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Phantasy [73]
2 years ago
6

Which substance increases in solubility as the temperature decreases?

Chemistry
2 answers:
Arlecino [84]2 years ago
7 0
NH₃ would.  This is because of the fact that ammonia is a molecular gas while the other options are ionic crystals.  Crystals increase in solubility when the temperature increases while gases increase in solubility while temperatures deceases.

I hope this helps.  Let me know in the comments if anything is unclear or if you want me to explain anything is more detail.
Alex73 [517]2 years ago
7 0

Answer:

The correct answer is 2.

Explanation:

The NH3 solubility curve shows that when the temperature increases, the solubility of the compound decreases. The molecule is not flat but has a tetrahedral shape with a vacant vertex, and this is due to the formation of hybrid sp³ orbitals. Unlike other compounds, the NH3 solubility curve has been experimentally tested and is attached for better understanding.

Have a nice day!

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2 years ago
Which of the following would be SI unit to use in measuring the temperature of hot liquid
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why do you think Kool Aid comes in packets that has a very finely ground powder (finely ground means very small particles).
marishachu [46]

Answer:

Answer 1:

When you pour the kool-aid into water, the little crystals go straight to the bottom because they are heavier than the water. If you left them there without stirring, and came back a few days later, you wouldn't see any crystals on the bottom. That's because the stuff in kool-aid can DISSOLVE in water, which means that each little molecule of kool-aid gets suspended between the molecules of water. When that happens, you can't see the kool-aid anymore...it's trapped between the water molecules. When you stir kool-aid, you help DISSOLVE the kool-aid in water by keeping all of the crystals off the bottom and in the water. So you see, stirring kool-aid speeds up the dissolving,

Answer 2:

Are you referring to Koolaid in the granular form?If so the koolaid grains sink in water because the grains have a greater density than that of water. Once your stir the grains dissolve and go into solution where they remain because the dissolved koolaid is miscible with water unlike oil (floats) or gasoline (sinks). How long did you let the koolaid remain in the water before you stirred it? I would think that if you left it undisturbed for a long time (days) it would eventually mix on its own.

Answer 3:

I'm not a chemist, but I think I can answer your question about Kool-Aid. Kool-Aid is mostly sugar, which is heavier than water, so when you pour it in it sinks to the bottom. When you stir it up the sugar (and flavoring) dissolves so that you don't have any solid particles any more. Stuff that is dissolved in water will not sink because it is no longer a physically separate thing. It becomes part of the water (or water-sugar-flavor solution). What happens if you pour the Kool-Aid in but don't stir it? Will it eventually dissolve? You may have to wait a long time, like over night. Try it and let me know what you find!

Answer 4:

It all has to do with the rate at which kool-aid crystals (basically its SUGAR!!) dissolves in water relative to the rate at which the sugar crystals sink. If you just dump the stuff in, it sinks because it is denser than the water. As it sinks it dissolves. But when you stir the water, the rate of dissolution becomes greater than the rate of sinking and so the crystals dissolve before they reach the bottom. So it all has to do with the comparison between the rate of sinking versus the rate of dissolution.

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You may have to borrow your mom's mixing machine because you will get tired of stirring after 10 minutes!!!!

If you do the experiment let me know how it turns out. Actually, you should set up a control. Make two batches of Jello...with one, put it in the refrigerator and dont stir; with the other, keep stirring it (in the refrigerator), if you can figure how to arrange that without your mom or dad getting mad!!!

8 0
3 years ago
Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin
Ainat [17]

Answer:

See explaination

Explanation:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

so

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

8 0
3 years ago
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