Write the name of the branched alkane next to the drawing of the molecule.

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 $C_2H_2O_4$

3 0

3 years ago

Sort these isotopes by whether they are most likely to undergo fusion or fission.

Maslowich

Answer:

Uranium-233- fission

Plutonium-239- fission

Plutonium-241- fission

Hydrogen-3 fusion

Hydrogen-1 fusion

Helium-3 fusion

Explanation:

In nuclear fission, heavy nuclear disintegrate into smaller nuclei when bombarded with particles such as neutrons. Fission reaction is common among nuclei having a high atomic number such as plutonium and uranium.

Fusion occurs between two light nuclei such as hydrogen or helium. It involves the combination of two lighter elements to give a heavier element with the release of tremendous amount of energy.

4 0

2 years ago

Another one please help

ladessa [460]

Possibly C or D i wanna say sorry if i’m wrong

4 0

2 years ago

Thirty grams of lead oxide and fifteen grams of ammonia react completely to produce solid lead, nitrogen gas, and liquid water.

Kitty [74]

45 g Thirty grams of lead oxide and fifteen grams of ammonia react completely to produce solid lead, nitrogen gas, and liquid water.

8 0

3 years ago

Read 2 more answers

a mass of mercury occupies 0.350 l. what volume would an equal mass of ethanol occupy? the density of mercury is 13.546 g/ml and

Fed [463]

The density told you the mass covered by one unit of volume of that substance.

Density defines how perfectly the molecules of a substance are packed in a unit of volume.

In the given problem the substances are expressed in g/ml which means that a unit of volume will be 1 mL.

Given ,

Density of mercury = 13.546 g/ml

It means 1mL of mercury will have a mass of 13.546 g, Similarly 1 mL of ethanol will have a mass of 0.789 g.

0.350 L * 10^3 ml/1 L* 13.546/1 mL = 4,741.1 g

The volume of ethanol will have an equal mass is

4741.1 g* 1 mL/0.789g = 6,008.9 mL

6,008.9 mL* 1L/10^3mL = 6.0081 L

Hence the answer is 6.0081 L

To know more about density occupies here :

brainly.com/question/18518241?referrer=searchResults

#SPJ4

8 0

1 year ago