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frutty [35]
3 years ago
5

Which of the following is true about phase changes? Evaporation occurs when a solid changes into a gas. Condensation occurs when

a gas changes into a liquid. Sublimation occurs when a solid changes to a liquid. If energy is removed from a liquid it will change into a gas.
Chemistry
2 answers:
MaRussiya [10]3 years ago
6 0

Answer:

Condensation occurs when a gas changes into a liquid.

Explanation:

Condensation is when a gas becomes a liquid. It happens when a gas, like water vapor, cools down. Matter can exist in three different states: solid, liquid or gas.

luda_lava [24]3 years ago
4 0
When a gas changes into a liquid
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What is the minimum cation-to-anion radius ratio (rC/rA) for an octahedral interstitial lattice site?
mash [69]

Answer:

0. 414

Explanation:

Octahedral interstitial lattice sites.

Octahedral interstitial lattice sites are in a plane parallel to the base plane between two compact planes and project to the center of an elementary triangle of the base plane.

The octahedral sites are located halfway between the two planes. They are vertical to the locations of the spheres of a possible plane. There are, therefore, as many octahedral sites as there are atoms in a compact network.

The Octahedral interstitial void ratio range is 0.414 to 0.732. Thus, the minimum cation-to-anion radius ratio for an octahedral interstitial lattice site is 0. 414.

3 0
3 years ago
Calculate the hydrogen-ion concentration (H+] for the aqueous solution in which [OH-1 is 1 x 10-12 mol/L.
Artemon [7]

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And here is the solution - as you can see it is an acidic one :

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Explanation:

7 0
3 years ago
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Burka [1]

Answer:

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3 0
3 years ago
An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the
san4es73 [151]

Answer:

the ionic radius of the anion r^- = 312.52 \ pm

Explanation:

From the diagram shown below :

The anion Cl^- is located at the corners

The cation Cs^+ is located at the body center

The Body diagonal length =  \sqrt{3 \ a }

∴ 2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a}  \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a

Given that :

\frac{r^+}{r^-} =0.84   (i.e the  ratio of the ionic radius of the cation to the ionic radius of

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0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a  \\ \\  1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a

Also ; a =  664 pm

Then :

r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm

Therefore,  the ionic radius of the anion r^- = 312.52 \ pm

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