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mel-nik [20]
3 years ago
13

Please help. Why does the temperature change when a powder is dissolved in water?

Chemistry
1 answer:
Schach [20]3 years ago
5 0

Answer:

In order for the powder to dissolve, each powder molecule must separate from the other powder molecules and be surrounded by water molecules. This shift in arrangement either absorbs or releases energy depending on the situation. It is due to the exchange of energy that the temperature of the solution fluctuates.

Explanation:

:)

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Ammonium Iodide dissociates reversibly to ammonia and hydrogen iodide:
NeTakaya

Answer:

The partial pressure of ammonia at equilibrium when a sufficient quantity of ammonium iodide is heated to 400°C Is 0.103 atm.

The correct option is A.

Explanation;

NH4I(s) ⇋ NH3(g) + HI(g)Kp = 0.215 at 400°C

NH4I(s)= 0.215

NH3(g)=0.103

HI(g)Kp=0.112

Therefore = 0.103 +0.112= 0.215

Therefore the partial pressure of ammonia at equilibrium is 0.103 atm

7 0
3 years ago
AL
Helga [31]
Metallic I’m pretty sure. :)
4 0
3 years ago
Please help me ASAP!!!
DENIUS [597]

Answer: Patrick was so sure he was a thief because why would you knock if it was your own room you would open the door. And he didn’t have a key to open the door.

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3 years ago
Define the term. Theory
Sonja [21]

Answer:

a supposition or a system of ideas intended to explain something, especially one based on general principles independent of the thing to be explained.

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Explanation:

6 0
3 years ago
Cu(s) + 4 HNO3 (aq) --> Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O(l)
GREYUIT [131]

Answer:

The percent by mass of copper in the mixture was 32%

Explanation:

The ammount of HNO₃ used is:

mol HNO₃ = volume * concentration

mol HNO₃ = 0.015 l * 15.8 mol/l

mol HNO₃ = 0.237 mol

According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.

Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.

1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:

0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.

Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:

100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g

Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%

6 0
3 years ago
Read 2 more answers
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