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BARSIC [14]
3 years ago
13

The proposed mechanism for a reaction is: Step 1: A + B X (fast) Step 2: X + C Y (slow) Step 3: Y D (fast) What is the overall r

eaction? A. A + B + C D B. A + X Y + D C. A + B Y D. A + Y D
Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
7 0

Answer:

A. A + B + C --> D

Explanation:

Step 1: A + B --> X (fast)

Step 2: X + C --> Y (slow)

Step 3: Y --> D (fast)

To obtain the overall reaction, we have to sum up the reactants and products of all step and eliminate the intermediates.

Reactants:

A + B + X + C + Y

Products:

X + Y + D

So we have;

A + B + X + C + Y  --> X + Y + D

Upon elimination of intermediates, we have;

A + B + C --> D

The correct option is A.

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Ca(s) + 2HCl(aq)  -> CaCl2(aq) +H2(g) ↑ + heat

The clues that it is a chemical reaction could be:
- formation of a new substance, gaseous hydrogen
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As silver is below hydrogen in the electrochemical series, it will not be expected to react with dilute hydrocloric acid. (however, it dissolves in oxidizing acid such as nitric acid, but not displacing hydrogen as a product).

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3 years ago
Which of these is the correct electron configuration
sdas [7]

Answer:

1s2,2s2,2p6,3s2,3p6,4s2

Explanation:

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6 0
2 years ago
Which one of the following molecules is polar? PBr5 CCl4 BrF5 XeF2 XeF4
Natasha_Volkova [10]
Polar molecules are characterized by unequal sharing of electrons in a molecule forming a partial negative and positive charges. For this case, the correct answer is the BrF5. It is has a free pair of electrons and the forces are not balanced.
6 0
3 years ago
Questions 3-6 refer to rhe solutions below:
Yanka [14]

Under room temperature where \text{pK}_w = 14:

3.) (A), (B), and (E).

4.) (D).

5.) (B).

<h3>Explanation</h3>

What makes a buffer solution? For a solution to be a buffer, it needs to contain large amounts of a weak acid and its conjugate base ion. Alternatively, the solution may contain large amounts of a weak base and its conjugate acid ion.  

Not every one of the five solutions is a buffer solution.

<h3>(A)</h3>

Ethanoic acid CH₃COOH (a.k.a. acetic acid) is a weak acid. pKa = 4.756. CH₃COONa is a salt. It dissolves to produce CH₃COO⁻, which is the conjugate base ion of CH₃COOH. The solution in (A) contains equal number of CH₃COOH and CH₃COO⁻, both at 1.0 M.

Refer to the Henderson-Hasselbalch equation for buffers of weak acids.

\displaystyle \text{pH} = \text{pK}_a + \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Acid}]}}.

\displaystyle \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Acid}]}} =\ln{1} = 0.

The pH of the solution in (A) will be the same as the pKa of CH₃COOH. pH = 4.746.

<h3>(B)</h3>

Consider the hydrogen halides:

  • HF: weak acid.
  • HCl: strong acid.
  • HBr: strong acid.

The radius of halogen atoms increases down the group, and hydrogen-halogen bond becomes weaker. It becomes easier for water to break those bonds. As a result, the strength of hydrogen halides increases down the group. HF is the only weak acid among the common hydrogen halides.

Mixing HBr and KBr at equal ratio will be similar to mixing HCl and KCl at the same ratio. All HBr in the solution breaks down into H⁺ and Br⁻. The pH of the solution will depend only on the concentration of HBr.

\displaystyle [\text{H}^{+}] = [\text{HBr}] \\\phantom{[\text{H}^{+}]}= \frac{n}{V} \\\phantom{[\text{H}^{+}]}= \frac{c(\text{HBr})\cdot V(\text{HBr})}{V(\text{HBr})+V(\text{KBr})}\\\phantom{[\text{H}^{+}]}=\frac{0.100\;\text{L}\times 1.0\;\text{mol}\cdot\text{L}^{-1}}{0.100\;\text{L}+0.100\;\text{L}} \\\phantom{[\text{H}^{+}]}= 0.50\;\text{mol}\cdot\text{L}^{-1}.

\text{pH} = -\log{[\text{H}^{+}] = -\log{0.50} \approx {\bf 0.30}.

<h3>(C)</h3>

Similarly to HCl and HBr, HI is also a strong acid. Mixing HI and NaOH at equal ratio will produce a solution of NaI, which is similar to NaCl. The final solution will be neutral. pH = 7 if pKw = 14.

<h3>(D)</h3>

NH₃ is a weak base. NH₄Cl dissolves completely to produce NH₄⁺ and Cl⁻. NH₄⁺ is the conjugate acid of NH₃. The final solution will contain an equal number of NH₃ and NH₄⁺. pKb = 4.75 for ammonia NH₃.

Apply the Henderson-Hasselbalch equation for buffers of weak bases:

\displaystyle \textbf{pOH} = \text{pK}_b + \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Base}]}}= 4.75 + \log{1} = 4.75.

Note that what this equation gives for buffers of weak bases is the pOH of the solution. pH = pKw - pOH. Assume that pKw = 14. pH = 14 - 4.75 = 9.25.

<h3>(E)</h3>

The solution in (E) will contain about 1.0 M of CH₃COOH. The volume of the solution will be 200 mL.

n(\text{CH}_3\text{COO}^{-}) = n(\text{NaOH}] = c\cdot V = 0.10\;\text{mol}.

\displaystyle [\text{CH}_3\text{COO}^{-}] = \frac{n}{V} = {0.10}{0.10 + 0.10} = 0.50 \;\text{mol}\cdot\text{L}^{-1}.

There's nearly no conjugate base of CH₃COOH. As a result, the solution will not be a buffer, and the Henderson-Hasselbalch Equation will not apply. Refer to an ICE table:

\begin{array}{c|ccccccc}\text{R}&\text{CH}_3\text{COO}^{-} &+&\text{H}_2\text{O}&\rightleftharpoons &\text{CH}_3\text{COOH}&+&\text{OH}^{-}\\\text{I}&0.50\\\text{C}& -x &&&& +x &&+x\\\text{E} &0.50 - x &&&&x&&x\end{array}

The value of pKa is large. Ka will be small. the value of x will be much smaller than 0.50 such that 0.50-x \approx 0.50.

The pKa of a weak acid is the same as pKw divided by the pKb of its conjugate base.

\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]} = \text{K}_b(\text{CH}_3\text{COO}^{-}) \\\phantom{\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]} }= \frac{\text{K}_w}{\text{K}_a(\text{CH}_3\text{COOH})} \\\phantom{\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]}} = \frac{10^{-14}}{1.75\times 10^{-5}} = 5.71\times 10^{-10}.

\displaystyle \frac{x^{2}}{0.50} =5.71\times 10^{-10}.

[\text{OH}^{-}] = x \approx 1.69\times 10^{-5}\;\text{mol}\cdot\text{L}^{-1}.

\text{pH} = \text{pK}_w + \log{[\text{OH}^{-}]} = 9.23.

7 0
3 years ago
Which represents the ionization of a strong electrolyte?
olga55 [171]
The answer is K3PO4(s) → 3K+(aq) + PO43–(aq) since water-soluble ionic tripotassium phosphate dissociates completely into K+ and PO43– ions when dissolved, that is, no K3PO4 remains in the solution. Carbonic acid H2CO3 and acetic acid CH3COOH are weak electrolytes since they are weak acids that do not completely ionize, while nonelectrolyte CH3OH do not dissociate into ions.
5 0
3 years ago
Read 2 more answers
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