Answer:
6 electrons
Explanation:
The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s.
Missing question: volume of <span>solution on the left is 10 mL.
V</span>₁(solution) = 10 Ml.
c₁(solution) = 0.2 M.<span>
V</span>₂(solution)
= ?.<span>
c</span>₂(solution)
= 0.04 M.<span>
c</span>₁ -
original concentration of the solution, before it gets diluted.<span>
c</span>₂
- final concentration of the solution, after dilution.<span>
V</span>₁
- <span>volume to
be diluted.
V</span>₂ - <span>final volume after
dilution.
c</span>₁ · V₁ = c₂ · V₂<span>.
</span>10 mL · 0.2 M = 0.04 M · V₂.
V₂(solution) = 10 mL · 0.2 M ÷ 0.04 M.
V₂(solution) = 50 mL.<span>
</span>
It is important to take note of th temperature in determining the density of a substance because this will set as a basis and will likely be a variable in the experiment because this will also contribute on the effects of the experiment and a basis of how the experiment has turned to be that way.
Unsaturation (IHD) 2 hydrogen Needed
IHD = [(2n+2) -H]/2
(H: X=1, N=-1, O= zero)
Unsaturation:
Double bonds = 1
Rings = 1
Triple Bonds = 2
The degrees of unsaturation in a molecule are additive — a
molecule with one double bond has one degree of unsaturation, a molecule with
two double bonds has two degrees of unsaturation, and so forth.
% error = 3.4 %
Percent error = |accepted value - experimental value|/accepted value × 100%
∴ % error = |355 mL – 343 mL|/355 mL × 100 % = |12|/355 × 100 % = 3.4 %
