Answer:
(a) 1s² 2s² 2p⁶ 3s² 3p⁴
(b) 1s² 2s² 2p⁶ 3s² 3p⁵
(c) sp³
(d) No valence orbital remains unhybridized.
Explanation:
<em>Consider the SCl₂ molecule. </em>
<em>(a) What is the electron configuration of an isolated S atom? </em>
S has 16 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁴.
<em>(b) What is the electron configuration of an isolated Cl atom? </em>
Cl has 17 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁵.
<em>(c) What hybrid orbitals should be constructed on the S atom to make the S-Cl bonds in SCl₂? </em>
SCl₂ has a tetrahedral electronic geometry. Therefore, the orbital 3s hybridizes with the 3 orbitals 3 p to form 4 hybrid orbital sp³.
<em>(d) What valence orbitals, if any, remain unhybridized on the S atom in SCl₂?</em>
No valence orbital remains unhybridized.
Answer is: the discovery of sub atomic particles like electrons.
J. J. Thomson discovered the electron in 1897.
His "plum pudding" model (1904) suggested: the electrons are embedded in the positive charge.
With this model, he abandoned his earlier hypothesis (the atom was composed of immaterial vortices).
J.J. Thomson placed two oppositely charged electric plates around the cathode ray. He did experiments using different metals as electrode materials and found that the properties of the cathode ray remained constant no matter what cathode material he used.
Tomson concluded that atoms are divisible and that the corpuscles are their building blocks (atoms are made up of smaller particles).
Answer:
C. Kidneys filter wastes from the bloodstream and produce urine
Explanation:
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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Density is Mass divided by volume.
