A ________ has wavy or wrinkled foliations. These rocks usually have a pseudo-metallic luster from the abundance of mica.

C2H5OH(aq) + MnO− 4 (aq) → Mn2+(aq) + CH3COOH(aq) of acetic acid from ethanol by the action of permanganate ion in acidic soluti

Tems11 [2777]

Response:

Ethanol serves as a reducing agent. $(C_2H_5OH)$

The lowest integer coefficient for $MnO_4^-$ in the balanced equation is 4

Clarification:

A redox reaction, or oxidation-reduction reaction, is defined as one in which oxidation and reduction processes occur simultaneously.

Oxidation is the process whereby a substance loses electrons, resulting in an increase in oxidation state. In contrast, reduction entails a substance gaining electrons, leading to a decrease in oxidation state.

Reducing agents enable other substances to undergo reduction while becoming oxidized themselves.

Conversely, oxidizing agents facilitate other substances' oxidation while being reduced in the process.

For the chemical reaction in question, the half-reaction would be:

Oxidation:

Reduction:

To balance the oxygen atoms on both sides

$C_2H_5OH(aq)+MnO_4^-(aq)\rightarrow CH_3COOH(aq)+Mn^{2+}(aq)$

For oxidation:

For reduction:

$C_2H_6O\rightarrow C_2H_4O_2$

$MnO_4^-\rightarrow Mn^{2+}$ Then balance the hydrogen atoms on both sides.

For oxidation:

For reduction:

$C_2H_6O+H_2O\rightarrow C_2H_4O_2$

Lastly, balance the charge. $MnO_4^-\rightarrow Mn^{2+}+4H_2O$

For oxidation: For reduction:

$C_2H_6O+H_2O\rightarrow C_2H_4O_2+4H^+$

To balance the electrons, multiply the oxidation reaction by 5 and the reduction by 4 before combining both equations for a balanced redox reaction.

$MnO_4^-+8H^+\rightarrow Mn^{2+}+4H_2O$ Oxidation:

Reduction:

The finalized balanced chemical equation in acidic conditions will be:

$C_2H_6O+H_2O\rightarrow C_2H_4O_2+4H^++4e^-$

$MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O$

In this redox process, ethanol acts as the reducing agent while the permanganate ion serves as the oxidizing agent.

4 0

2 months ago

Given two half reactions as follows: A2+ → 2 A2+ + 3 e− 4 e− + B → B4− What would you multiply each half-reaction by, to cancel

Tems11 [2777]

To achieve the cancellation of electrons, the oxidation half-reaction needs to be multiplied by 4 while the reduction half-reaction must be multiplied by 3. Explanation: The oxidation reaction accounts for the loss of electrons, increasing the oxidation state, while the reduction implies gaining electrons, leading to a decrease in oxidation state. The respective half-reactions illustrate this, confirming that multiplying the oxidation by 4 and the reduction by 3 achieves the desired effect.

5 0

3 months ago

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2

eduard [2782]

Answer:- 64015 J

Solution: The calorimeter contains 4250 mL of water, which is at a temperature of 22.55 degrees Celsius.

The water's density is 1 gram per mL.

Thus, the mass of water = $4250mL(\frac{1g}{1mL})$ = 4250 grams.

After introducing the hot copper bar, the final temperature of the water reaches 26.15 degrees Celsius.

Thus, $\Delta T$ for the water = 26.15 - 22.55 = 3.60 degrees Celsius.

The specific heat capacity of water is 4.184 $\frac{J}{g.^0C}$ .

To determine the heat absorbed by the water, we can use the following formula:

$q=mc\Delta T$

where q represents heat energy, m refers to mass, and c indicates specific heat.

Now let's substitute the values into the equation to perform the calculations:

$q=4250g*\frac{4.184J}{g.^0C}*3.60^0C$

q = 64015 J

Therefore, the water absorbs 64015 J of heat.

5 0

3 months ago

An alkene with the molecular formula C8H16 undergoes ozonolysis to yield a mixture of (CH3)2C=O and (CH3)3CCHO. The alkene is:

eduard [2782]

The ozonolysis of 2,4,4-trimethyl-2-pentene produces a combination of $(CH_{3})_{2}C=O$ and $(CH_{3})_{3}CHO$

Explanation: In ozonolysis (where a reducing agent like Zn is involved during hydrolysis), a pi bond cleaves to generate ketones or aldehydes.

Ketones arise from the double bond's disubstituted side, whereas aldehydes come from the monosubstituted side of the same bond.

Notably, ozonolysis comprises two steps: (1) the formation of an ozonide, followed by (2) the hydrolysis of the ozonide.

Hydrolysis can transpire with or without a reducing agent. When it occurs without a reducing agent, carboxylic acid, carbon dioxide, or ketones can be produced.

In this case, 2,4,4-trimethyl-2-pentene yields a mixture of $(CH_{3})_{2}C=O$ and $(CH_{3})_{3}CHO$

The reaction process is illustrated below.

8 0

3 months ago