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marshall27 [118]
2 years ago
8

Noble gases have a blank balance she’ll with blank electrons. Because their balance she’ll is full they blank need to borrow or

gain electrons
Chemistry
1 answer:
FromTheMoon [43]2 years ago
4 0
Noble gases have a FULL balance she’ll with 8 electrons. Because their balance she’ll is full they DONT need to borrow or gain electrons

fyi it’s actually:
Noble gases have a full valence shell with eight electrons. Because their valence shell is full they don’t need to borrow or gain electrons.
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An aqueous solution of sodium fluoride is slowly added to a water sample that contains barium ion (2.75×10-2M ) and calcium ion
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Answer : The remaining concentration of the first ion to precipitate when the second ion begins to precipitate is, 1.07\times 10^{-6}M

Explanation :

The dissociation of barium fluoride is written as:

BaF_2\rightleftharpoons Ba^{2+}+2F^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ba^{2+}][F^-]^2

As we know that at room temperature 25^oC the K_{sp} of barium fluoride is, 1\times 10^{-6}.

Now put all the given values in this expression, we get:

1\times 10^{-6}=(2.75\times 10^{-2})\times [F^-]^2

[F^-]=6.03\times 10^{-3}M=0.00603M

The barium fluoride precipitate when fluoride ion is equal to 0.00603 M.

The dissociation of calcium fluoride is written as:

CaF_2\rightleftharpoons Ca^{2+}+2F^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ca^{2+}][F^-]^2

As we know that at room temperature 25^oC the K_{sp} of calcium fluoride is, 3.90\times 10^{-11}.

Now put all the given values in this expression, we get:

3.90\times 10^{-11}=(6.70\times 10^{-2})\times [F^-]^2

[F^-]=2.41\times 10^{-5}M=0.0000241M

The calcium fluoride precipitate when fluoride ion is equal to 0.0000241 M.

Since, the fluoride ion concentration in calcium fluoride is less then the fluoride ion concentration in barium fluoride. That means, calcium fluoride will precipitate first.

Thus, the concentration Ca^{2+} ion remaining at F^- concentration (0.00603 M) is calculated as:

K_{sp}=[Ca^{2+}][F^-]^2

Now put all the given values in this expression, we get:

3.90\times 10^{-11}=[Ca^{2+}]\times (0.00603)^2

[Ca^{2+}]=1.07\times 10^{-6}M

Therefore, the remaining concentration of the first ion to precipitate when the second ion begins to precipitate is, 1.07\times 10^{-6}M

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There are

Explanation:

There are 10 elements in CH3CH2COONa.

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Which Group is in the leftmost column on the periodic table?
svp [43]
Alkali metals. Hope this helps
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What is the charge of an atom's nucleus of an atom has 20 protons, 23 neutrons, and 18 electrons?
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Answer:

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Explanation:

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