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jenyasd209 [6]
3 years ago
5

This is chemistry: 2 questions having to do with particles

Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
6 0

Answer 1 :

Image 1 shows : Liquid phase

Image 2 shows : Gas phase

Image 3 shows : Solid phase

Image 4 shows : Solid phase

  • Solid phase : In this state, the molecules are closely and tightly packed. There is no vacant space present in it.
  • Liquid phase : In this state, the molecules are loosely packed and there is an intermediate vacant space present in it.
  • Gas phase : In this state, the molecules are very loosely packed and there is a lot of vacant space present in it.

Answer 2 :

Liquid to gas : Evaporation

Solid to gas : Sublimation

Gas to solid : Deposition

Solid to liquid : Melting

Liquid to solid : Freezing

Gas to liquid : Condensation

Explanation :

  • Evaporation : It is a type of vaporization process in which a liquid changes into gas phase by providing heat.
  • Sublimation : It is a process in which a solid changes directly into gas phase without passing through a liquid phase.
  • Deposition : It is a process in which a gas transforms directly into a solid phase without passing through a liquid phase.
  • Melting : It is a process in which a solid changes into liquid phase by providing heat.
  • Freezing : It is a process in which a liquid transform into a solid phase.
  • Condensation : It is a process in which a water vapor(gas) changes into liquid state.
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332 mL of phosphorus vapour weighs 0.155 g at 527 °C and 1bar pressure. What is the molar mass of phosphorus?
matrenka [14]

Answer:

124.98

Explanation:

The molar mass of phosphorus, M=PVwRT=1×34.05×10−30.0625×0.08314×819=124.98.

7 0
4 years ago
When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed?
Ad libitum [116K]

You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.

For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.

6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2

For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.

13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2

As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.

In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!

6 0
3 years ago
A 1 m ID spherical cooler made of 2 cm thick foam insulation (k = 0.15 Wm-1K-1) is used to store iced water at 0°C. The tank is
Leto [7]

The answer is unknown

3 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
7th grade science lol help​
Luden [163]

Answer:

C) Organic Weathering

Explanation:

Heat and Cold

And if some people start reporting and making fun of me in this answer don't listen to them.

They are hunting me down.

8 0
3 years ago
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