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3 years ago

What determines how many items make up a particular unit

2 answers:

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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3 years ago

Read 2 more answers

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

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FromTheMoon [43]

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No horse play

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