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Anna11 [10]
4 years ago
12

In the nuclear transmutation, 168O (?, α)137N, what is the bombarding particle? In the nuclear transmutation, O (?, )N, what is

the bombarding particle? an alpha particle a gamma photon a proton a beta particle a phosphorus nucleus.
Chemistry
1 answer:
xeze [42]4 years ago
5 0

Answer:

The bombarding particle is a Proton

Explanation:

A Nuclear transmutation reaction occurs when radioactive element decay, usually converting them from one element/isotope into another element. Transmutation is the process which causes decay, generally, alpha or beta.

¹⁶₈O(P,alpha) ¹³₇N, can be written as

¹⁶₈O + x goes to ¹³₇N + ⁴₂He

Where x can be anything, balancing the equation in order to give us the correct amount of proton number and nucleus number

16 + x = 13 + 4

x = 17 – 16 = 1, Hence we can say that x = ¹₁P

<u>¹⁶₈O + ¹₁P goes to ¹³₇N + ⁴₂He</u>

Here we can clearly see the bombarding particle is ¹₁P (proton). The ejected particle being ⁴₂He which is also known as an alpha particle

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Which one is the catalyst?
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The concentration of pb2+ in a commercially available standard solution is 1.00 mg/ml. what volume of this solution should be di
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3 0
3 years ago
on dr v’s bookshelf there are 23 science textbooks each with an average of 31 chapters of each chapter contains 5 pages how many
olga55 [171]

Answer:

There are 3600 pages to 2 significant figures

Explanation:

Working only with the information given, we have that there are 23 books each with about 31 chapters.

This means that if 1 book has 31 chapter, the 23 science books will have

31 X 23 chapters =713 chapters in total.

If each chapter contains 5 pages, for the 713 chapters we will have a total of 713 X 5 pages = 3565 pages.

Rounding off the answer to 2 significant figures , we are looking to have only two non-zero figures in our answer. To do this, we count two numbers from the left. This will be numbers 3 and 5.

Next,  we check the net number after 5 to see if it is greater than 5 if it is, we will have to add the number (1 ) to the 5. This shows that we are approximating upwards, to account for the fact that that number has passed the middle point.

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8 0
3 years ago
1.) Calculation: If 9.02 x 1024 particles of vinegar (HC2H3O2)HC2H3O2) are added to 16.5 moles of eggshell (CaCO3) and 6.35 mole
blsea [12.9K]

The theoretical yield of acetate is 2607 g. The actual yield of acetate is 1066.8 g. The percentage yield of acetate is 41%.

If 1 mole of vinegar contains 6.02  x 10^23 particles

x moles of vinegar contains 9.02 x 10^24 particles

x = 1 mole x 9.02 x 10^24 /6.02 x 10^23

x = 15 moles of vinegar

The reaction is as follows;

2HC2H3O2 + CaCO3 -----> Ca(C2H3O2)2 + H2O + CO2

Since 2 moles of vinegar reacts with 1 mole of carbonate

x moles of vinegar reacts with 16.5 moles of carbonate

x =  2 moles x 16.5 moles/ 1 mole

x = 33 moles of vinegar

We can see that the vinegar is the reactant in excess hence the carbonate is the limiting reactant.

Theoretical yield = 16.5 moles x 158 g/mol = 2607 g

Actual yield = 6.35 moles  x 158 g/mol = 1066.8 g

Percent yield = 1066.8 g/2607 g × 100/1

= 41%

Learn more: brainly.com/question/13440572?

7 0
3 years ago
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