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Tcecarenko [31]
3 years ago
8

Determine the pH of a 2.8 ×10−4 M solution of Ca(OH)2.

Chemistry
1 answer:
shepuryov [24]3 years ago
3 0

Answer:

pH = 10.75

Explanation:

To solve this problem, we must find the molarity of [OH⁻]. With the molarity we can find the pOH = -log[OH⁻]

Using the equation:

pH = 14 - pOH

We can find the pH of the solution.

The molarity of Ca(OH)₂ is 2.8x10⁻⁴M, as there are 2 moles of OH⁻ in 1 mole of Ca(OH)₂, the molarity of [OH⁻] is 2*2.8x10⁻⁴M = 5.6x10⁻⁴M

pOH is

pOH = -log 5.6x10⁻⁴M

pOH = 3.25

pH = 14-pOH

<h3>pH = 10.75</h3>
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Answer:

0.712 mol of NO₂ are formed .

Explanation:

For the reaction , given in the question ,

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From the above balanced reaction ,

2 mol of N₂O₅  reacts to give 4 mol of NO₂

Applying unitary method ,

1 mol of N₂O₅  reacts to give 4 / 2 mol of NO₂

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<u>now, using the above equation , to calculate the moles of the NO₂ , as follow -</u>

Since ,

1 mol of N₂O₅  reacts to give 4 / 2 mol of NO₂

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kipiarov [429]
Remark
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Answer:

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Explanation:

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