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diamong [38]
3 years ago
5

PLEASE ANSWER ARE THESE CORRECTTTTTT

Chemistry
2 answers:
Harman [31]3 years ago
6 0

Left Frame

The answer you have chosen is not correct. You have selected a physical property. Color is something that will not change as long as nothing chemical is done to the sample. If it is just going to sit there at room temperature in a flask and be admired, (particularly if the flask has a cork in it), then whatever you see is a physical property. The only one that isn't is E. When you burn something, there is a change. One chemical turns into something else. If you burn gasoline (for your car) then C8H18 + 13.5 02 ==> 8CO2 +  9H20 is the result.  That gasoline is not easily reversible. It has changed into something else (Water and Carbon Dioxide). That's Chemical.

Middle Frame

I've never encountered these terms before. Your two examples of extensive are correct. I'm not sure about the chemical reaction. I think it is intensive, but I wouldn't bet the farm on it. A chemical reaction is going to take place no matter how much is used. You could have too much sample or too much acid, but what happens does not depend on the amount. I think you are correct.

Answer: you have placed these in the right place.

Right frame.

All four are in the right category. Very nicely done.


Naddik [55]3 years ago
3 0

1. The answer is E.

2. I agree with your answers.

3. I also agree with your answers

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Agata [3.3K]

<u>Answer:</u> The molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

<u>Explanation:</u>

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

PbI_2\rightleftharpoons Pb^{2+}+2I^-

                s       2s

The expression for solubility constant for this reaction will be:

K_{sp}=[Pb^{2+}][I^-]^2

We are given:

K_{sp}=7.9\times 10^{-9}

Putting values in above equation, we get:

7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L

Hence, the molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

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