Answer: The correct answer is the option: B. An element with eight valence electrons is chemically unstable.
Explanation:
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We will analyze each of the options:
A. The group number of the element provides a clue to the number of valence electrons: it is correct, since it provides the number of valence electrons.
B. An element with eight valence electrons is chemically unstable: this is not correct, since elements with eight electrons in the valence shell cannot react because they already have the last complete shell. Therefore, they are chemically stable.
C. The points must be placed one at a time on each side of the chemical symbol: it is correct, because that is the way to make the point diagram.
D. An atom is chemically stable if all the points are paired: this is correct since this verifies that the point diagram has been done well.
We conclude that the correct answer is the option: B. An element with eight valence electrons is chemically unstable.
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Adding (S2O3)2- would affect the reaction mechanism that involves this ion. From the reaction mechanism given above, the equilibrium of step 2 would be affected. Adding the stock solution of (S2O3)2- would shift the equilibrium to the right thus making more products of the said mechanism. Also, the reaction rate of this step would occur faster than the original rate. This is based on Le Chatelier's Prinicple which states that a corresponding change would happen to the equilibrium of a reaction when pressure, concentration of the substances or temperature is changed. So, that after the addition, a color change would appear immediately because I3- would be removed slowly from solution, and would therefore be able to react with starch.
Answer:
The 1st and 2nd ones on the top
Explanation:
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Explanation:
The given data is as follows.
Concentration = 0.1 
= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions
= 
T =
= (30 + 273) K = 303 K
Formula for electric double layer thickness (
) is as follows.
= 
where,
= concentration = 
Hence, putting the given values into the above equation as follows.
=
=
=
m
or, =
= 1 nm (approx)
Also, it is known that
= 
Hence, we can conclude that addition of 0.1
of KCl in 0.1
of NaBr "
" will decrease but not significantly.