Chemistry, asked by rk4741769, 4 months ago

Take three iron nails and clean them by
Take two test tubes marked as (A) and
(B). In each test tube, take about 10 ml
Tie two iron nails with a thread and
immerse them carefully in the copper
sulphate solution in test tube B for
about 20 minutes (Fig. 1.8 (a). Keep one
cop
Activity 1.9
1.
rubbing with sand paper.
copper sulphate solution.
iron nail aside for comparison.
After 20 minutes, take out the iron nails
from the copper sulphate solution.
Compare the intensity of the blue colour
of copper sulphate solutions in test tubes
(A) and (B) (Fig. 1.8 (b)].
Also, compare the colour of the iron nails
dipped in the copper sulphate solution
with the one kept aside (Fig. 1.8 (b)].​

Answers

Answered by vanshikaj981
1

Cannot Understand The Question

Answered by akramqidwai
3

Answer:

we know the colour of copper sulphate solution is blue.

  • so, in test tube a no change in colour will happen solution will remain blue, because there is no iron nail present in it.

  • in test tube b iron preset in nails will displace copper from it solution and form iron sulphate solution( green in colour).

Fe(s) + cuso4 (aq) -----> FeSo4 (aq) + Cu (s)

hence it is displacement rection.

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