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Thepotemich [5.8K]
2 years ago
11

Will give brainliest

Chemistry
1 answer:
Vinvika [58]2 years ago
6 0

Answer:

I think its C

Explanation:

it's not A or B

hope this helps

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The ionization of pure water forms _____.
Evgesh-ka [11]

Answer: The ionization of pure water forms <u><em>hydroxide and hydronium ions.</em></u>

Explanation:

Ionization is a reaction in the pure water in which water breaks down into its constituting ions that hydronium ion and hydroxide ions.

H_2O+H_2O\rightleftharpoons H_3O^++OH^-

One molecule of water looses its proton to form hydroxide ion and l=the lost protons get associated with another water molecule to form hydronium ion.

4 0
3 years ago
If sodium arsenite is Na3AsO3, the formula for calcium arsenite would be
lara31 [8.8K]

Answer:

Ca₃(AsO₃)₂

Explanation:

Sodium arsenite, with the chemical formula Na₃AsO₃, is formed  by the cation Na⁺ and the anion AsO₃³⁻. For the molecule to be neutral, 3 cations Na⁺ and 1 anion AsO₃³⁻ are required.

Calcium arsenite would be formed by the cation Ca²⁺ and the anion AsO₃³⁻. For the molecule to be neutral, we require 3 cations Ca²⁺ and 2 anions AsO₃³⁻. The resulting chemical formula is Ca₃(AsO₃)₂.

4 0
3 years ago
How did Bohr solve the contradiction with physics with his model of an atom
daser333 [38]

Answer:

He used Velocity and Radius.

Explanation:

The uncertainty truths contradicts Bohr's thoughts of electrons.

4 0
3 years ago
The atomic radius of a silver atom is 1.44 x 10-10. Give the
professor190 [17]

Answer:

0.144 nm

Explanation:

Silver's electronic configuration is (Kr)(4d)10(5s)1, and it has an atomic radius of 0.144 nm.

5 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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