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elena-14-01-66 [18.8K]

3 years ago

11

What type of bond occurs between atoms having nearly equal attraction for the electrons?

1 answer:

SCORPION-xisa [38]3 years ago

6 0

It is called covalent

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The phase change in which a substance changes from a liquid to a solid is A. freezing. B. melting. C. sublimation. D. condensati

vovangra [49]

freezing. Think of what happens when you put water in the freezer

4 0

3 years ago

U ever get an answer

den301095 [7]

Yesn’t. they are either wrong or links

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2 years ago

Carbon burns in the presence of oxygen to give carbon dioxide. Which chemical equation describes this reaction?

Alex73 [517]

Your answer is B.
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3 years ago

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Describe how you would crystallise potassium nitrate from its aqueous liquid?

Rus_ich [418]

Just look here https://www.jiskha.com/search/index.cgi?query=describe+how+you+can+crystallize+potassium+nitrate+fro...

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3 years ago

On the basis of the information above, a buffer with a pH = 9 can best be made by using

telo118 [61]

Answer:

D H2PO4– + HPO42–

Explanation:

The acid dissociation constant for $\mathbf{H_3PO_4 , H_2PO^{-}_4 , HPO_4^{2-}}$ are $\mathbf{7\times 10^{-3}, \ \ 8\times 10^{-8} ,\ \ 5\times 10^{-13}}$ respectively.

$\mathbf{pka (H_3PO_4) = -log (7\times 10^{-3} )=2.2}$

$\mathbf{pka (H_2PO_4^-) = -log (8\times 10^{-8} )=7.1}$

$\mathbf{pka (HPO_4^{2-}) = -log (5\times 10^{-13} )=12.3}$

The reason while option D is the best answer is that, the value of pKa for both

$\mathbf{H_2PO^{-}_4 ,\ \& \ HPO_4^{2-}}$ lies on either side of the desired pH of the buffer. This implies that one is slightly over and the other is slightly under.

Using Henderson-Hasselbach equation:

$\mathbf{pH = pKa + log \Big( \dfrac{HPO_4^{2-}}{H_2PO_4^-} \Big)}$

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3 years ago