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balu736 [363]
2 years ago
15

I NEED HELP ASAP DON'T JUST ANSWER FOR POINTS OR YOU WILL GET REPORTED!

Chemistry
2 answers:
White raven [17]2 years ago
7 0

Answer:

The number of neutrons is entirely dependent on the Mass number of the particular atom. The standard mass for potassium is 39.

Potassium is element number 19, so it has 19 protons and 19 electrons in the neutral atom. It has therefore 39-19 = 20 Neutrons.

Explanation:

Lelu [443]2 years ago
4 0

Answer:

Number of Protons 19

Explanation:

Number of Neutrons 20

Number of Electrons 19

Melting Point 63.65° C

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Is there a way to dissolve more sugar into a solution that it should hold at that temperature? Explain how this is done and what
Darya [45]

The answer is: supersaturated solution.

A supersaturated solution contains more of the dissolved substance than could be dissolved by the solvent under normal circumstances.

A way to dissolve more sugar into a solution is heating a solution.

The more heat is added to a system, the more soluble a substance (in this example sugar) becomes.

The solution will become supersaturated if this solution is suddenly cooled at a rate faster than the rate of precipitation.

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Pure substances are _______ in appearance ​
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Answer:

constant appearance

Explanation:

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How is a compound different from mixtures
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Answer: a compound is a substance that is made up of more than one type of atoms bonded together, mixtures are a mix of multiple elements or compounds which haven't bonded together.  

Explanation:

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2 years ago
What do calories have to do with combustion
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Answer:

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3 years ago
Determine the number of moles of gas contained in a 1.25 L container at 25 degrees Celsius and 800 torr
tiny-mole [99]
You would have to use the ideal gas law: PV = nRT where,

P = pressure
V = volume
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T = temperature in Kelvin

First, convert Celsius to kelvin by adding 273 to 25, which gives you 298K

Now plug in your variables to find n:

(800 torr)(1.25 L) = n(62.364)(298K)

1000 = 18584.472n

Now divide 1000 by 18584.472 to get n:

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3 years ago
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