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mario62 [17]
3 years ago
5

A volume of 90.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What

was the molarity of the KOH solution if 21.2 mL of 1.50 M H2SO4 was needed? The equation is
2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

Express your answer with the appropriate units.
Chemistry
1 answer:
Gemiola [76]3 years ago
7 0

Answer:

The answer to your question:  0.7 M

Explanation:

Data

V of KOH = 90 ml

[KOH] = ?

V H2SO4 = 21.2 ml

[H2SO4] = 1.5 M

                       2KOH(aq)  +  H₂SO₄(aq)   →   K₂SO₄(aq)  +  2H₂O(l)

Molarity = moles / volume

moles of H₂SO₄ = (1.5) (21.2)

                           = 31.8

                    2 moles of KOH --------------  1 mol of H₂SO₄

                   x                           --------------  31.8 mol of H₂SO₄

                    x = (31.8)(2) / 1

                    x = 63.8 moles of KOH

Molarity = 63.8 / 90

             = 0.7 M

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2 years ago
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Answer: The element expected to be most electronegative is Ca.

The element expected to be least electronegative is K.

Explanation:

Electronegativity is defined as the property of an element to attract a shared pair of electron towards itself.

Down the group:

The size of an atom increases as we move down the group because a new shell is added and electron gets added up.

As, the size of an element increases, the valence electrons gets away from the nucleus. So, the attraction between the nucleus and the shared pair of electrons decreases

Hence, electronegativity decreases moving from top to bottom down a group

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As, the size of an element decreases, the valence electrons come near to the nucleus. So, the attraction between the nucleus and the shared pair of electrons increases.

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7 0
3 years ago
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Answer:

λ=2167.6 nm

The wavelength of light emitted is 2167.6 nm.

Explanation:

We recall that Eₙ=\frac{-2.18*10^{-18} J}{n^{2} }

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λ=2167.6*10^-9 m

λ=2167.6 nm

The wavelength of light emitted is 2167.6 nm.

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