Students performed a procedure similar to part ii of this experiment (analyzing juices for vitamin c content) as described in the procedure section. three 10.00ml samples of juice were titrated with dcp that had a standardized concentration of 9.98x10-4m. the three titrations took an average of 16.34ml of dcp. calculate the mass (in mg) in 50.00ml of juice. (mm ascorbic acid = 176.124 g/mol)

Explanation:

Hello,

In this case, the titration is based of the equality between the moles of the titrant and the analyte, then, in terms of molarity we define it as shown below:

As the ascorbic acid is the substance reacting with the DCP, we infer that the 16.34 mL of DCP completely neutralized the ascorbic acid; in such a way, solving for the molarity of the neutralized acid, one obtains:

Finally, by using the molar MASS of the ascorbic acid and applying the proper units conversion, the result is shown below:

Therefore, 14.36 mg of ascorbic acid are contained into 50.00mL of juice.

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answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

e

catya that's not even a b or c