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MAXImum [283]
3 years ago
7

Compare and contrast and give the definitions of: Partial + charge Partial - charge

Chemistry
2 answers:
sineoko [7]3 years ago
5 0
When two elements of different electronegativity forms a compound, -ve charge is shifted to more electronegative element & +ve towards less electronegative.

So, the charge on less electronegative element in polar compound is known as "Partial +ve charge" and charge on more electronegative element in Polar compound is known as "Partial -ve charge"

Hope this helps!
Aloiza [94]3 years ago
3 0
Partial charges are formed in molecules when electronegativity difference between is small. 

Partial charges formed can be permanent as well temporary. Inductive effect creates permanent partial charge in the molecule, whereas electromeric effect creates temporary partial charge in the molecule.

Inductive effect: when a covalent bond is formed between the atoms of different electronegatives, the electron density is more towards the more electronegative(EN) atom of the bond which results in the formation of a polar covalent bond. This polarity developed in molecule creates dipole moment.


Electromeric effect: It is temporary delocalisation of 
π-electron in compound containing multiple covalent bond( double and triple bond).

Partial negative charge is formed when more EN element is attached to the molecule, whereas partial negative charge is formed when less EN element is attached to the molecule.

Definitions:

1.) Partial positive charge: The positive charge developed in a molecule which is non-integral is called partial positive charge. Denoted by +
δ.
2.) Partial negative charge: The negative charge developed in a molecule which is non-integral is called partial negative charge. Denoted by -δ.
Hope this helps!






 


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Answer:

Explanation:

As it was stated in the comments by other user, the question is incomplete but luckily it was posted the rest of the question, so, I'm gonna answer it with the data of that. If you have another questions there, please submit it again or put it in the comments.

<u>a) Which barium salt will precipite first?</u>

In order to know this, we need to take a look at the Ksp of both salts. The given Ksp are:

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Ksp 2: 1.3x10^-6 (BaC2O4)

Now, we can see that Ksp1 < Ksp 2, but what's this Ksp value means? a Ksp value means that an aqueous solution will form a precipite (So the solid formed, it's not soluble in water), As Ksp 1 is a smaller value than Ksp2, means that the concentrations of Ba and CrO4 are too small, and therefore, it takes more time to precipite. <em>Therefore, the BaC2O4 will precipite first.</em>

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In this, we know that the reaction taking place is the following:

BaCrO4(s) <--------> Ba^2+ + CrO4^2-   Ksp = 2.1x10^-10

The expression for Ksp is:

Ksp = [Ba][CrO4]

We know the concentration of CrO4 cause this comes from the K2CrO4 so, replacing here, we solve for Ba:

[Ba^2+] = Ksp / [CrO4^2-]

[Ba^2+] = 2.1x10^-10 / 0.0351

[Ba^2+] = 5.98x10^-9 M

<u>c) Concentration of Ba^2+ to reduce 10% oxalate concentration</u>

The 10% concentration of oxalate is:

[C2O4] = 0.0537 * 0.1 = 0.00537 M

So, we do the same thing we did in part b) but solving for C2O4:

BaC2O4(s) <--------> Ba^2+ + C2O4^2-   Ksp = 1.3x10^-6

[Ba^2+] = 1.3x10^-6 / 0.00537

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In this case, we calculate concentration of CrO4 and C2O4 with the value of Ba and it's respective Ksp, and then, calculate the ratio:

[CrO4^2-] = Ksp1 / [Ba^2+] = 2.1x10^-10 / 0.0050 = 4.2x10^-8 M

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So the ratio is:

[C2O4^2-] / [CrO4^2-] = 2.6x10^-4 / 4.2x10^-8

[C2O4^2-] / [CrO4^2-] = 6.19x10^3

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