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Ronch [10]
3 years ago
14

If 31.6 g of KMnO4 is dissolved in enough water to give 160 mL of solution, what is the molarity?

Chemistry
1 answer:
Zina [86]3 years ago
5 0

Answer:

A. 1.25M

B. 19.98g

Explanation:

A. Data obtained from the question include the following:

Mass of KMnO4 = 31.6 g

Volume = 160 mL

Molarity =..?

We'll begin by calculating the number of mole KMnO4 in the solution. This is can be obtained as follow:

Mass of KMnO4 = 31.6 g

Molar mass of KMnO4 = 39 + 55 + (16x4) = 158g/mol

Number of mole of KMnO4 =..?

Mole = mass /Molar mass

Number of mole of KMnO4 = 31.6/158 = 0.2 mole

Now, we can obtain the molarity of the solution as follow:

Volume = 160 mL = 160/1000 = 0.16L

Mole of KMnO4 = 0.2 mole

Molarity = mole /Volume

Molarity = 0.2/0.16 = 1.25M

B. Data obtained from the question include the following:

Volume = 300mL

Molarity = 0.74 M

Mass of H2C2O4 =..?

First, we shall determine the number of mole H2C2O4. This is illustrated below:

Volume = 300mL = 300/1000 = 0.3L

Molarity = 0.74 M

Mole of H2C2O4 =?

Mole = Molarity x Volume

Mole of H2C2O4 = 0.74 x 0.3

Mole of H2C2O4 = 0.222 mole

Now, we can easily find the mass of H2C2O4 by converting 0.222 mole to grams as shown below:

Number of mole of H2C2O4 = 0.222 mole

Molar mass of H2C2O4 = (2x1) + (12x2) + (16x4) = 2 + 24 + 64 = 90g/mol

Mass of H2C2O4 =..?

Mass = mole x molar mass

Mass of H2C2O4 = 0.222 x 90

Mass of H2C2O4 = 19.98g

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Consider the redox reaction below.
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Explanation:

Let us consider the complete redox reaction:

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This is a redox reaction because, both oxidation and reduction is simultaneously taking place.

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Here Zn(s) is undergoing oxidation from OS 0 to +2

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39. Analyze What subscripts would you most likely use if
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Based on their valencies, the subscripts of the ionic compounds formed will be:

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  • 2 and 1
  • 1 and 2
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<h3>What subscripts would you most likely use if the following substances formed an ionic compound?</h3>

A. An alkali metal and a halogen.

An alkali metal and a halogen both have valencies of one.

Therefore the subscripts would be 1.

B. An alkali metal and a non-metal from group 16.

An alkali metal has a valency of 1 and a group 16 non-metal has a valency of 2.

By exchange of valencies, the subscript would be 2 and 1.

C. An alkaline earth metal and a halogen.

An alkaline earth metal has a valency of 2 and a halogen has a valency of 1.

By exchange of valencies, the subscripts would be 1 and 2.

D. An alkaline earth metal and a non-metal from group 16.

An alkaline earth metal has a valency of 2 and a non-metal from group 16 has a valency of 2.

By exchange of valencies, the subscripts would be 2 and 2.

Therefore, the subscripts of the ionic compounds formed will be:

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