Copper has 29 electrons, and its electron configuration is:
1s2 2s2 2p6 3s2 3p6 3d10 4s1.
Therefore, copper has 1 valence electron.
Hope this helps~
Answer is: the percent composition of Hg in the compound is 71.5%.
Balanced chemical reaction: Hg + Br₂ → HgBr₂.
m(Hg) = 60.2 g; mass of the mercury.
m(Br₂) = 24.0; mass of the bromine.
m(HgBr₂) = m(Hg) + m(Br₂).
m(HgBr₂) = 60.2 g + 24 g.
m(HgBr₂) = 84.2 g; mass of the compound.
ω(Hg) = m(Hg) ÷ m(HgBr₂) · 100%.
ω(Hg) = 60.2 g ÷ 84.2 g · 100%.
ω(Hg) = 71.5%.
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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Answer:
a) = 0.704%
b) = 1.30%
c) = 2.60%
Explanation:
Given that:
= 
For Part A; where Concentration of A = 0.270 M
Percentage Ionization(∝) 



percentage% (∝) = 
= 0.704%
For Part B; where Concentration of B =
M



percentage% (∝) = 0.0130 × 100%
= 1.30%
For Part C; where Concentration of C= 



percentage% (∝) = 0.02608 × 100%
= 2.60%
The acronym is RNA. Ribo Nucleic Acid