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Romashka [77]
2 years ago
15

The main advantage of solar energy is:

Chemistry
1 answer:
Colt1911 [192]2 years ago
3 0

Answer:

I think ithe answer is it is free

hope this helps

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It means 3p (p being your variable) to the third power, or cubed
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How do plants release energy
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no but oxygen

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There are always attractive forces between a collection of atoms or molecules which may not be negligible as suggested by the ki
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Negligible

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According to the kinetic theory of gases, the degree of intermolecular interaction between gases is minimal and gas molecules tend to spread out and fill up the volume of the container.

If the attraction between gas molecules increases, then the volume of the gas decreases accordingly. This is because, gas molecules become highly attracted to each other.

This intermolecular attractive force may be so strong, such that the actual volume of the gas become negligible compared to the volume of the container.

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Convert 1.580 hectoliters to deciliters.<br><br> liters
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Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs
saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of Ag^+ when CuBr just begins to precipitate is, 1.34\times 10^{-6}M

Percent of Ag^+ remains is, 0.0018 %

Explanation :

K_{sp} for CuBr is 4.2\times 10^{-8}

K_{sp} for AgBr is 7.7\times 10^{-13}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

CuBr\rightleftharpoons Cu^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Cu^+][Br^-]

4.2\times 10^{-8}=0.073\times [Br^-]

[Br^-]=5.75\times 10^{-7}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgBr\rightleftharpoons Ag^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][Br^-]

7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M

[Ag^+]=1.34\times 10^{-6}M

Now we have to calculate the percent of Ag^+ remains in solution at this point.

Percent of Ag^+ remains = \frac{1.34\times 10^{-6}}{0.073}\times 100

Percent of Ag^+ remains = 0.0018 %

3 0
2 years ago
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