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goblinko [34]
3 years ago
8

Due to the small and highly electronegative nature of fluorine, the oxyacids of the this element are much less common and less s

table than those of the other halogens. Bonding theory, however, does allow one to propose structures for these acids and use formal charges for the evaluation of these structures. For a molecule of fluorous acid, the atoms are arranged as HOFO. (Note: In this oxyacid, the placement of fluorine is an exception to the rule of putting the more electronegative atom in a terminal position.) What is the formal charge on each of the atoms? Enter the formal charges in the same order as the atoms are listed.
Chemistry
1 answer:
steposvetlana [31]3 years ago
4 0
We are told we have an oxyacid of the formula HOFO. We will assume the atoms are in this order and will draw a proper lewis structure for this compound by first drawing bonds between each of the 4 atoms and then place the remaining electron pairs on each atom:
      ..    ..    ..
H - O - F - O:
      ··   ··    ··
We can calculate the formal charge of an atom using the following formula:

Formal charge = [# of valence electrons] - [# of non-bonded electrons + # of bonds]

H: Formal charge = [1]-[0+1] = 0

O: Formal charge = [6]-[4+2] = 0

F: Formal charge = [7]-[4+2] = +1

O: Formal charge = [6]-[6+1] = -1

As we can see the overall charge of the molecule is neutral since the fluorine as a +1 charge and the oxygen a -1 charge.
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ΔHfo stands for <span>standard heat of formation for a compound.

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3 0
3 years ago
Suppose you want to make an acetic acid/acetate buffer to a pH of 5.00 using 10.0 mL of 1.00 M acetic acid solution. How many mi
olasank [31]

Answer:

Explanation:

Molarity of NaOAc needed

Using the Henderson-Hasselbalch Equation calculate base molarity needed given [HOAc] = 1.00M and pKa(NaOAc) = 4.75 and [HOAc] = 1.00m.

pH = pKa + log [NaOAc]/[HOAc]

5.00 = 4.75 + log[NaOAc]/[1.00M]

5.00 - 4.75 = log [NaOAc] - log[1.00M]

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Given 10ml of HOAc, how much (ml) 1.78M NaOAc to obtain a buffer pH of 5.00.

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(M·V)acid = (M·V)base => V(base) = (M·V)acid / (M)base

Vol (NaOAc) needed = (1.00M)(0.010L)/(1.78M) = 0.0056 liter = 5.6 ml.

Checking Results:

5.00 = 4.75 + log [1.78M]/[1.00M] = 4.75 + 0.25 = 5.00    QED.

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vekshin1

Given :

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