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2 years ago

7

The reduction of ketone groups, such as those present on coq, produces alcohols. in contrast, oxidation of aldehydes produces wh

ich functional group?

1 answer:

nevsk [136]2 years ago

7 0

On oxidation of aldehydes produces carboxylic acid functional group.

The product of oxidation of aldehydes depends upon whether the reaction occur in acidic medium or alkaline condition.

If oxidation of aldehydes occurs under acidic condition the product is carboxylic acid but if oxidation of aldehydes occurs under alkaline condition then reduce as well as oxidized product obtained which is known as disproportional product.

The oxidation of aldehydes occur through potassium dichromate, potassium permanganate or many more. The oxidation of aldehydes in the presence of base is known as cannizzaro's reaction.

learn about oxidation

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Calculate AHræn for the following reaction

HACTEHA [7]

Answer:

60ax 70by 67ay ah251

Explanation:

8 0

3 years ago

A chemist titrates 190.0 mL of a 0.8125 M ammonia (NH) solution with 0.3733 M HCl solution at 25 °C. Calculate the pH at equival

stealth61 [152]

Answer:

Approximately $4.92$ .

Explanation:

Initial volume of the solution: $V = 190.0\; \rm mL = 0.1900\; \rm L$ .

Initial quantity of $\rm NH_3$ :

$\begin{aligned} n({\rm NH_3}) &= c({\rm NH_3}) \cdot V({\rm NH_3}) \\ &= 0.3733\; \rm mol \cdot L^{-1} \times 0.1900\; \rm L \\ &\approx 0.154375\; \rm mol\end{aligned}$ .

Ammonia $\rm NH_3$ reacts with hydrochloric $\rm HCl$ acid at a one-to-one ratio:

$\rm NH_3 + HCl \to NH_4 Cl$ .

Hence, approximately $n({\rm HCl}) = 0.154375\; \rm mol$ of $\rm HCl\!$ molecules would be required to exactly react with the $\rm NH_3\!$ in the original solution and hence reach the equivalence point of this titration.

Calculate the volume of that $0.3733\; \rm mol \cdot L^{-1}$ $\rm HCl$ solution required for reaching the equivalence point of this titration:

$\begin{aligned}V({\rm HCl}) &= \frac{n({\rm HCl})}{c({\rm HCl})} \\ &\approx \frac{0.154375\; \rm mol}{0.3733\; \rm mol \cdot L^{-1}} \approx 0.413541\; \rm L\end{aligned}$ .

Hence, by the assumption stated in the question, the volume of the solution at the equivalence point would be approximately $0.413541\; \rm L + 0.1900\; \rm L \approx 0.6035\; \rm L$ .

If no hydrolysis took place, $0.154375\; \rm mol$ of $\rm NH_4 Cl$ would be produced. Because $\rm NH_4 Cl\!$ is a soluble salt, the solution would contain $0.154375\; \rm mol\!$ of $\rm {NH_4}^{+}$ ions. The concentration of $\rm {NH_4}^{+}\!$ would be approximately:

$\begin{aligned}c({\rm {NH_4}^{+}}) &= \frac{n({\rm {NH_4}^{+}})}{V({\rm {NH_4}^{+}})}\\ &\approx \frac{0.154375\; \rm mol}{0.6035\; \rm L} \approx 0.255782\; \rm mol \cdot L^{-1}\end{aligned}$ .

However, because $\rm NH_3 \cdot H_2O$ is a weak base, its conjugate $\rm {NH_4}^{+}$ would be a weak base.

$\begin{aligned}pK_{\rm a}({{\rm NH_4}}^{+}) &= pK_{\rm w} - pK_{\rm b}({\rm NH_3})\\ &\approx 13.99 - 4.75 = 9.25\end{aligned}$ .

Hence, the following reversible reaction would be take place in the solution at the equivalence point:

$\rm {NH_4}^{+} \rightleftharpoons NH_3 + H^{+}$ .

Let $x\; \rm mol \cdot L^{-1}$ be the increase in the concentration of $\rm H^{+}$ in this solution because of this reversible reaction. (Notice that $x \ge 0$ .) Construct the following $\text{RICE}$ table:

$\begin{array}{c|ccccc} \textbf{R}& \rm {\rm NH_4}^{+} & \rightleftharpoons & {\rm NH_3}& + & {\rm H}^{+}\\ \textbf{I} & 0.255782 \; \rm M \\ \textbf{C} & -x \;\rm M & & + x\;\rm M & & + x\; \rm M \\ \textbf{E} & (0.255782 - x)\; \rm M & & x\; \rm M & & x\; \rm M\end{array}$ .

Thus, at equilibrium:

Concentration of the weak acid: $[{\rm {NH_4}^{+}}] \approx (0.255782 - x) \; \rm M$ .
Concentration of the conjugate of the weak acid: $[{\rm NH_3}] = x\; \rm M$ .
Concentration of $\rm H^{+}$ : $[{\rm {H}^{+}}] \approx x\; \rm M$ .

$\displaystyle \frac{[{\rm NH_3}] \cdot [{\rm H^{+}}]}{[{ \rm {NH_4}^{+}}]} = 10^{pK_\text{a}({\rm {NH_4}^{+}})}$ .

$\displaystyle \frac{x^2}{0.255782 - x} \approx 10^{-9.25}$

Solve for $x$ . (Notice that the value of $x\!$ is likely to be much smaller than $0.255782$ . Hence, the denominator on the left-hand side $(0.255782 - x) \approx 0.255782$ .)

$x \approx 1.19929 \times 10^{-5}$ .

Hence, the concentration of $\rm H^{+}$ at the equivalence point of this titration would be approximately $1.19929 \times 10^{-5}\; \rm M$ .

Hence, the $pH$ at the equivalence point of this titration would be:

$\begin{aligned}pH &= -\log_{10}[{\rm {H}^{+}}] \\ &\approx -\log_{10} \left(1.19929 \times 10^{-5}\right) \approx 4.92\end{aligned}$ .

5 0

3 years ago

‼️SHOW ALL STEPS‼️ If 14.4 grams of iron reacts with hydrochloric acid, how many grams of hydrogen gas will be produced?

Alex Ar [27]

Answer: 0.771 g of $H_2$ will be produced

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Fe=\frac{14.4g}{56g/mol}=0.257moles$

The balanced chemical reaction is:

$2Fe(s)+6HCl(aq)\rightarrow 2FeCl_3(aq)+3H_2(g)$

According to stoichiometry :

2 moles of $Fe$ produce = 3 moles of $H_2$

Thus 0.257 moles of $Fe$ will produce= $\frac{3}{2}\times 0.257=0.385moles$ of $H_2$

Mass of $H_2=moles\times {\text {Molar mass}}=0.385moles\times 2g/mol=0.771g$

Thus 0.771 g of $H_2$ will be produced

3 0

3 years ago

After distilling your crude methyl benzoate, you set aside 4.83 grams of the purified ester. You then prepare the grignard reage

Ber [7]

Answer:

95.6 %

Explanation:

For this question, we will have <u>2 reactions</u>, the formation of the <u>grignard reagent</u> and the <u>formation of the alcohol</u>. The first step then is the calculation of the <u>maximum amount</u> of the grignard reagent. For this, we have to convert the grams to moles and check the smallest value. To do this we have to take into account the <u>following conversion ratios</u>:

Molar mass of Mg = 24 g/mol

Molar mass of phenylmagnesium bromide ( $C_6H_5Br$ )= 157 g/mol

Density of bromobenzene= 1.5 g/mL

Molar ratio between Mg and $C_6H_5Br$ = 1:1

$2.3~g~Mg\frac{1~mol~Mg}{24~g~Mg}=0.096~mol~Mg$

$9.45~mL~\frac{1g}{1.5mL}\frac{1~mol~C_6H_5Br}{157~g}=0.0402~mol~C_6H_5Br$

The smallest value is the mol of bromobenzene therefore <u>0.0402 mol</u> of phenylmagnesium bromide would be produced.

The next step is repite the same steps for the reaction of <u>formation of the alcohol</u>. Therefore we have to find the moles of methyl benzoate, so:

Molar mass of methyl benzoate: 136.14 g/mol

$4.83~g~\frac{1~mol}{136.14~g}=0.35~mol$

The we have to <u>divide by the coefficient</u> of each reactive in the balance reaction. So:

$\frac{0.35~mol}{1}=0.35$

$\frac{0.0402~mol}{2}=0.0201$

Therefore the <u>limiting reagent</u> would be the phenylmagnesium bromide. Now, the <u>molar ratio</u> between the phenylmagnesium bromide and triphenyl carbinol is <u>2:1</u>, so the amount of alcohol produced is 0.0201 mol triphenyl carbinol. The next step is the conversion from mol to <u>grams of triphenyl carbinol</u>:

Molar mass of triphenyl carbinol= 260.33 g/mol

$0.0201~mol\frac{260.33~g}{1~mol}=5.23~g~triphenyl carbinol$

Finally, we have to <u>divide</u> the obtanied solid by the calculated one:

$Percentage=\frac{5}{5.23}*100=95.6\%$

5 0

3 years ago

What will be the end product for each electrode after electrolysis if the solution is concentrated aqueous sodium chloride ?

Eddi Din [679]

Answer:

Cathode: Hydrogen gas

Anode: Chlorine gas

Explanation:

① Write down the ions present in the electrolyte

Cations: Na⁺, H⁺

Anions: Cl⁻ , OH⁻

② Decide which ions are preferentially discharged.

These are the factors:

For the discharge of cations,

- Reactivity series

(The lower the position of the cation in the reactivity series, the easier it is to be discharged)

For the discharge of anions,

- Concentration effect (look at this first)

(The more concentrated the ion, the easier for it to be discharged)

- If solution is not concentrated (dilute), look at the position of the anion on the electrochemical series.

(The lower the position of the anion on the electrochemical series, the easier of it to be discharged)

In this case:

For cations, H⁺ ions are selectively discharged at the cathode as its position is lower than Na⁺ in the reactivity series.

Anion: Cl⁻ ions, being more concentrated, are selectively discharged at the anode.

☆For electrolysis,

Cation at the cathode (-ve terminal)

Anion at the anode (+ve terminal)

In summary, here's what happened at each electrode:

<u>C</u><u>athode</u>

- H⁺ selectively discharged

- ionic half equation: 2H⁺ (aq) +2e⁻ → H₂ (g)

<u>Anode</u>

- Cl⁻ ions selectively discharged

- ionic half equation: 2Cl⁻ (aq) → Cl₂ (g) + 2e⁻

3 0

3 years ago