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Serhud
2 months ago
5

If the standard free energy change for the conversion of fructopyranose to fructofuranose is 1.7 kJ/mol, what fraction of the to

tal fructose in solution is fructopyranose?
Chemistry
1 answer:
alisha [2.9K]2 months ago
7 0
The connection between Gibb's free energy and temperature can be described as follows. The value for \Delta G is specified as 1.7 kJ/mol. Additionally, we know that k = \frac{[product]}{[substrate]}

= \frac{\text{[fructofuranose]}}{\text{[fructopyranose]}}. Since k has a value of 0.50357 at a temperature of 298 K, we can conclude that

\frac{\text{[fructofuranose]}}{\text{[fructopyranose]}} = 0.50357, leading to \frac{\text{[fructofuranose]}}{\text{[fructopyranose]}} + 1 = 1.50357, which equals \frac{\text{[total fructose solution]}}{\text{[fructopyranose]}} \frac{1}{1.50357} = 0.665. Therefore, we can deduce that 0.665 of the total fructose in the solution exists as fructopyranose.

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Given the connection between Aw and K (Aw=2k) could you use the ideal gas law and derive the Boltzmann constant. Water freezes a
lions [2927]

Answer:

Explanation:

The relationship between the new temperature scale and the absolute temperature scale is defined as follows

Aw = 2 K

for K = 273.15 (the freezing point of water on the absolute scale)

Aw = 2 x 273.15 = 546.3 K

Each division of the new scale is equivalent to half that of each division on the absolute scale

each division of the new scale is minimal.

The value of R = 8.314 J per mole per K

Here, per K corresponds to 2Aw

Hence, the value of R in the new scale = 8.314/2 J per mole per Aw

= 4.157 J per mole per Aw

k = R / N

= 4.157 / 6.02 x 10²³

= .69 x 10⁻²³

= 6.9 x 10⁻²⁴ J per molecule per Aw .

7 0
4 months ago
Read 2 more answers
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_2OOxidation:

  • 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
The volume of a gas at 6.0 atm is 2.5 L. What is the volume of the gas at 7.5 atm at the same temperature?
castortr0y [3046]

Greetings!

The result is:

The new volume is: 2L

Rationale:

Because the temperature remains constant, we can apply Boyle's Law to solve this issue.

Boyle's Law stipulates that:

P_{1}V_{1}=P_{2}V_{2}

Where,

P is the gas's pressure.

V is the gas's volume.

According to the information provided:

V_{1}=2.5L\\P_{1}=6.0atm\\P_{2}=7.5atm

Let's put the values into the equation:

2.5L*6.0atm=7.5atm*V_{2}

2.5L*6.0atm=7.5atm*V_{2}\\\\V_{2}=\frac{2.5L*6.0atm}{7.5atm}=\frac{15L.atm}{7.5atm}=2L

Consequently, the new volume is: 2L

Wishing you a lovely day!

7 0
3 months ago
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