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emmainna [20.7K]
2 years ago
15

. Write the two resonance hybrids for the carbocation that would be formed by protonation at C-1 of 2-methyl-1,3-pentadiene. Wit

hout doing a calculation, would you expect C-2 or C-4 (the two end carbons of the allylic cation) to have the most positive charge on it
Chemistry
1 answer:
never [62]2 years ago
8 0

Answer:

Answer is explained in the explanation section below.

Explanation:

This question is requires the diagram, which I have drawn and it is attached in the attachment below. Please for this answer, refer to the diagram attached in the attachment below.

Referring to the diagram, attached. As we know that, Allylic secondary carbocations are more stable than Allylic tertiary carbocations.

Hence,

C2 will have a more positive charge since a tertiary carbocation (C2) is more stable than a secondary carbocation (C4). Therefore, the resonance structure will favor the positive charge at C2.

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Using VSEPR theory which of the following would be the correct shape for nitrogen trifluoride?
RUDIKE [14]

Answer:

trigonal pyramidal

Explanation:

In NF3, the nitrogen atom is sp3 hybridized. Now we must remember that according to the VSEPR theory, the number of electron pairs in the valence shell of the central atom in a molecule determines its shape.

Here, the nitrogen atom is the central atom and its outermost shell is surrounded by four electron pairs - one lone pair and three bond pairs. This means that it has a tetrahedral electron pair geometry.

However, due to the lone pair, the three fluorine atoms are arranged in a  trigonal pyramidal geometry. Hence the correct shape of the molecule is trigonal pyramidal.

6 0
2 years ago
Calculate the percent ionization of formic acid (hco2h) in a solution that is 0.311 m in formic acid and 0.189 m in sodium forma
ohaa [14]
<span>Answer: 0.094%


</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />

<span>Only the ionization of the formic acid is the important part.
</span><span />

<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />

<span>2) Mass balance:
</span><span />

<span>                   HCOOH(aq)     HCOO⁻(aq)     H⁺(aq).

Start             0.311                 0.189

Reaction       - x                      +x                   +x

Final             0.311 - x          0.189 + x            x


3) Acid constant equation:
</span><span />

<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />

<span>= (0.189 + x )x / (0.311 - x) = 0.000177


4) Solve the equation:


You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />

<span>With that approximation the equation to solve becomes:


</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M


5) With that number, the percent of ionization (alfa) is:
</span><span />

<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />
8 0
3 years ago
Sodium hydroxide +HCI ➙ sodium chloride +H2O<br><br> Is this a:
Alexus [3.1K]
Neutralization reaction??
4 0
3 years ago
What is the boiling point in c of a 0.743 m aqueous solution of KCI?
Maksim231197 [3]

Answer:

0.512^ is probably the answer

4 0
2 years ago
In a titration of HCl with NaOH, 100mL of the base was required to neutralize 20mL of 5.0 M HCl. What is the molarity of the NaO
nikklg [1K]

Answer:

1 M

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Volume of base (Vb) = 100mL

Volume of ac(Va) = 20mL

Molarity of acid (Ma) = 5M

Molarity of base (Mb) =...?

Step 2:

The balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H2O

From the above equation, the following were obtained:

Mole ratio of the acid (nA) = 1

Mole ratio of the base (nB) = 1

Step 3:

Determination of the molarity of the base.

The molarity of the base can be obtained as follow:

MaVa/MbVb = nA/nB

5 x 20 / Mb x 100 = 1

Cross multiply to express in linear form

Mb x 100 = 5 x 20

Divide both side by 100

Mb = (5 x 20)/100

Mb = 1 M

Therefore, the molarity of the base is 1 M

8 0
3 years ago
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