All categories

3 years ago

Describe the formation of a hydrogen bond and explain how it differs from a covalent or ionic bond

Chemistry

1 answer:

vlada-n [284]3 years ago

3 0

Answers:

Hydrogen bond is formed when positive end of one molecule attracted negative end of other molecule. The concept is similar to magnetic attraction where opposite poles attract each other.

Explanation:

While understanding hydrogen bond, two terms are important, one is electronegativity (tendency of atoms to attract electrons towards itself) and other is dipole (separation of positive and negative charge in a molecule). Hydrogen bond is always formed between hydrogen atom and other atoms having electronegativity different than hydrogen.

Differ from Ionic and Covalent Bonds:

Hydrogen bond is weaker than ionic and covalent bond. Ionic and covalent bonds are intramolecular (within the molecule) whereas hydrogen bond is intermolecular (between molecules).

Example:

Formation of hydrogen bonding in water.

You might be interested in

3. At a pressure of 405 kPa, the volume of a gas is 6.00 cm

monitta

The pressure gets increased to 486 kPa from 405 kPa, when the volume is decreased from 6 cm³ to 4 cm³.

Explanation:

In the present problem, the temperature is said to remain at constant and there is change in the pressure. So according to Boyle's law, the relationship between pressure and volume of any gaseous objects are inversely related to each other. In other words, the pressure attained by gas molecules in a container will be inversely proportional to the volume of the gas molecules occupied in the container, at constant temperature.

$V=\frac{1}{P}$

So, if two volumes V₁ and V₂ are considered, then their respective pressure will be represented as P₁ and P₂. Then, as per Boyle's law,

$V_{1}P_{1}=V_{2}P_{2}$

So let us consider, V₁ = 6 cm³ and V₂ = 4 cm³ and pressure P₁ = 405 kPa and we have to determine P₂.

Then, $6*405=5*P_{2}\\ \\P_{2}=\frac{2430}{5} =486 kPa$

So, the pressure at new volume of 4 cm³ is 486 kPa. It can be seen that as there is decrease in the volume, there is an increase in the pressure. So it satisfied the Boyle's law.

Thus, the pressure gets increased to 486 kPa from 405 kPa, when the volume is decreased from 6 cm³ to 4 cm³.

6 0

3 years ago

Question 2 of 15

guapka [62]

Answer:

B. The products are nuclei of elements that are different from the original elements.

6 0

1 year ago

A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution

Dmitrij [34]

Answer:

The change in internal energy is - 1.19 kJ

Explanation:

Step 1: Data given

Heat released = 3.5 kJ

Volume calorimeter = 0.200 L

Heat release results in a 7.32 °C

Temperature rise for the next experiment = 2.49 °C

Step 2: Calculate Ccalorimeter

Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C

Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C

Step 3: Calculate energy released

Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ

Step 4: Calculate change in internal energy

ΔU = Q + W W = 0 (no expansion)

Qreac = -Qcal = - 1.19 kJ

ΔU = - 1.19 kJ

The change in internal energy is - 1.19 kJ

4 0

3 years ago

Question 5

pantera1 [17]

Answer: The coefficient for the diatomic oxygen (O2) is 3.

Explanation:

To know the coefficient for the diatomic Oxygen, we need to balance the equation.

Fe + O2 -------> Fe2O3

LHS of the equation; Fe = 1 , O2 = 1

RHS of the equation; Fe = 2 , O = 3

∴ Multiply 'Fe' on the LHS of the equation by 4 and O2 by 3

Doing that will give the balance equation which is;

4 Fe + 3 O2 --------> 2 Fe2O3

The coefficient for the diatomic oxygen (O2) as seen from the equation is 3.

7 0

3 years ago

What’s number 1, 2, 3 Major grade 6th grade

Maslowich

Answer:

1 is B

2 is A

3 is A

Explanation:

Hope you pass

8 0

2 years ago