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Mariulka [41]
3 years ago
12

What process requires water to gain heat energy from the environment

Chemistry
1 answer:
olganol [36]3 years ago
5 0
Evaporation is the answer to your question. If this helps please mark as Brainest it really helps.
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Calculate the mean of 2, 5 and 3. write your answer to 2 decimal places. please help me on this
makvit [3.9K]

Hello.

The answer is: 3.3333

To get the answer add 2,5 and 3 that is 10 then divide by 3 to get 3.3

Have a nice day

6 0
3 years ago
Any one from Michigan on here????
Andrew [12]

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dont go to any taco bells in Michigan!! There have been numerous cases of Ligma that have been traced back to Taco Bells across Michigan

Explanation:

7 0
2 years ago
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During dehydration synthesis, an enzyme removes a ____________ functional group from the monomers of one organic molecule and a
Reika [66]

Linking monomers together to form a polymer .This chemical reaction also forms water molecules.

<h3>What is Polymerization?</h3>

This is a type of reaction which involves the linking of two or more monomers to form a polymer.

Dehydration reaction forms water molecules as part of the product thereby making it the most appropriate choice.

Read more about Dehydration here brainly.com/question/1301665

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3 0
1 year ago
How many liters are equivalent to 43 milliliters?
marysya [2.9K]

Answer:

The answer is B) 0.043 L

Explanation:

Hope this helps :))

8 0
3 years ago
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I performed an experiment and mixed copper nitrate and potassium iodide. When they reacted, they formed a precipitate, even thou
Alexus [3.1K]

Answer:

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

Explanation:

The reaction performed in the experiment is;

2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2

The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.

The oxidation - reduction equation is as follows;

2Cu2^+ + 2e ----> 2Cu^+ reduction half equation

2I^- ----> I2 + 2e. Oxidation half equation

Balanced redox reaction equation;

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

3 0
3 years ago
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