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3 years ago

What type(s) of intermolecular forces are expected between ch3ch2ch2ch2oh molecules?

Chemistry

2 answers:

Alexxx [7]3 years ago

8 0

Answer:

1. Hydrogen bonding

2. Dipole dipole interaction

3. London dispersion force

Explanation:

Intermolecular forces refers to the force of attraction or force of repulsion between two molecules of same or other type.

In this case three types of Intermolecular forces acting:

1. Hydrogen bonding between O and H atom of different molecules.

2. Dipole dipole interaction between C and O atom because of great electronegative difference.

3. London dispersion force it is between two $-CH_3$ group of different molecules. It is also known as induced dipole force.

Hence, this is the required solution.

12345 [234]3 years ago

5 0

Answer : Hydrogen-bonding, Dipole-dipole attraction and London-dispersion force.

Explanation :

The given molecule is $CH_3CH_2CH_2CH_2OH$ .

Three types of inter-molecular forces are present in this molecule which are Hydrogen-bonding, Dipole-dipole attraction and London-dispersion force.

Hydrogen-bonding : when the partial positive end of hydrogen is bonded with the partial negative end of another molecule like, oxygen, nitrogen, etc.

Dipole-dipole attraction : When the partial positively charged part of the molecule is interact with the partial negatively charged part of the molecule. For example : In case of HCl.

London-dispersion force : This force is present in all type of molecule whether it is a polar or non-polar, ionic or covalent. For example : In case of Br-Br , F-F, etc

Hydrogen-bonding is present between the oxygen and hydrogen molecule.

Dipole-dipole forces is present between the carbon and oxygen molecule.

London-dispersion forces is present between the carbon and carbon molecule.

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3 years ago

Difference between dipole dipole and hydrogen bonding

riadik2000 [5.3K]

Hydrogen bonds are stronger than the dipole dipole attraction force present in any molecule.

<h3>What is bonding in molecules?</h3>

Bonding is a type of attraction force which is present between the different atoms or elements of any substance.

Dipole dipole attraction force is a weak force as compared to the hydrogen bonding and present between any two oppositely charged atoms.
Hydrogen bond is present between the hydrogen atom and more electronegative atoms like O, S, N and F.

Hence main difference is that hydrogen bond is only present between the hydrogen atom and more electronegative.

To know more about dipole-dipole force, visit the below link:
brainly.com/question/24197168

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2 years ago

in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

dimulka [17.4K]

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{12}{4}=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{1}{4}=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{6}{4}=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{10}{4}=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_{H_2O}=X_{H_2O}\times p_T$

where,

$p_{H_2O}$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_{H_2O}$ = mole fraction of water vapor gas = 0.345

Now put all the given values in the above formula, we get:

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.345\times 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

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