Question

Composition of Solutions Big Beaker Small Beaker Trials Distilled H2O Buffer 0.3M KI Starch 0.02 M Na2S2O3 0.1M H2O2 1 123 mL 25 mL 10 mL 2 mL 10 mL 30 mL 2 113 mL 25 mL 20 mL 2 mL 10 mL 30 mL 3 103 mL 25 mL 30 mL 2 mL 10 mL 30 mL 4 83 mL 25 mL 30 mL 2 mL 10 mL 50 mL 5 63 mL 25 mL 30 mL 2 mL 10 mL 70 mL

          Composition of Solutions
Big Beaker
Small Beaker
Trials
Distilled H2O Buffer
0.3M KI Starch 0.02 M Na2S2O3
0.1M H2O2
1
123 mL
25 mL
10 mL
2 mL
10 mL
30 mL
2
113 mL
25 mL
20 mL
2 mL
10 mL
30 mL
3
103 mL
25 mL 30 mL
2 mL
10 mL
30 mL
4
83 mL
25 mL 30 mL
2 mL
10 mL
50 mL
5
63 mL
25 mL 30 mL
2 mL
10 mL
70 mL

Show more…

Composition of Solutions
Big Beaker
Small Beaker
Trials
Distilled H2O Buffer
0.3M KI Starch 0.02 M Na2S2O3
0.1M H2O2
1
123 mL
25 mL
10 mL
2 mL
10 mL
30 mL
2
113 mL
25 mL
20 mL
2 mL
10 mL
30 mL
3
103 mL
25 mL 30 mL
2 mL
10 mL
30 mL
4
83 mL
25 mL 30 mL
2 mL
10 mL
50 mL
5
63 mL
25 mL 30 mL
2 mL
10 mL
70 mL

Added by Cody J.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Dj Tan

09/03/2022

Step 1

To calculate the moles of H2O2, we need to use the formula: moles = concentration (M) x volume (L) Given that the concentration of H2O2 in the small beaker is 0.1 M and the volume is 30 mL (which is equivalent to 0.03 L), we can calculate the moles of H2O2 as Show more…

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determine the initial concentration of H2O2 in each solution (determine how many moles of H2O2 is used in each solution and divide by totalsolution volume (L) after the beakers are mixed). Composition of Solutions Big Beaker Trials Distilled H2O Buffer 0.3M KI Starch 0.02 M Na2S2O3 1 123 mL 25 mL 10 mL 2 mL 10 mL 2 113 mL 25 mL 20 mL 2 mL 10 mL 3 103 mL 25 mL 30 mL 2 mL 10 mL 4 83 mL 25 mL 30 mL 2 mL 10 mL 5 63 mL 25 mL 30 mL 2 mL 10 mL Small Beaker 0.1M H2O2 30 mL 30 mL 30 mL 50 mL 70 mL

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Transcript

00:01 So to solve for this, to solve first for the mass of h2o2, we need to use the value of 3 % weight over weight.

00:16 So we're going to use that information.

00:18 So we need to assume that we have 100ml of the solution.

00:25 All right.

00:26 And then from this, we will solve for the mass of h2o2.

00:29 So we know that we have 3 % by weight, we will divide us by 100%.

00:35 And this is equivalent to the weight of h22 divided by the volume, which we have assumed 100 ml.

00:44 Right? since we have here, the density of the solution will just multiply this.

00:49 One gram per ml, that's the density.

00:51 So as you can see, you'll be able to cancel ml.

00:53 And the remaining unit now is grams.

00:56 So that means the weight or the mass of h3.

00:59 H202 is 3 grams.

01:02 Now we need to solve for the number of moles of h202.

01:08 So we have 3 grams, right? and then molar mass of h202 is 34 .02 grams per mole.

01:16 So that needs cancel the unit again.

01:19 And you should have 0 .082.

01:22 This is mole of h202.

01:26 And then let's solve first for the initial concentration.

01:29 So m1 are the initial molarity...

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