How do you separate aluminum and steel cans for recycling?
2 answers:
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Answer:
Sulfur: -1
Carbon: 0
Nitrogen: 0
Explanation:
The thiocyanate ion SCN- can have only two resonance structures, which are:
S - C ≡ N <--------> S = C = N
In the first structure, we have one single bond and one triple bond, in this case, the negative charge is located in the sulfur. This is because Sulfur have 6 electrons and those electrons are present in the atom, (see picture below), and counting the electron that is sharing with the Carbon, the total electrons that sulfur has is 7 (It has one more than usual). Carbon and nitrogen are already stable with 0 of formal charge, because carbon can only have 4 electrons which 1 is sharing with sulfur and the other 3 with the nitrogen, and nitrogen have 5 electrons, three sharing with carbon and the other two kept it for itself.
In the second structure, the negative charge of the sulfur is transfered to the nitrogen, meaning that it has 6 electrons the nitrogen (formal charge -1) and carbon and sulfur with 4 and 6 electrons respectively.
Between these two structures, the most stable is the first one basically because Sulfur is a better nucleophile than the Nitrogen, and can form stronger hydrogen bond in acid, giving more stable structure.
1.66 M is the concentration of the chemist's working solution.
<h3>What is molarity?</h3>Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.
In this case, we have a solution of Zn(NO₃)₂.
The chemist wants to prepare a dilute solution of this reactant.
The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.
We want to know the concentration of this diluted solution.
As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:
=
(1)
and we also know that:
n = M x
If we replace this expression in (1) we have:
x
=
x
Where 1, would be the stock solution and 2, the solution we want to prepare.
So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is :
4.93 x 210 = 620 x
= 1.66 M
This is the concentration of the solution prepared.
Learn more about molarity here:
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