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Archy [21]
3 years ago
9

Positrons are spontaneously emitted from the nuclei of

Chemistry
2 answers:
Fudgin [204]3 years ago
6 0

Answer: Option (1) is the correct answer.

Explanation:

  • A positron is a small particle which contains a +1 charge and its mass is equal to the mass of an electron, that is, 9.109 \times 10^{-31} kg.

A positron is represent by the symbol ^{0}_{+1}\beta.

For example, ^{37}_{19}K \rightarrow ^{37}_{18}Ar + ^{0}_{+1}\beta

  • Whereas nitrogen-16 does not decay to give a positron because nitrogen-16 is the daughter isotope formed due to decay of oxygen-16.

The decay reaction is as follows.

           ^{16}_{8}O \rightarrow ^{16}_{7}N + ^{0}_{+1}\beta

  • And, on radioactive decay of thorium-232 there will be formation of radium-228 along with emission of an alpha particle.

The reaction will be as follows.

            ^{232}_{90}Th \rightarrow ^{228}_{88}Ra + ^{4}_{2}\alpha

Therefore, we can conclude that out of the given options positrons are spontaneously emitted from the nuclei of potassium-37.

Leto [7]3 years ago
4 0
Positrons are spontaneously emitted from the nuclei of potassium -37.
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Answer:

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Explanation:

The Santa Ana wind occurs in early autumn. They bring hot and dry weather to areas around the South-West coast. They move at high speeds, affecting most parts of Southern California.

Monsoon is a seasonal wind that blows across Southern Asia. It usually blows in summer. It is a a seasonally changing wind.

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in an experiment 3.5g of element A reacted with 4.0g of element G to form a compound Calculate the empirical formula for this co
kolezko [41]

Additional information

Relative atomic mass(Ar) : A=7, G=16

The empirical formula : A₂G

<h3>Further explanation</h3>

Given

3.5g of element A

4.0g of element G

Required

the empirical formula for this compound

Solution

The empirical formula is the smallest comparison of atoms of compound forming elements.

The empirical formula also shows the simplest mole ratio of the constituent elements of the compound

mol of element A :

\tt mol=\dfrac{mass}{Ar}\\\\mol=\dfrac{3.5}{7}=0.5

mol of element G :

\tt mol=\dfrac{4}{16}=0.25

mol ratio A : G = 0.5 : 0.25 = 2 : 1

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