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xeze [42]
3 years ago
10

during her science fair project, mary discovered that solutions of ionic compounds in water conduct electricity. What type of in

vestigation could mary perform to discover which type of ions conduct electricity best?
Chemistry
1 answer:
Brums [2.3K]3 years ago
8 0

Mary could do two types of experiments

<em>Type 1</em>. She could <em>hold the cation constant and vary the anion</em>.

She could try solutions of NaCl, NaBr, NaI, Na₂SO₄, Na₃PO₄, etc.

These experiments would tell her which <em>anions</em> are the best conductors.

<em>Type </em>2. She could <em>hold the anion constant and vary the cation</em>.

She could try solutions of NaNO₃, KNO₃, Mg(NO₃)₂, Ca(NO₃)₂, Al(NO₃)₃, etc.

These experiments would tell her which <em>cations</em> are the best conductors.

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Which is an example of a compound?
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2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
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In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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3 years ago
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