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goldfiish [28.3K]
3 years ago
14

A group of students forgot to add thiosulfate to one test tube> what will they observe?

Mathematics
1 answer:
anzhelika [568]3 years ago
6 0

Answer:

The solution will experience an instant colour change because thiosulphate which is meant to be used as a clocking or Redox reagent which would have to react with iodine is absent.

Step-by-step explanation:

Thiosulphate is a chemical reagent that is used in chemical reactions such as titrations.

Thiosulphate is used as in the chemical reaction called Iodometry. This is a reaction whereby Thiosulphate is used as reactant and it is titrated against iodine in order to obtain an endpoint.

If Thiosulphate is not added to the test tube, there would be a colour change that occurs instantly because Thiosulphate did not react with iodine.

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Please Help!! These are 3 seperate questions I need help with.
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Answer:

  1. A.  (x1, x2) = (1, 4)
  2. C.  There is no solution.
  3. B.  The solutions are of the form: (x1, x2, x3) = (4t+4, t-3, t)

Step-by-step explanation:

1.  Subtract the first equation from twice the second to eliminate x2.

  2(2x1 -x2) -(x1 -2x2) = 2(-2) -(-7)

  3x1 = 3 . . . . . simplify

  x1 = 1 . . . . . . .divide by 3

Substitute this value into the second equation.

  2·1 -x2 = -2

  4 -x2 = 0 . . . . . add 2

  4 = x2 . . . . . . . add x2

The solution is (x1, x2) = (1, 4).

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2. Multiply the first equation by 3 and add the second equation.

  3(-5x1 -3x2) +(15x1 +9x2) = 3(7) +(2)

  0 = 23 . . . . . not true;

There is no solution.

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3. There are two equations in 3 unknowns, so there cannot be a unique solution. The equations are not dependent, so there will be an infinite number of solutions that can be written in terms of a single parameter (t).

Let x3 = t. Then the system of equations can be rewritten as

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  (2x1 -5x2) -2(x1 -4x2) = (23 +3t) -2(16)

  3x2 = 3t -9 . . . simplify

  x2 = t -3 . . . . . divide by 3

Substituting this into the second equation above, we have ...

  x1 -4(t -3) = 16

  x1 = 16 +4t -12 . . . . . . add 4(t-3)

  x1 = 4t +4 . . . . . . . . simplify

The solutions are (x1, x2, x3) = (4t+4, t-3, t).

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The attachment shows the first system solved by graphical means. This validates our answer.

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