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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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8 0

1 year ago

What is the percent composition by mass of sulfur in the compound MgSO4 (gram formula mass = 120 g/mol)?

slavikrds [6]

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate. Explanation: Molar mass of compound = 120 g/mol.

7 0

2 years ago

Plsssss helppppp asapppppppppp

Marianna [84]

Answer:

Under the concept of popular sovereignty, the people of each territory would decide whether or not slavery would be permitted.

Explanation:

hopes this help 3> D:

6 0

2 years ago

During combustion, methane yields carbon dioxide and water. The unbalanced equation for this reaction is:CH4(g)+O2(g) → CO2(g)+

almond37 [142]

Answer:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Mole ratios for the balanced equation be:

1:2, 2:1, 1:1, 1:2, 2:2

Explanation:

CH4(g)+O2(g) → CO2(g)+ H2O(l)

To balance a chemical equation, you must have the same mole of each element in both sides of the reaction (reactant side and product side)

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

2 C

8 H

8 O

In both side.

It takes 1 mole of methane to react with 2 mole of oxygen in order to produce 1 mol of dioxide and 2 mole of water.

Mole ratios for the balanced equation be:

1:2, 2:1, 1:1, 1:2, 2:2 - We should compare each compound

1 mol methane → 2 mole of oxygen

2 mole of oxygen → 1 mol of methane

2 mole of oxygen → 1 mol of dioxide

1 mole of dioxide → 1 mol of Methane

2 mole of water → 2 mole of oxygen (the same as opposite)

1 mol methane → 2 mole of water

2 mol of water → 1 mol of methane

4 0

2 years ago

1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?

garri49 [273]

Answer:

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

2HI(aq) + BaCO3(s) ⇒ BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

197 g of BaCO3 ----------------- 1 mol

5.97 g ----------------- x

x = (5.97 x 1) /197

x = 0.03 mol of BaCO3

2 moles of HI ---------------- 1 mol of BaCO3

x ---------------- 0.03 mol of BaCO3

x = (0.03 x 2) / 1

x = 0.060 mol of HI

Molarity = moles / volume

volume = moles / molarity

volume = 0.060 / 0.401

volume = 0.151 l or 151 ml

2.-

V = ? NaoH 0.757 M

Co⁺² Volume = 167 ml 0.548 M

CoSO4(aq) + 2NaOH(aq) ⇒ Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x volume

Moles of Co = 0.548 x 0.167

Moles of Co = 0.092

1 mol of CoSO4 -------------- 2 moles of NaOH

0.092 moles --------------- x

x = (0.092 x 2) /1

x = 0.183 moles of NaOH

Volume of NaOH = moles / molarity

= 0.183 / 0.757

= 0.241 l or 241 ml of NaOH

6 0

3 years ago

What values can you draw from this lesson?​

What values can you draw from this lesson?