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xenn [34]
3 years ago
13

What is emitted when the radioactive nucleus of 131 53i decays to form the stable isotope of xenon 131 54xe?

Chemistry
1 answer:
raketka [301]3 years ago
3 0
It is a beta particle that is being emitted when the radioactive nucleus of 131 / 53 I decays to form the stable isotope of xenon 131 / 54 Xe. It is written as 0/-1 e-. It is a high speed and high energy positron or electron during beta decay. An unstable nucleus containing an excess of neutrons would unergo a beta decay wherein the neutron is transformed into a proton, and electron and an antiparticle. The electrons emitted are not the electrons from the shells  but are being produced when a neutron splits firming a proton and an electron. These particles would be able to pass in paper but can be blocked by an aluminum foil.
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Which of the following is a substance that is found between the cell membrane and the nuclear us which primarily consist of wate
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3 0
2 years ago
The transuranium synthesis process involves creating a transuranium element through _____.
raketka [301]
The transuranium synthesis process involves creating a transuranium element through the transmutation of a lighter element.
So, the answer would be the C the transmutation of a lighter element.
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6 0
2 years ago
A 7.74 L balloon is filled with water at 3.88 atm. If the balloon is squeezed into a 0.23 L beaker and does NOT burst, what is t
Vladimir [108]

Answer:

131 atm

Explanation:

To find the new pressure, you need to use Boyle's Law:

P₁V₁ = P₂V₂

In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the new pressure (P₂) by plugging the given values into equation and simplifying.

P₁ = 3.88 atm                       P₂ = ? atm

V₁ = 7.74 L                           V₂ = 0.23 L

P₁V₁ = P₂V₂                                                      <----- Boyle's Law

(3.88 atm)(7.74 L) = P₂(0.23 L)                       <----- Insert values

30.0312 = P₂(0.23 L)                                      <----- Simplify left side

131 = P₂                                                          <----- Divide both sides by 0.23

6 0
1 year ago
Menstruation occurs on a ______ interval from puberty until menopause
dmitriy555 [2]
D. One Month Interval
8 0
2 years ago
How many atoms of oxygen are present in 7.51 grams of<br> glycine with formula C₂H5O2N?
Blizzard [7]

1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.

How to calculate number of atoms?

The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

However, the number of moles of oxygen in glycine can be calculated using the following expression:

Molar mass of C₂H5O2N = 75.07g/mol

Mass of oxygen in glycine = 32g/mol

Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine

Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles

Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms

Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.

Learn more about number of atoms at: brainly.com/question/8834373

#SPJ1

3 0
1 year ago
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