Answer:
40mL of glycerol are needed to make a 20% v/v solution
Explanation:
This problem can be solved with a simple rule of three:
20% v/v is a sort of concentration. In this case, 20 mL of solute are contained in 100 mL of solution.
Therefore, in 100 mL of solution you have 20 mL of solvent (glycerol)
In 200 mL, you would have, (200 .20)/ 100 = 40 mL
6.0 will the pH at the neutralization point of 0.00812 m Ba(OH)2 be when titrated with HCl.
<h3><u>What are </u><u>
titration reactions ?</u></h3>
A titrant/titrator, a standard solution with fixed volume and concentration, is prepared as part of the titration procedure. When an endpoint or equivalence point is reached, the titrant is made to continue reacting with the analyte, and at that point, the amount of titrant consumed may be used to calculate the analyte's concentration. Alternately, titration is the application of the stoichiometry principle to determine an unknown solution's concentration.
The technique begins by adding a very little quantity of analyte to a beaker or Erlenmeyer flask. Under a calibrated burette or chemical pipetting syringe containing the titrant, a little quantity of an indicator (such as phenolphthalein) is inserted.
To view more about titration reactions, refer to:
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Answer:
the picture is too blury to see lol.
Explanation:
The % mass/mass concentration of solute in the seawater sample is calculated as below
% mass = mass of the solute /mass of the solvent(sea water) x100
mass of the solute =1.295 g
mass of the solvent(sea water_) = 25.895 g
there the % mass = 1.295/25.895 x100 = 5.001 %
