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NemiM [27]
2 years ago
13

Determine what type of compound is shown below:

Chemistry
1 answer:
34kurt2 years ago
3 0

Answer:

Polar

Explanation:

Methanol is not electrostatic

It's not metallic as it has no metals in it

It's not ionic as coordinate bonding is made.

It's polar as a lone pair is present over OH-

You might be interested in
N2+3H2—> 2NH3 What volume of NH3 at STP is produced if 25.0g of N2 is reacted with an excess of H2 ?
Butoxors [25]

Answer:

The answer to your question is 40 L of NH₃

Explanation:

Data

Volume of NH₃ = x

mass of N₂ = 25 g

mass of H₂ = excess

Balanced chemical reaction

                     N₂  +  3H₂   ⇒  2NH₃

Process

1.- Find the molar mass of N₂ and NH₃

N₂ = 14 x 2 = 28g

2NH₃ = 2[ 14 + 3] = 34 g

2.- Write a proportion to solve this problem

                    28 g of N₂ --------------- 34 g of NH₃

                     25 g of N₂ -------------   x

                     x = (25 x 34)/28

                     x = 30.36 g of NH₃

3.- Calculate the volume of NH₃

                     17 g of NH₃ -------------- 22.4 L

                     30.36 g of NH₃ --------  x

                     x = (30.36 x 22.4) / 17

                     x = 40 L

6 0
3 years ago
A) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO2 and 1.37 mg of H2O. If the compound contains o
Mumz [18]

<u>Answer:</u>

<u>For a:</u> The empirical formula for the given compound is CH

<u>For b:</u> The empirical and molecular formula for the given organic compound are C_{10}H_{20}O

<u>Explanation:</u>

  • <u>For a:</u>

The chemical equation for the combustion of hydrocarbon follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

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

We are given:

Conversion factor used:  1 g = 1000 mg

Mass of CO_2=5.86mg=5.86\times 10^{-3}g

Mass of H_2O=1.37mg=1.37\times 10^{-3}g

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

So, in 5.86\times 10^{-3}g  of carbon dioxide, \frac{12}{44}\times 5.86\times 10^{-3}=1.60\times 10^{-3}g of carbon will be contained.

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

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

So, in 1.37\times 10^{-3}g of water, \frac{2}{18}\times 1.37\times 10^{-3}=0.152\times 10^{-3}g of hydrogen will be contained.

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{1.60\times 10^{-3}g}{12g/mole}=0.133\times 10^{-3}moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.152\times 10^{-3}g}{1g/mole}=0.152\times 10^{-3}moles

  • <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 0.133\times 10^{-3} moles.

For Carbon = \frac{0.133\times 10^{-3}}{0.133\times 10^{-3}}=1

For Hydrogen = \frac{0.152\times 10^{-3}}{0.133\times 10^{-3}}=1.14\approx 1

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

The ratio of C : H = 1 : 1

Hence, the empirical formula for the given compound is CH

  • <u>For b:</u>

The chemical equation for the combustion of menthol 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  = 0.2829 g

Mass of H_2O = 0.1159 g

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

So, in 0.2829  g of carbon dioxide, \frac{12}{44}\times 0.2829=0.077g of carbon will be contained.

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

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

So, in 0.1159 g of water, \frac{2}{18}\times 0.1159=0.013g of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.013) = 0.105 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{0.077g}{12g/mole}=0.0064moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.013g}{1g/mole}=0.013moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0105g}{16g/mole}=0.00065moles

  • <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 0.00065 moles.

For Carbon = \frac{0.0064}{0.00065}=9.84\approx 10

For Hydrogen = \frac{0.013}{0.00065}=20

For Oxygen = \frac{0.00065}{0.00065}=1

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

The ratio of C : H : O = 10 : 20 : 1

The empirical formula for the given compound is C_{10}H_{20}O

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 = 156 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

n=\frac{156g/mol}{156g/mol}=1

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

C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O

Hence, the empirical and molecular formula for the given organic compound are C_{10}H_{20}O

3 0
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How many grams are there in 4.60 x 10 24atoms of silver?
Snezhnost [94]

Answer:

820 g (i think)

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What is a segment of dna that contains a single unit of information?
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This segment is called a gene
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Answer:

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