# All of the following are true of London dispersion forces except (2 points) a.) they occur between all molecules and particles. b.) they are the strongest of all the intermolecular forces. c.) they get stronger as the number of electrons increases. d.) they are the only force of attraction between noble gas atoms.

Answer is: b.) they are the strongest of all the intermolecular forces.

Intermolecular forces are the forces between molecules or particles.

The London dispersion force is the weakest intermolecular force.

Dispersion force is also called an induced dipole-induced dipole attraction.

The London dispersion force (intermolecular force) is a temporary attractive force between molecules.

There are several types of intermolecular forces: hydrogen bonding, ion-induced dipole forces, ion-dipole forces andvan der Waals forces.

There are two kinds of Van der Waals forces: weak London dispersion forces and stronger dipole-dipole forces.

Answer: B... i think not totally sure

## Related Questions

A balloon is filled with 35.0 L of helium in the morning when the temperature is 303 degrees kelvin. By 3:00 p.m., the temperature has risen to 333 degrees kelvin. What is the new volume of the balloon? 2882 L

38.5 L

31.8 L

65.0 L

38.5 L

Explanation:

For this question, we can use

V1/T1=V2/T2

where

V1 (initial volume)= 35 L

T1 (initial temperature in Kelvin)= 303 K

V2(final volume)= ?

T2 (final temperature in Kelvin)= 333K

Since we are trying to find the final volume, we can rearrange the equation to say

V2= (V1/T1) × T2

V2= (35/303)×333

V2= 38.5 L

The average propane cylinder for a residential grill holds approximately 18 kg of propane. How much energy (in kJ) is released by the combustion of 14.50 kilograms of propane in sufficient oxygen?

Combustion is a chemical reaction between a fuel and an oxidant, oxygen, to give off combustion products and heat. Complete combustion results when all of the fuel is consumed to form carbon dioxide and water, as in the case of a hydrocarbon fuel. Incomplete combustion results when insufficient oxygen reacts with the fuel, forming soot and carbon monoxide.

The complete combustion of propane proceeds through the following reaction:
+  -->  +

Combustion is an exothermic reaction, which means that it gives off heat as the reaction proceeds. For the complete combustion of propane, the heat of combustion is (-)2220 kJ/mole, where the minus sign indicates that the reaction is exothermic.

The molar mass of propane is 44.1 grams/mole. Using this value, the number of moles propane to be burned can be determined from the mass of propane given. Afterwards, this number of moles is multiplied by the heat of combustion to give the total heat produced from the reaction of the given mass of propane.

14.50 kg propane  x  1000 g  x   1 mole propane     x   2220 kJ
1 kg              44.1 g                     1 mole

= 729,931.97 kJ

Aluminum metal reacts with hydrochloric acid to form aluminum chloride according to the reaction below. If 30.8 grams of aluminum react with 84 grams of HCl, how many grams of aluminum chloride can theoretically be formed? 2Al + 6HCl --> 2AlCl3 + 3H2 You Answered

Answer: The Theoretical yield of Aluminum Chloride = 152.19 g

Explanation:

In order to determine the theoretical yield of Aluminum Chloride we first determine the limiting reagent,

Al = 27g/mol

HCl = 1 + 35.5 = 36.5g/mol

number of moles= mass/molar mass

for Al:

number of moles of Al = 30.8/27 = 1.14 moles

number of moles of HCl = 84/36.5 = 2.30moles

Hence, Al is the limiting reagent

Therefore,

since from the balanced reaction,

2moles of Al react with HCl to give 2moles of Aluminum Chloride

Therefore 1.14moles Al will give 1.14moles of Aluminum Chloride.

molar mass of Aluminum Chloride = 27 + (35.5)*3 = 133.5g/mol

Theoretical yield of Aluminum Chloride = number of moles x molar mass = 1.14 x 133.5 = 152.19 g

The amount in grams of aluminum chloride that can theoretically be formed from the 84 g of HCl and 30.8 g aluminium is 102.4 grams

Explanation:

Aluminum metal reacts with hydrochloric acid to form aluminum chloride according to the reaction below. If 30.8 grams of aluminum react with 84 grams of HCl, how many grams of aluminum chloride can theoretically be formed? 2Al + 6HCl --> 2AlCl3 + 3H2

Hydrochloric acid / Molar mass

36.46 g/mol

Aluminium chloride / Molar mass

133.34 g/mol

Aluminium / Atomic mass

26.981539

Hydrogen / Atomic mass

1.00784 u

Number of moles of aluminium = mass/molar mass = 30.8 g/ 26.98 g/mol

= 1.142 moles.

Number of moles of HCl = mass/molar mass = 84/36.46 = 2.303 moles

From the reaction, 2 moles of Al react with 6 moles of HCl, therefore, 1.142 moles of aluminium react with (1.142 × 6/2) moles or 3.425 moles of HCl. Since the calculated number of moles of required HCl is more than the available moles, then HCl is the limiting reagent.

6 moles of HCl react with 2 moles of aluminium, then 2.303 moles will react with (2/6×2.303) or 0.76796 moles of aluminium to produce 0.76796 moles of aluminum chloride

Therfore mass of aluminum chloride produced = (Number of moles of aluminium chloride produced) × (molar mass of aluminium chloride)

0.76796 moles × 133.34 g/mol = 102.4 g

J.J. Thompson's experiments with the cathode ray tube lead to the invention of _____. a. The radio
c. Television
b. Video games
d. Microwave ovens