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s344n2d4d5 [400]
3 years ago
14

Classify each of the following forces as intermolecular or intramolecular. Those preventing O2 in air from forming O atoms Those

allowing fog to form on a cool, humid evening Those preventing oil from evaporating at room temperature Those responsible for the low boiling point of hexane Those preventing butter from melting in a refrigerator
Chemistry
1 answer:
Angelina_Jolie [31]3 years ago
6 0

Answer: Those preventing O_2 in air from forming O atoms : Intramolecular

Those allowing fog to form on a cool, humid evening : Intermolecular

Those responsible for the low boiling point of hexane : Intermolecular

Those preventing butter from melting in a refrigerator :  Intermolecular

Explanation:

Intermolecular forces are forces which exist within different molecules. Example: Forces between water and salt

Intramolecular forces are forces which exist within the same molecules. Example: forces within water

a) When oxygen is dissociated to form atoms , the bonds between the oxygen atoms is broken and hence the forces are intramolecular.

b) Those allowing fog to form on a cool, humid evening : strong interactions between the molecules cause the gaseous water molecules to condense and hence is intermolecular

c) Those responsible for the low boiling point of hexane : the weak dispersion forces among the molecules of hexane are responsible and hence is intermolecular.

d)  Those preventing butter from melting in a refrigerator : The attractions between the fat molecules are broken and hence is  intermolecular.

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mol/dm³ is measure for molarity
3 0
3 years ago
What are the differences between simple, compound, and electron microscopes? WILL GET BRAINIEST!
timurjin [86]
I think the answer is= simple/uses surrounding light source and is restricted in magnification.
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5 0
3 years ago
g Select the correct statements. I. Reduction is the loss of electrons, and oxidation is the gain of electrons II. Reduction is
Karolina [17]

<u>Answer:</u> The correct answer is Option D.

<u>Explanation:</u>

Reduction reaction is defined as the reaction in which a substance gains electrons. Here, the oxidation state of the substance decreases.

X^{n+}+ne^-\rightarrow X

Oxidizing agents are the agents that helps in the oxidation of other substance and itself gets reduced. These agents undergoes reduction reactions.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. Here, oxidation state of the substance increases.

X\rightarrow X^{n+}+ne^-

Reducing agents are the agents that helps in the reduction of the other substance and itself gets oxidized. These agents undergoes reduction reactions.

Oxidation state is the number which is given to an atom when it looses or gains electron. It is written as a superscript. In a compound, the total charge is equal to the sum of the charges of all atoms in that compound. <u>For Example:</u> In MnO_4^-, manganese has +7 oxidation number and oxygen has -2.

So, the charge on the compound = [=7+(4\times (-2))]=-1

Hence, the correct answer is Option D.

6 0
3 years ago
A chemistry student is given 600. mL of a clear aqueous solution at 37.° C. He is told an unknown amount of a certain compound X
barxatty [35]

Answer:

  • <u>Yes, it is 14. g of compound X in 100 ml of solution.</u>

Explanation:

The relevant fact here is:

  • the whole amount of solute disolved at 21°C is the same amount of precipitate after washing and drying the remaining liquid solution: the amount of solute before cooling the solution to 21°C is not needed, since it is soluble at 37°C but not soluble at 21°C.

That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.

Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.

With that, the solubility can be calculated from the followiing proportion:

  • 84. g solute / 600 ml solution = y / 100 ml solution

      ⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.

The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.

<u>The answer is 14. g of solute per 100 ml of solution.</u>

7 0
3 years ago
Identify the major ionic species present in an aqueous solution of C12H22O11 (sucrose).
Marina86 [1]
<h3><u>Answer;</u></h3>

No ions present

<h3><u>Explanation;</u></h3>
  • Ionic compounds are compounds made up of ions. These ions are atoms that gain or lose electrons, giving them a net positive or negative charge.
  • Atoms that gain electrons and therefore have a net negative charge are known as anions. Conversely, atoms that lose electrons have a net positive charge are called cations.
  • C12H22O11 (sucrose) is not an ionic compound, and therefore does not have any ions. Sucrose is a molecular compound.
4 0
4 years ago
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