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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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4 0
3 years ago
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<h3><u>Answer;</u></h3>

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5 0
3 years ago
A container of gas has a volume of 140.0 cm3 at a temperature of 27.0°C. If the
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Answer:

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Answer:

\huge 3.322 \times  {10}^{ - 5}  \:  \: moles  \\

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

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L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{2.00 \times  {10}^{19} }{6.02 \times  {10}^{23} }  \\  \\  = 3.322 \times  {10}^{ - 5}  \:  \: moles

Hope this helps you

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Describe the location of the positive charge of the atom in j. j. thomson’s plum pudding model.
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Thompson proposed a model (the plum pudding model) whereby the negatively charged corpuscles were distributed in a uniform sea of positive charge. In other words the positive charge was like the pudding and the negative charge were like raisins on the pudding.
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