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zalisa [80]
2 years ago
9

Identify the nuclide produced when thorium-234 decays by beta emission:234 90Th→ 0−1e + ?

Chemistry
1 answer:
aleksandrvk [35]2 years ago
7 0
<h3>Answer: 234 90Th → 0 -1e + 234 91Pa  Explanation: Given that : Thorium 234 decays by beta emission  Beta emission = 0, - 1 e 234 90Th → 0 -1e + 234 91Pa  Hence, when Thorium 234 90 undergoes a beta Decay ; mass number remains the same at 234 while atomic number increases by 1, becoming (90 + 1) = 91 to form a protactinium nuclide. </h3>

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first, we should have the balanced equation of the reaction:
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Second, we start to convert mass to moles
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6 0
3 years ago
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3 0
3 years ago
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7 0
3 years ago
Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili
Katen [24]

<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}

\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M

\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

                         H_2+I_2\rightleftharpoons 2HI

<u>Initial:</u>                0.1    0.1

<u>At eqllm:</u>          0.1-x   0.1-x   2x

Evaluating the value of 'x'

\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M

The expression of K_c for above equation follows:

K_c=\frac{[HI]^2}{[H_2][I_2]}

[HI]_{eq}=2x=(2\times 0.079)=0.158M

[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M

[I_2]_{eq}=0.0210M

Putting values in above expression, we get:

K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61

Hence, the value of equilibrium constant for the given reaction is 56.61

6 0
3 years ago
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