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dalvyx [7]
4 years ago
14

What is the chemical name of the covalent compound P4O9?

Chemistry
1 answer:
Lunna [17]4 years ago
3 0

P4O9 is an oxide of phosphorous

so it will named accordingly

There are four phosphorous atoms and nine oxygen atoms

four will be represented as "tetra"

Nine will represented as "non"

so the name of the compound will be

tetraphosphorous nonoxide


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The decay of radioactive elements occurs at a fixed rate. The half-life of a radioisotope is the time required for one half of t
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Answer:

100 grams of C-14 decays to 25 grams in 11,460 years.

The C-14 isotope is only useful for dating fossils up to about 50,000 years old

If an ancient bone contains 6.25% of its original carbon, then the bone must be 22,920 years old.

Explanation:

We already know that the half life of C-14 is 5,730 years. After the first half life, we have 50 grams remaining. This takes 5,730 years. After the second half life (11,460 years now gone) we have 25 grams of C-14  left.

If a fossil material is  older than 50,000 years an undetectable amount of 14C is left in the sample hence Carbon-14 is no longer suitable for dating the sample.

From;

0.693/5730 = 2.303/t log (No/0.0625No)

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3 years ago
The smell of dirty gym socks is caused by the compound caproic acid(contains H, C, O). Combustion of 0.844 g of caproic acid pro
adoni [48]

<u>Answer:</u> The molecular formula for the given organic compound is C_2H_{32}O_4

<u>Explanation:</u>

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=0.784g

Mass of H_2O=1.92g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 0.784 g of carbon dioxide, \frac{12}{44}\times 0.784=0.214g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in 1.92 g of water, \frac{2}{18}\times 1.92=0.213g of hydrogen will be contained.

Mass of oxygen in the compound = (0.844) - (0.214 + 0.213) = 0.417 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.214g}{12g/mole}=0.0134moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.213g}{1g/mole}=0.213moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.417g}{16g/mole}=0.0261moles

  • <u>Step 2:</u> 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 0.0134 moles.

For Carbon = \frac{0.0134}{0.0134}=1

For Hydrogen = \frac{0.213}{0.0134}=15.89\approx 16

For Oxygen = \frac{0.0261}{0.0134}=1.95\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 16 : 2

The empirical formula for the given compound is CH_{16}O_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 = 116.2 g/mol

Mass of empirical formula = 60 g/mol

Putting values in above equation, we get:

n=\frac{116.2g/mol}{60g/mol}=1.94\approx 2

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

C_{(1\times 2)}H_{(16\times 2)}O_{(2\times 2)}=C_2H_{32}O_4

Hence, the molecular formula for the given organic compound is C_2H_{32}O_4

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