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Harrizon [31]
3 years ago
12

What determines how many items make up a particular unit

Chemistry
2 answers:
Ierofanga [76]3 years ago
3 0
The Avogadro's number determines how many items make up a particular unit, such that 1 mole of a substance contains 6.022 × 10^23 items.
These items may be particles such as molecules, atoms, ions , etc
Therefore, 1 mole of anything has 6.022×10^23 items.
This number is called the Avogadro's constant and applies to all substances, and is normally used by researchers and scientists to determine the number of moles of elements and compounds.
mel-nik [20]3 years ago
3 0

Answer:

Avogadro's number determines how many elements make up a particular unit.

Explanation:

According to the International System, the mole is defined as the amount of substance that contains as many entities (atoms, molecules, ions, electrons or other entities) as the number of atoms in 0.012 kg of pure carbon-12. According to this definition, the parameters are in their lowest state of energy, at rest and without interacting with others.

On the other hand, the Avogadro Number or Avogadro Constant is called the number of particles that have a substance (usually atoms or molecules) and that can be found in the amount of one mole of that substance. So Avogadro's number represents the amount of atoms in 0.012 kg of carbon 12. Its value is 6.023 * 1023 particles per mole. Avogadro's number represents a quantity without an associated physical dimension, so it is considered a pure number that allows describing a physical characteristic without dimension or explicit unit of expression. Avogadro's number applies to any substance.

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Answer:

4066 decay/s

Explanation:

Given that:-

The weight of the person is:- 190 lb

Also, 1 lb = 453.592 g

So, weight of the person = 86182.6 g

Also, given that carbon is 18% in the human body. So,

\%\ Carbon=\frac{18}{100}\times 86182.6\ g=15512.868\ g

Carbon-14 is 1.6\times 10^{-10}\ \% of the carbon in the body. So,

\%\ Carbon-14=\frac{1.6\times 10^{-10}}{100}\times 15512.868\ g=2.48\times 10^{-8}\ g

Also,

14 g of Carbon-14 contains  6.023\times 10^{23} atoms of carbon-14

So,  

2.48\times 10^{-8}\ g of Carbon-14 contains  \frac{6.023\times 10^{23}}{14}\times 2.48\times 10^{-8} atoms of carbon-14

Atoms of carbon-14 =  1.07\times 10^{15}

Given that:

Half life = 5730 years

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k=\frac {ln\ 2}{t_{1/2}}

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The rate constant, k = 0.00012 years⁻¹

Also, 1 year = 3.154\times 10^7 s

So, The rate constant, k = \frac{0.00012}{3.154\times 10^7} s⁻¹ = 3.8\times 10^{-12}\ s^{-1}

Thus, decay events per second = K\times atoms decayed = 3.8\times 10^{-12}\times 1.07\times 10^{15}\ decay/s = 4066 decay/s

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