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Vlada [557]
3 years ago
12

Concentrated hydrochloric acid, HCl, comes with an approximate molar concentration of 12.1 M. If you are instructed to prepare 3

50.0 mL of a 0.975 M HCl solution, how many milliliters of the stock (concentrated) HCl solution will you use
Chemistry
1 answer:
Semmy [17]3 years ago
6 0

Answer:

28.20 mL of the stock solution.

Explanation:

Data obtained from the question include the following:

Molarity of stock solution (M1) = 12.1 M

Volume of diluted solution (V2) = 350.0 mL

Molarity of diluted solution (M2) = 0.975 M

Volume of stock solution needed (V1) =..?

The volume of stock solution needed can be obtained by using the dilution formula as shown below:

M1V1 = M2V2

12.1 x V1 = 0.975 x 350

Divide both side by 12.1

V1 = (0.975 x 350)/12.1

V1 = 28.20 mL.

Therefore, 28.20 mL of the stock solution will be needed to prepare 350.0 mL of 0.975 M HCl solution.

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Answer:

The products are SnPO4 and LiC2H3O2

Explanation:

The reactants are LiPO4 + Sn(C2H3O2)2

This is a double replacement reaction

So what you do is switch the elements the other way around.

To do that, all you have to do switch Sn with PO4 since Sn is a cation and PO4 is an anion.

Then you switch Li with C2H3O2 because Li is a cation and C2H3O2 is an anion.

After that, check the charges. PO4 has -3 charge

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In word form, the product would be Tin(III) Phosphate

C2H3O2 has a -1 charge Li has a +1 charge

So leave both of them the way it is without any subscripts.

In word form, the product would be Lithium Acetate

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<h3>What is hydrocarbon?</h3>

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