frac{sin(\frac{4 \pi}{4})}{4 \pi})    

P = \frac{1}{8} (1 - \frac{sin (\pi)}{\pi})    

P = \frac{1}{8}

I hope it helps you!

7 0
3 years ago
1. How much energy (in KJ) is required to convert 50.0g of ice at – 30 ˚C to<br> steam at130˚C.
Vitek1552 [10]

Answer:

How much heat energy required to convert following?

How much heat energy, in kilojoules, is required to convert 47.0 g of ice at -18.0 C to water at 25.0 C ?

Specific Heat of Ice - 2.09 j/g * c

This is how I did it and the answer is wrong...Please check and correct me

Q = m * Cice * Change in Temp

Q = (47.0 g)(2.09 J/g*c)(43) = 4222.6 J * 0.001 kj / j = 4.22 kj

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2 years ago
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Double-replacement and decomposition. (C)
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