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hammer [34]
3 years ago
15

if you obtain the powder of a mineral by rubbing a mineral on a porcelain plate, which property have u observed?

Chemistry
2 answers:
Nina [5.8K]3 years ago
8 0
The property of the rock is the streak
charle [14.2K]3 years ago
7 0
you have observed streak :)
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When the compound BaCl2 forms , what happens to the Ba and Cl ions
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Ba stays as Ba+2 and Cl stays as Cl-
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When does the given chemical system reach dynamic equilibrium?
AVprozaik [17]
Once it becomes balanced. 
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A more electronegative atom A) will have more attraction to the electrons in a chemical bond. B) is more likely to lose an elect
DIA [1.3K]

Answer: A more electronegative atom will have more attraction to the electrons in a chemical bond.

Explanation:

An atom that is able to attract electrons or shared pair of electrons more towards itself is called an electronegative atom.

For example, fluorine is the most electronegative atom.

Due to its high electronegativity it is able to attract an electropositive atom like H towards itself. As a result, both fluorine and hydrogen will acquire stability by sharing of electrons.

Thus, we can conclude that a more electronegative atom will have more attraction to the electrons in a chemical bond.

6 0
3 years ago
What is Charles’s law? State the definition of the law in words. What are the assumptions of Charles’s law? Write mathematical e
Leviafan [203]

Answer:

The volume of a given mas of a gas is directly proportional to the temperature if the pressure remains constant

V is directly proportional to T

V=1/T

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Explanation:

4 0
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)}

3 0
3 years ago
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