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vitfil [10]
3 years ago
13

Which of the following expands to fill the container it is in but can be compressed to a smaller volume when pressure is applied

?
A.solid iron
B. liquid bromine
C. neon gas
D. sugar crystals
Chemistry
1 answer:
goblinko [34]3 years ago
8 0

Answer: C. neon gas

Explanation:  There are three states of matter:

A. Solid iron: The molecules in solid are very tightly packed as inter molecular forces are very high. There is very less space between the particles and thus the molecules can not be compressed. They have fixed shape and volume.

B. liquid bromine: The molecules in liquid are less tightly packed as inter molecular forces are not very high. There is less space between the particles and thus the molecules can not be compressed. They have fixed volume and take the shape of the container.

C. neon gas : The molecules in gas are very less tightly packed as inter molecular forces are very low. There is large vacant space between the particles and thus the molecules can be compressed. They neither have fixed volume nor a fixed shape.They occupy complete volume of the container in which it is put.

D. sugar crystals: The molecules in solid are very tightly packed as inter molecular forces are very high. There is very less space between the particles and thus the molecules can not be compressed. They have fixed shape and volume.

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3 years ago
True or False. Solutions for which water is the solvent are called aqueous solutions.
Stolb23 [73]
This is true. Water is the solvent in aqueous solutions
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3 years ago
(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW
Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M  InF₃. </em>Thus:

[In⁺³] = 2.3x10⁻³M  InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M  In⁺³

[F⁻] = 2.3x10⁻³M  InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M  In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>

<em></em>

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

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Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, <em>solubility is 6.31x10⁻⁶M</em>

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6 0
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K=㏑2/ t1/2 and when we have this given (missing in question):
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