Answer:
In the final solution, the concentration of sucrose is 0.126 M
Explanation:
Hi there!
The number of moles of solute in the volume taken from the more concentrated solution will be equal to the number of moles of solute in the diluted solution. Then, the concentration of the first solution can be calculated using the following equation:
Ci · Vi = Cf · Vf
Where:
Ci = concentration of the original solution
Vi = volume of the solution taken to prepare the more diluted solution.
Cf = concentration of the more diluted solution.
Vf = volume of the more diluted solution.
For the first dillution:
26.6 ml · 2.50 M = 50.0 ml · Cf
Cf = 26.6 ml · 2.50 M / 50.0 ml
Cf = 1.33 M
For the second dilution:
16.0 ml · 1.33 M = 45.0 ml · Cf
Cf = 16.0 ml · 1.33 M / 45.0 ml
Cf = 0.473 M
For the third dilution:
20.0 ml · 0.473 M = 75.0 ml · Cf
Cf = 20.0 ml · 0.473 M / 75.0 ml
Cf = 0.126 M
In the final solution, the concentration of sucrose is 0.126 M
Answer:
Explanation:
When a salt is dissolved , it increases the boiling point . Increase in boiling point depends upon number of ions . So it is a colligative property .
.19 m AgNO₃ . Each molecule will ionize into two ions . So effective molar concentration is 0.19 x 2 = .38 m
0.17 m CrSO4.Each molecule will ionize into two ions . So effective molar concentration is 0.17 x 2 = .34 m
0.13 m Mn(NO₃)₂. Each molecule will ionize into three ions . So effective molar concentration is 0.13 x 3 = .39 m
0.31 m Sucrose(nonelectrolyte). Molecules will not ionize . So effective molar concentration is 0.31 x 1 = .31 m
Higher the molar concentration , greater the depression in boiling point .
So lowest boiling point is 0.13 m Mn(NO₃)₂.
second highest boiling point is 0.19 m AgNO3.
Third lowest boiling point is 0.17 m CrSO4
Highest boiling point or lowest depression 0.31 m Sucrose.
a . 4
b . 1
c . 2
d . 3
<h2>Answer : Option B) The hematite particles rearrange to form a new substance.</h2><h3>Explanation :</h3>
Hematite particles when gets rearranged to form a new substance, is an example of a chemical change.
As a chemical changes is usually a change where a substance undergoes a chemical change and forms a new substance; which cannot be easily reversed into reactants by any simple physical methods.
In this example the hematite ore forms a new substance by an irreversible chemical change.
Answer:
it is possible to remove 99.99% Cu2 by converting it to Cu(s)
Explanation:
So, from the question/problem above we are given the following ionic or REDOX equations of reactions;
Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V
Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V
In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:
Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.
Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.
Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.
Thus, ΔG° = - 92640.
This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.
