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Stells [14]
3 years ago
9

Toluene is subjected to the action of the following reagents in the order given: (1) KMnO4,OH-, heat; then H3O (2) HNO3, H2SO4 (

3) Br2, FeBr3 What is the final product of this sequence?
Chemistry
1 answer:
Shkiper50 [21]3 years ago
5 0

Answer:

See image attached

Explanation:

The reaction of toluene with alkaline potassium permanganate in the presence of heat leads to the oxidation of the -CH3 to give benzoic acid.

Reaction benzoic acid with HNO3/H2SO4 yields the nitronium ion (NO2+).

Recall that -COOH is a metal director and deactivated the ring towards electrophilic substitution hence the m-nitrobenzoic acid is formed.

Reaction with FeBr3/Br2 yields the product shown in the image attached.

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Which of the following best explains why bonds form between atoms?
ANEK [815]

Answer:

D

Explanation:

I believe the answer is D because atoms are always seeking to fill up their outer electron shell/valence shell and want to gain a full octet.

6 0
3 years ago
How can the requirements of Na (sodium) solve the requirements of Cl (chloride) instability achievement?
Paul [167]

Na releases 1 electron to be stable

Cl requires 1 electron to be stable

both are ionic bonded to be stable

8 0
2 years ago
Help plz already know 1&2
Aleonysh [2.5K]
3. A
4. B
5. A
6. E
7. A
8. C
9. A
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3 0
3 years ago
Enter the appropriate symbol for an isotope of potassium-39 corresponding to the isotope notation AZX.
muminat
The notation is actually written in this way (shown in the attached picture). The X is the symbol of the element. So, for potassium, the symbol is K. The A represents the mass number of the isotope. In this case, that would be equal to 39. The Z represents the atomic number of potassium, which is 19. Therefore, the symbol for the isotope is:  ³⁹₁₉K

6 0
3 years ago
Read 2 more answers
If a buffer contains 1.05M B and 0.750M BH+ has the pH of 9.5. What would be the pH after 0.005mol of HCL is added to 0.5L of so
Leviafan [203]

Answer:

Final pH: 9.49.

Round to two decimal places as in the question: 9.5.

Explanation:

The conjugate of B is a cation that contains one more proton than B. The conjugate of B is an acid. As a result, B is a weak base.

What's the pKb of base B?

Consider the Henderson-Hasselbalch equation for buffers of a weak base and its conjugate acid ion.

\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}}.

\text{pOH} = \text{pK}_w - \text{pH}.

\text{pK}_w = 14.

\text{pOH} = 14 - 9.5 = 4.5

\displaystyle \text{pK}_b = \text{pOH} -\log{\frac{[\text{Salt}]}{[\text{Base}]}}\\\phantom{\text{pK}_b} = 4.5 - \log{\frac{0.750}{1.05}} \\\phantom{\text{pK}_b} =4.64613.

What's the new salt-to-base ratio?

The 0.005 mol of HCl will convert 0.005 mol of base B to its conjugate acid ion BH⁺.

Initial:

  • n(\text{B}) = c\cdot V = 1.05 \times 0.5 = 0.525\;\text{mol};
  • n(\text{BH}^{+}) = c\cdot V = 0.750 \times 0.5 = 0.375\;\text{mol}.

After adding the HCl:

  • n(\text{B}) = 0.525 - 0.005 = 0.520\;\text{mol};
  • n(\text{BH}^{+}) = 0.375+ 0.005 = 0.380\;\text{mol}.

Assume that the volume is still 0.5 L:

  • \displaystyle [\text{B}] = \frac{n}{V} = \frac{0.520}{0.5} = 1.04\;\text{mol}\cdot\text{dm}^{-3}.
  • \displaystyle [\text{BH}^{+}] = \frac{n}{V} = \frac{0.380}{0.5} = 0.760\;\text{mol}\cdot\text{dm}^{-3}.

What's will be the pH of the solution?

Apply the Henderson-Hasselbalch equation again:

\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}} = 4.64613 + \log{\frac{0.760}{1.04}} = 4.50991

\text{pH} = \text{pK}_w - \text{pOH}= 14 - 4.50991 = 9.49.

The final pH is slightly smaller than the initial pH. That's expected due to the hydrochloric acid. However, the change is small due to the nature of buffer solutions: adding a small amount of acid or base won't significantly impact the pH of the solution.

3 0
3 years ago
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