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guajiro [1.7K]
3 years ago
15

If a radioactive isotope of thorium (atomic number 90, mass number 232) emits 6 alpha particles and 4 beta particles during the

course of radioactive decay, what is the mass number of the stable daughter product?
Chemistry
1 answer:
kolbaska11 [484]3 years ago
5 0

Answer:

The mass number of the stable daughter product is 208

Explanation:

First thing's first, we have to write out the equation of the reaction. This is given as;

²³²₉₀Th → 6 ⁴₂α  +  4 ⁰₋₁ β + X

In order to obtain the identity of X, we have to obtain it's mass numbers and atomic number.

There is conservation of matter so we expect the mass number to remain the same in both the reactant and products.

Mass Number

Reactant = 232

Product = (6* 4 = 24) + (4 * 0 = 0) + x = 24 + x

since reactant = product

232 = 24 + x

x = 232 - 24 = 208

Atomic Number

Reactant = 90

Product = (6* 2 = 12) + (4 * -1 = -4) + x = 8 + x

since reactant = product

90 = 8 + x

x = 90 - 8 = 82

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In the periodic table hydrogen is placed in Group 1A (group 1) and helium is placed in Group 8A (group 18). The most likely reas
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Answer:

This question appears incomplete

Explanation:

This question appears incomplete because of the absence of options. However, hydrogen is placed in group 1 because it has just one electron in it's outermost shell (which happens to be the only shell it has) just like every other group 1A/group 1 element. While helium is placed in group 8A/group 18 because it has a completely filled outermost shell (which is also the only shell it has) just like every other element in group 8A/group 18.

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3 years ago
wo reactions and their equilibrium constants are given. A + 2 B − ⇀ ↽ − 2 C K 1 = 2.57 2 C − ⇀ ↽ − D K 2 = 0.226 A+2B↽−−⇀2CK1=2.
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<u>Answer:</u> The value of equilibrium constant for the net reaction is 11.37

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  A+2B\xrightarrow[]{K_1} 2C

<u>Equation 2:</u>  2C\xrightarrow[]{K_2} D

The net equation follows:

D\xrightarrow[]{K} A+2B

As, the net reaction is the result of the addition of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

K=K_1\times \frac{1}{K_2}

We are given:

K_1=2.57

K_2=0.226

Putting values in above equation, we get:

K=2.57\times \frac{1}{0.226}=11.37

Hence, the value of equilibrium constant for the net reaction is 11.37

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