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3 years ago

Which of the following compounds exhibit hydrogen bonding?

Chemistry

1 answer:

hodyreva [135]3 years ago

7 0

Answer:

a,b,e,f,g

Explanation:

Hydrogen bonding is an intermolecular force of attraction between two molecules. It is a special type of dipole-dipole attraction.

In hydrogen bonding, the hydrogen on one molecule binds with an electronegative atom on another molecule usually oxygen, nitrogen and fluorine.

The simple electrostatic attraction leads to strong intermolecular interaction between two molecules.

For the formation of hydrogen bond, a hydrogen atom must be bonded to more electronegative specie usually N, O and F

In H₂O and H₂O₂ ; hydrogen bonds with Oxygen

CH₃NH₂; hydrogen is attached to nitrogen

HF; hydrogen is attached to fluorine

CH₃OH; hydrogen is attached to oxygen

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Name the type of chemical bonding in potassium bromide

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The bond between Potassium and Bromide is considered iconic, Because of how they lose electrons and become ions when they bond together.

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3 years ago

4. If the heat of reaction is written on the reactant side of a chemical equation, what does that indicate about the reaction?

Licemer1 [7]

Answer:

B is the correct anwer

Explanation:

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3 years ago

Compare the root mean square speeds of O2 and UF6 at 65 degree Celsius

gayaneshka [121]

<h3>Answer:</h3>

The root mean square speeds of O₂ and UF₆ is 513m/s and 155 m/s respectively.

<h3>Solution and Explanation:</h3>

To find how fast molecules or particles of gases move at a particular temperature, the root mean square speed is calculated.
Root mean square speed of a gas is calculated by using the formula;

$Root mean square=\sqrt{ \frac{3RT}{M}$

Where R is the molar gas constant, T is the temperature and M is the molar mass of gas in Kg.

<h3>Step 1: Root mean square speed from O₂</h3>

Molar mass of Oxygen is 32.0 g/mol or 0.032 kg/mol

Temperature = 65 degrees Celsius or 338 K

Molar gas constant = 8.3145 J/k.mol

$Root mean square speed = \sqrt\frac{(3)(8.3145)(338K)}{0.032}$

$= 513.289 m/s$

<h3>Step 2: Root mean square speed of UF₆ </h3>

The molar mass of UF₆ is 352 g/mol or 0.352 kg/mol

$Root mean square speed = \sqrt\frac{(3)(8.3145)(338K)}{0.352}$

$= 154.762m/s$

Therefore; the root mean square speeds of O₂ and UF₆ is 513m/s and 155 m/s respectively.

7 0

3 years ago

For the following reaction, 8.70 grams of benzene (C6H6) are allowed to react with 13.7 grams of oxygen gas. benzene (C6H6) (l)

artcher [175]

Answer:

Maximum amount of carbon dioxide that can be formed → 7.52 g

Limiting reactant → O₂

Amount of the excess reagent, after the reaction occurs → 12.9 g

Explanation:

We determine the reaction. This is a combustion:

2C₆H₆ (l) + 15O₂ (g) → 6CO₂(g) + 6H₂O (g)

We need to determine the limting reactant so we convert the mass to moles:

8.70 g. 1mol / 78g = 0.111 moles of benzene

13.7 g . 1mol / 32g = 0.428 moles of oxygen

Ratio is 2:15. 2 moles of benzene react with 15 moles of O₂

Then, 0.111 moles of benzene may react with (0.111 .15) /2 = 0.832 moles of O₂

We have 0.428 moles but we need 0.832 moles for the complete reaction, so there are (0.832 - 0.428) = 0.404 moles remaining. Oxygen is the limiting reactant. We work now, with the reaction:

15 moles of O₂ can produce 6 moles of CO₂

So, 0.428 moles of O₂ may produce (0.428 . 6)/ 15 = 0.171 moles of CO₂

We convert the moles to mass → 0.171 mol . 44g / 1mol = 7.52 g

This is the maximum amount of carbon dioxide that can be formed

We convert the mass of the limiting reactant that remains after the reaction is complete → 0.404 mol . 32g / 1mol = 12.9 g of O₂

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3 years ago

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Akimi4 [234]

Answe5r:

Explanation:

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4 years ago