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Lubov Fominskaja [6]
3 years ago
8

Which is a polar molecule?

Chemistry
1 answer:
fenix001 [56]3 years ago
8 0
A) nitrogen tricloride
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If 12.4 mol of Ne gas occupies 122.8 L, how many mol of Ne would occupy 339.2 L under the same temperature and pressure? Record
8090 [49]

Answer:

3.43×10¹ mol

Explanation:

Given data:

Initial number of  moles = 12.4 mol

Initial volume = 122.8 L

Final number of moles = ?

Final volume = 339.2 L

Solution:

The number of moles and volume are directly proportional to each other at same temperature and pressure.

V₁/n₁  =  V₂/n₂

122.8 L/ 12.4 mol  =  339.2 L / n₂

n₂ = 339.2 L× 12.4 mol  / 122.8 L

n₂ = 4206.08 L.mol /122.8 L

n₂ = 34.3mol

In scientific notation:

3.43×10¹ mol

7 0
3 years ago
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2 years ago
The stronger the wind, the larger the particles it erodes.
Fittoniya [83]
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7 0
3 years ago
Which statement is true of the particles that make up a substance?
victus00 [196]

Particles in a gas have more energy than particles in a liquid. Because in gaseous state particles are free to move around due to which kinetic energy of molecule or gas increases and hence overall energy increases

8 0
2 years ago
If 6. 000 g of sugar is mixed with 9. 000 g of water, what is the concentration in weight percent?.
lbvjy [14]

The concentration in weight percent when 6 g of sugar is mixed with 9 g of water is 40%.

There are several ways to denote the concentration of a solution like

  • Molarity
  • Molality
  • Mass percent
  • Mole Fraction

The formula for calculating mass percent is as follows

Mass per cent = (Mass of solute/Mass of solute + Mass of solvent) x 100%

In the given situation sugar is the solute and water is the solvent.

Putting the given values in the above formula

Mass per cent = (6/6+9) x 100%= 6/15 x 100% = 40%

Hence, the concentration in weight percent when 6 g of sugar is mixed with 9 g of water is 40%.

To know more about "concentration of a solution", refer to the link given below:

brainly.com/question/26579666?referrer=searchResults

#SPJ4

7 0
11 months ago
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