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Semenov [28]
3 years ago
15

Sarah added sugar to her iced tea but it is not dissolveing very quickly which can Sarah do to make the sugar dissolve faster ?

Chemistry
1 answer:
Aliun [14]3 years ago
7 0
Stir it,
Or as warmer water makes solutes dissolve faster Sarah can do that
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The sodium-potassium exchange pump
trapecia [35]

Answer: The correct answer is option E

Explanation:

Sodium/potassium pump is a mechanism that involves the movement of sodium ions (Na+) out of a cell and potassium ions (K+) into a cell, thereby regulating concentration of ions on both sides of a typical cell membrane.

In this situation, the sodium-potassium pump is usually helps in the establishment of the resting potential. The potassium voltage channels normally closes before the membrane potential is brought to a resting level.

In summary, sodium/potassium pump helps to maintain a balance in the system.

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3 years ago
Balance the equation and state the limiting reagent in the following reaction:
romanna [79]
Hgffvvcbchcgivigcigxgsgixigdfhcgcc fcfuc
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PLEASE ANSWER AS QUICKLY AS POSSIBLE THANK YOU SO MUCH
Harlamova29_29 [7]

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2 years ago
Which best describes a single atom of an element?
Anettt [7]

Answer:

I believe the answer is A: *It is the simplest form of matter" not 100% sure but I think that's correct

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4 0
3 years ago
Read 2 more answers
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
3 years ago
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