When drawing lewis dot diagram structure we consider the number of valence electrons of the atom.Se has six valence electrons since it is in group six thus it Lewis dot diagram is as follows
..
: Se:
Hydrogen is in group one hence has one valence electron.The lewis dot diagram for 2H is therefore
H:H
Answer:
D. The electron-withdrawing fluorine atoms pull electron density from the oxygen in trifluoroacetate. The negative charge is more stabilized in trifluoroacetate by this effect.
Explanation:
The structures of trifluoroacetate and acetic acid are both shown in the image attached.
The trifluoroacetate anion (CF3CO2-), just like the acetate anion has in the middle, two oxygen atoms.
However, in the trifluoroacetate anion, there are also three electronegative fluorine atoms attached to the nearby carbon atom attached to the carbonyl, and these pull some electron density through the sigma bonding network away from the oxygen atoms, thereby spreading out the negative charge further. This effect, called the "inductive effect" stabilizes the anion formed,the trifouoroacetate anion is thus more stabilized than the acetate anion.
Hence, trifluoroacetic acid is a stronger acid than acetic acid, having a pKa of -0.18.
<span>Molds are created to achieve a
specific design of a material. These materials either came from a process of
having a higher or lower temperature. Therefore, the molder must have thermal
resistant properties. Low melting points means that the material to be shaped
came from a cooler process. Wood and metal have higher thermal conductivity and
therefore can easily be cooled. The wax can turn really hard and can be
unbreakable when present in colder materials due to the lipids present in it. Clay
however can become a mold because of its low melting point.</span>
11.
С7Н16 + 11О2 --> 8Н2О + 7СО2
Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
