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Kruka [31]
3 years ago
14

Repulsion of electrons within two interacting molecules produces changes in electron distribution. This change in electron distr

ibution creates temporary dipole moments.Which of the following does this explain? Choose one or more: A. This explains how two noble gases' molecules can have an attractive force between them. B. This explains why ammonia and nitrogen gas exhibit an attractive force between them.C. This explains why long hydrocarbon chains have relatively high boiling points. D. This explains how the molecules hydrogen fluoride and methanol can exhibit uncharacteristically strong intermolecular forces. E. This explains why the dipole-dipole attractive force between dimethyl ether and acetone does not entirely account for the attractive force between these molecules.
Chemistry
1 answer:
ycow [4]3 years ago
6 0

Answer:

D

Explanation:

This explains how two noble gases molecules can have an attractive force between them.

This force is called as van dar Waals forces.

It plays a fundamental role in fields in as diverse as supramolecular chemistry structural biology .

If no other forces are present, the point at which the force becomes repulsive rather than attractive as two atoms near one another is called the van der Waals contact distance. This results from the electron clouds of two atoms unfavorably coming into contact.[1] It can be shown that van der Waals forces are of the same origin as the Casimir effect, arising from quantum interactions with the zero-point field.[2] The resulting van der Waals forces can be attractive or repulsive.[3] It is also sometimes used loosely as a synonym for the totality of intermolecular forces.[4] The term includes the force between permanent dipoles (Keesom force), the force between a permanent dipole and a corresponding induced dipole (Debye force), and the force between instantaneously induced dipoles

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Answer:

Coating a material with metal

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The half life for the decay of carbon- is years. Suppose the activity due to the radioactive decay of the carbon- in a tiny samp
EleoNora [17]

The question is incomplete. The complete question is:

The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.

Answer:

570 years

Explanation:

The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.

6 0
3 years ago
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777dan777 [17]

Answer:

glucose and oxygen

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Glucose and oxygen are needed for cellular respiration to occur:)

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Answer:

<u>Nucleus</u> - The central part of the atom containing protons and neutrons.

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