Is mercury chloride ionic or covalent

5. Azulene is a beautiful blue hydrocarbon. If 0.106 g of the compound is burned in oxygen, 0.364 g of CO2 and 0.0596 g of H2O a

max2010maxim [7]

Answer: The empirical and molecular formula for the given organic compound is CH and $C_{10}H_{10}$ and it is not an alkane.

Explanation:

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

$C_xH_y+O_2\rightarrow CO_2+H_2O$

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

We are given:

Mass of $CO_2=0.364g$

Mass of $H_2O=0.0596g$

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 0.364 g of carbon dioxide, $\frac{12}{44}\times 0.364=0.0993g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.0596 g of water, $\frac{2}{18}\times 0.0596=0.0067g$ of hydrogen will be contained.

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.0993g}{12g/mole}=0.00828moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0067g}{1g/mole}=0.0067moles$ ]

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 0.0067 moles.

For Carbon = $\frac{0.00828}{0.0067}=1.23\approx 1$

For Hydrogen = $\frac{0.0067}{0.0067}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is CH

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 = 128.2 g/mol

Mass of empirical formula = 13 g/mol

Putting values in above equation, we get:

$n=\frac{128.2g/mol}{13g/mol}=9.86\approx 10$

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

$C_{(1\times 10)}H_{(1\times 10)}=C_{10}H_{10}$

The general formula of an alkane is $C_nH_{(2n+2)}$ , where n = any natural number

Here, n = 10 and it does not satisfy being an alkane

Hence, the empirical and molecular formula for the given organic compound is CH and $C_{10}H_{10}$ and it is not an alkane.

7 0

4 years ago

A transverse wave moves a medium at an abtuse angle to the wave , true or false ?

skelet666 [1.2K]

Answer:

True! A transverse wave does move a medium at an obtuse angle to the wave!!

4 0

3 years ago

Show your calculation by uploading a picture. Calculate the molar mass of ammonia, NH3

Cloud [144]

Answer:

17.04 g/mol

Explanation:

Molar Mass of NH₃

we know that

Nitrogen has 14.01 gram/mol

And Hydrogen has 1.01 gram/mol

but we have 3 Hydrogens So we multiply

1.01 by 3 i.e., 3.03

Now, add

14.01

+ 3.03

17.04

So, The molar mass of ammonia, NH₃ is

17.04 g/mol

-TheUnknownScientist

5 0

3 years ago

gypsum is insoluble in water. you are asked to purify a sample of gypsum that is contaminated with a soluble salt.

katen-ka-za [31]

Answer:

a. The apparatus required to purify gypsum sample are: Bunsen burner, beaker, Filter Funnel, stirring rod, the filter paper.

b. Gypsum is a sulfate mineral that is made up of calcium sulfate dihydrate. Step-by-step instruction to purify gypsum sample is as follows:

1. Add water to the gypsum sample in a beaker.

2. Use the stirring rod to mix the mixture well.

3. Filter off the excess solid from the mixture using the filter paper and filter funnel.

4. Put the filtered mixture over the bunsen burner and evaporate the excess water from the mixture.

5. Allow the hot liquid to cool down and filter it again through the filter paper to get the pure gypsum.

7 0

3 years ago

How might the biodiversity of a mowed lawn compare to that of huge weedy field?

Dennis_Churaev [7]

Answer: The mowed lawn is the one from where the grasses are removed by using the machines or tools.

Explanation:

The mowed lawn is expected to have low number of species as the grasses may be few or scanty thus can support the population of few species like insects, mice, birds, and small number of grazing animals. On the other hand the weedy field can be hub of insects, reptiles like snakes, small mammals, and large mammals. Large weed field can provide food, and habitat to the large number of species. This will support the increase in biodiversity as compared to the mowed lawn.

5 0

3 years ago