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kondaur [170]
1 year ago
8

A decay series starts with the synthetic isotope ²³⁹₉₂U. The first four steps are emissions of a β⁻ particle, another β⁻, an a p

article, and another α. Write a balanced nuclear equation for each step. Which natural series could start by this sequence?
Chemistry
1 answer:
VladimirAG [237]1 year ago
7 0

<u>Thorium series</u> could start by this sequence.

<h3>Brief explanation</h3>

To write balanced equations for nuclear decay processes. It's important to remember that the mass number and the atomic numbers must be balanced. And so what that means is that if we look at an elements nuclear symbol, the atomic number is the bottom number and the top number, the superscript, is the mass number, and so when we add them up on both sides, they have to be equal. There are two different ways in which decay can occur.

In this, series one is through beta decay, which means that the following particle is produced. The other is Alpha Decay, which produces this particle. Both are products. So if we start off with uranium to 39 you read it in nuclear notation, which means we have to find the atomic number just 92 and it undergoes beta decay.

So that means that it produces this particle find the second particle we used the atomic number, so 92 equals minus one plus x, where X equals 93 which is Neptune IAM. The mass number of our new isotope is zero plus X equals to 39 where X equals to 39. This product becomes the reactant in my next decay, which is also a beta decay. And to find the unknown element we do the same here.

Except for that it's 93 equals minus one plus x, where X is 94 which is P u plutonium, and the mass number is zero plus X equals to 39 or to 39. The next decay starts with the isotope that we just form to 39 p. U. This time it's an Alpha decay. So we produce this particle to find the unknown. Element 94 equals two plus x, where X equals 92 which takes us back to uranium.

Find the mass number of this isotope 2 39 equals four plus X, where X equals to 35. Finally, for the last decay, you have another Alpha decay starting with uranium to 35 making an alpha particle. The atomic number will be 90 which is T H and the top is 2 31 For the mass number. This begins the natural decay, series of thorium .

Learn more about chemical decay

brainly.com/question/1898040

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Calculate the pH after 10 mL of 1.0 M sodium hydroxide is added to 60 mL of 0.5M acetic acid.
baherus [9]

Explanation:

It is given that the total volume is (10 mL + 60 mL) = 70 mL.

Also, it is known that M_{1}V_{1} = M_{2}V_{2}

Where,    V_{1} = total volume

               V_{2} = initial volume

Therefore, new concentration of CH_{3}COOH = \frac{M_{2}V_{2}}{V_{1}}

                                        = \frac{60 \times 0.5}{70}

                                        = 0.43 M

New concentration of NaOH = \frac{M_{2}V_{2}}{V_{1}}

                                               = \frac{10 \times 1.0}{70}

                                               = 0.14 M

So, the given reaction will be as follows.

              CH_{3}COOH + OH^{-} \rightarrow CH_{3}COO^{-} + H_{2}O

Initial:             0.43          0.14                     0

Change:          -0.14        -0.14                    0.14

Equilibrium:    0.29          0                       0.14

As it is known that value of pK_{a} = 4.74

Therefore, according to Henderson-Hasselbalch equation calculate the pH as follows.

           pH = pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}

                 = 4.74 + log \frac{0.14}{0.29}

                 = 4.74 + (-0.316)

                 = 4.42

Therefore, we can conclude that the pH of given reaction is 4.42.

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