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andreev551 [17]
3 years ago
6

A laboratory experiment requires 2.25 L of a 1.0 M solution of phosphoric acid (H3PO4), but the only available H3PO4 is a 9.0 M

stock solution. How could you prepare the solution needed for the lab experiment? Show all the work used to find your answer.
Chemistry
2 answers:
Ksenya-84 [330]3 years ago
6 0
The solution needed is prepared  as below

by use of the   M1V1 =M2 V2  formula  where
M1 = 2.25 L
v2 = 1.0M
M2 = 9.0 M
V2 =? l

make V2  the subject of the formula  V2 =M1V1/M2

= 2.25 L x  1.0M/9.0 M  = 0. 25 L

therefore  the solution  need  0.25 L  of   9.0M H3PO4   and  dilute it a final volume  of 2.25 l 
Semenov [28]3 years ago
3 0
We know that, M1V1   =   M2V2
                        (Initial)      (Final)

Given: M1 = 9.0 M
            M2 = 1.0 M
            V2 = 2.25 L

Therefore we have, 9.0 x V1 = 1.0 x 2.25
                                         V1 = 0.25 L

Thus, 0.25 L of stock solution (9.0 M), when diluted till 2.25 L, the resultant solution will have concentration 1.0 M.
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Calculate the volume of chlorine molecules produced at room temperature and pressure, when 234g of sodium chloride are electroly
rewona [7]

The volume of chlorine molecules produced at STP would be 96 dm³.

<h3>Stoichiometric problem</h3>

Sodium chloride ionizes during electrolysis to produce sodium and chlorine ions as follows:

NaCl --- > Na^+ + Cl^-

This means that 1 mole of sodium chloride will produce 1 mole of sodium ion and 1 mole of chlorine ion respectively.

Recall that: mole = mass/molar mass

Hence, 234 g of sodium chloride will give:

                         234/58.44 = 4.00 moles.

Thus, the equivalent number of moles of chlorine produced by 234 g of sodium chloride will be 4 moles.

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4 moles of chlorine = 4 x 24 = 96 Liters or 96 dm³.

More on stoichiometric problems can be found here: brainly.com/question/14465605

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otez555 [7]

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2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

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Mass of fat = 2.4 kg = 2.4 × 1000 g = 2400 g

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M = 57 × 12 g/mol + 110 × 1 g/mol+ 6 × 16 g/mol = 890 g/mol[/tex]

Moles of fat = \frac{2400 g}{890 g/mol}=2.6966 mol

According to reaction , 2 moles of fat gives 110 moles of water. Then 2.6966 moles of fat will give  ;

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Mass of 148.31 moles of water ;

148.31 mol × 18 g/mol = 2,669.58 g

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