Answer:
<em>C. The electron-withdrawing fluorine atoms pull electron density from the oxygen in trifluoroacetate. The negative charge is more stabilized in trifluoroacetate by this effect.</em>
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Explanation:
<em>The structures of trifluoroacetate and acetic acid are both shown in the image attached.</em>
<em>The trifluoroacetate anion (CF3CO2-), just like the acetate anion has in the middle, two oxygen atoms.</em>
<em>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.</em>
<em>Hence, trifluoroacetic acid is a stronger acid than acetic acid, having a pKa of -0.18.</em>
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<u><em>Hope this helps!</em></u>
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You would be most likely to use a slicing machine if you were using the <u>icebox </u>method to produce cookies.
In the icebox method a type of cookie in which the dough is made, rolled into a stick, and refrigerated until the dough hardens. The dough can be removed from the refrigerator, cut into individual pieces, and then baked. The rest of the dough is returned to the refrigerator until needed.
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Learn more about the Icebox method here,
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The given statement is True.
This is single replacement reaction.
As reactivity decreases down the group,
Iodine is less reactive than Chlorine.
And so Iodine cannot replace chlorine from FeCl2.
I2 + FeCl2 ---> no reaction (Iodine is less reactive than chlorine)
NaOH is a strong base and complete dissociation into Na⁺ and OH⁻ ions.
Therefore [NaOH] = [OH⁻]
To calculate the [OH⁻], we can first find the pOH as NaOH is a basic solution.
pH + pOH = 14
Since pH = 11.50
pOH = 14 - 11.50
pOH = 2.50
We can calculate [OH⁻] by knowing pOH
pOH = -log[OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 3.2 x 10⁻³ M
therefore [NaOH] = 3.2 x 10⁻³ M
