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Keith_Richards [23]
3 years ago
5

Given the balanced ionic equation representing a reaction:

Chemistry
1 answer:
Assoli18 [71]3 years ago
8 0
The answer should be (4)

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Answer:

B

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Exergonic reactions are also called spontaneous reactions, because they can occur without the addition of energy. ... In this case, the products, or final state, have more free energy than the reactants, or initial state. Endergonic reactions are non-spontaneous, meaning that energy must be added before they can proceed.

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In a water molecule, what type of bond forms between the oxygen and hydrogen atoms
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7 0
3 years ago
Given the following unbalanced chemical reaction: As + NaOH Na3AsO3 + H2 What would be the coefficient of the NaOH molecule in t
Lady_Fox [76]
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6 0
3 years ago
Read 2 more answers
Preparation of Standard Buffer for Calibration of a pH Meter The glass electrode used in commercial pH meters gives an electrica
lana66690 [7]

Answer:

  • Mass of NaH₂PO₄·H₂O = 8.542 g
  • Mass of Na₂HPO₄ = 5.410 g

Explanation:

Keeping in mind the equilibrium:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺

We use the Henderson-Hasselbalch equation (H-H):

pH = pka + log\frac{[A^{-}]}{[HA]}

For this problem [A⁻] = [HPO₄⁻²] and [HA] = [H₂PO₄⁻]

From literature we know that pka = 7.21, from the problem we know that pH=7.00 and that

[HPO₄⁻²] + [H₂PO₄⁻] = 0.100 M

From this equation we can <u>express [H₂PO₄⁻] in terms of [HPO₄⁻²]</u>:

[H₂PO₄⁻] = 0.100 M - [HPO₄⁻²]

And then replace [H₂PO₄⁻] in the H-H equation, <u>in order to calculate [HPO₄⁻²]</u>:

7.00=7.21+log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]} \\-0.21=log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\10^{-0.21} =\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\0.616*(0.100M-[HPO4^{-2}])=[HPO4^{-2}]\\0.0616 M = 1.616*[HPO4^{-2}]\\0.03812 M =[HPO4^{-2}]

With the value of  [H₂PO₄⁻],<u> we calculate [HPO₄⁻²]</u>:

[HPO₄⁻²] + 0.0381 M = 0.100 M

[HPO₄⁻²] = 0.0619 M

Finally, using the concentrations, the volume, and the molecular weights; we can calculate the weight of each substance:

  • Mass of NaH₂PO₄·H₂O = 0.0619 M * 1 L * 138 g/mol = 8.542 g
  • Mass of Na₂HPO₄ = 0.0381 M * 1 L * 142 g/mol = 5.410 g

8 0
4 years ago
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