Can someone tell me what I fill in the blanks

A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure

ycow [4]

Answer: The molecular formula for the compound is $C_6H_6$

Explanation:

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles$

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 7.68 moles.

For Carbon = $\frac{7.68}{7.68}=1$

For Hydrogen = $\frac{7.74}{7.68}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is $CH$

Calculating the molar mass of the compound:

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{m}{M}RT$

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = $62.3637\text{ L. torr }mol^{-1}K^{-1}$

T = temperature of the gas = $100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol$

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 = 78.10 g/mol

Mass of empirical formula = 13 g/mol

Putting values in above equation, we get:

$n=\frac{78.10g/mol}{13g/mol}=6$

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

$C_{(1\times 6)}H_{(1\times 6)}=C_6H_6$

Hence, the molecular formula for the compound is $C_6H_6$

8 0

3 years ago

Which change in oxidation number indicates oxidation? A) Ito +2 C +2 to -3 B) Ito -2 D) +3 to +2

tatuchka [14]

As there is the increase in Oxidation number from +1 to +2. Hence, the correct Option (A) i.e, 1 to +2,

<h3>What is Oxidation ?</h3>

Oxidation doesn't necessarily have anything to do with oxygen.

Oxidation occurs when an atom, molecule, or ion loses one or more electrons in a chemical reaction.
When oxidation occurs, the oxidation state or oxidation number of the chemical species increases.
Oxidation doesn't necessarily involve oxygen! Originally, the term was used when oxygen caused electron loss in a reaction.

Hence, Option (A) indicates the oxidation as there is the increase in Oxidation number from +1 to +2.

Learn more about Oxidation here ;

brainly.com/question/13110055

#SPJ1

3 0

2 years ago

Some one help me please I'm stuck Show using two conversion factors how you would convert from 0.020 kg into mg.

julia-pushkina [17]

Answer:

$=2.0x10^4mg$

Explanation:

Hello there!

In this case, when performing units conversions involving two proportional factors we need to make sure we first convert to the base unit and then to the target one; thus, since 1 kg = 1000 g and 1 g = 1000 mg, we set up the following expression:

$=0.020kg*\frac{1000g}{1kg} *\frac{1000mg}{1g} \\\\=2.0x10^4mg$