The simplest and most helpful visual aid for this experiment would be if the student sets rulers of some sort or maybe pieces of wood that are properly marked with certain measuring unite. If they are set right next to each plant, the student will be able to monitor constantly how are the plants growing, and the smaller the measuring unites the better, as the student will be able to monitor the situation in higher detail.
Over time, the plants will most surely develop differently, so the student will have it easier to notice the changes. Considering that it is an experiment in question, the student should use proper measurements and to be able to have detailed numbers for it.
From 15 mol of Silver nitrate, the moles of silver chloride produced have been 15 mol. Thus, option B is correct.
The balanced chemical reaction for the synthesis of silver chloride has been:

From the balanced equation, since there has presence equal moles of silver nitrate and sodium chloride, the moles of silver chloride formed has been equivalent. Thus, 1 mole of silver nitrate gives 1 mole of silver chloride.
The moles of silver nitrate available are, 
The moles of silver nitrate produced can be given as:

Thus, the moles of silver chloride produced have been 15 mol. Thus, option B is correct.
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We are given the number of moles of solid magnesium supplied for the reaction which is 0.02 moles while hydrochloric acid is supplied in excess thus we can say that the reaction proceeds to completion. Calculation is as follows:
0.020 mol Mg ( 1 mol H2 / 1 mol Mg ) = 0.020 mol H2 gas is produced
To convert the number of moles to volume, we use the conditions at STP of 1 mol of a substance is equal to 22.4 L. Thus,
0.020 mol H2 (22.4 L / 1 mol) (1000 mL / 1 L) = 448 mL
At equilibrium the concentrations of:
[HSO₄⁻] = 0.10 M;
[SO₄²⁻] = 0.037 M;
[H⁺] = 0.037 M;
There is initially very little H+ and no SO₄²⁻ in the solution. A salt is KHSO₄⁻. All KHSO₄⁻ will split apart into K⁺ and HSO₄⁻ ions. [HSO₄⁻] will initially be present at a concentration of 0.14 M.
HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid. HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.
R
⇄ 
I 
C

E

×
for
. As a result,
![\frac{[H^+]. [SO_4^2^-]}{HSO_4^-} = K_a](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%5E%2B%5D.%20%5BSO_4%5E2%5E-%5D%7D%7BHSO_4%5E-%7D%20%3D%20K_a)
is large. It is no longer valid to approximate that
at equilibrium is the same as its initial value.

×
× 
Solving the quadratic equation for
since
represents a concentration;

Then, round the results to 2 significant figure;
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Answer: Pauli exclusion principle: only two electrons can occupy the same orbital and they must have opposite spin directions
Explanation: