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Marina CMI [18]
3 years ago
7

Electrophilic aromatic substitution of furan favors addition to the 2-position because the intermediate cation is more stable th

an addition to the 3-position. Draw the mechanism arrows for the Friedel-Crafts acylation of furan. In steps 1-3 draw the curved arrows to show the formation of the acyliumion electrophile and acylation of furan. In steps 4 and 5 draw all resonance structures for the intermediate cation, then showing reaction of the last resonance structure with FeCl4 to generate the final product. Make sure to draw lone pairs of electrons for all species and nonzero formal charges.
Chemistry
1 answer:
alex41 [277]3 years ago
4 0

Answer:

See picture below

Explanation:

In the attached picture, you have the mechanism of all this reaction.

In the first step, as this is Friedel Craft acylation, we need to use either AlCl₃ or FeCl₃. So, this compound (I will use FeCl₃) will attach to the Chlorine group of the CH₃COCl.

In the second step, the whole CH₃CO is attached to the furan, because of the attack of the double bond of the ring.

After this, it occurs several resonance structure, to give a more stable structure, and the Hydrogen in the 2 position is substracted to give the final product of acylation.

Hope this helps.

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B. Heat is a form of energy so boiling it would increase energy. (I guess)
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(Brainliest) Please Help! Please do not give me a random, gibberish answer or I will report you and give you a low rating theref
nataly862011 [7]

Answer:

3.10g

Explanation:

Please see the attached picture for the full solution.

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Se construye una pila galvánica con una barra de cobre sumergida en una disolución 1M de cationes Fe+2 y una barra de plata sume
pashok25 [27]

Answer:

El potencial celular estándar, E_{cell} is +0.46 V

Explanation:

Las reacciones de media célula son;

Media reacción del ánodo Cu²⁺ + 2e⁻ ↔ Cu, E ° = 0.34 V

Media reacción catódica 2Ag + 2e⁻ ⁻ 2Ag, E ° = 0.80 V

Sin embargo tenemos para hierro Fe²⁺ + 2e⁻ ↔ Fe, E ° -0.44 V

y Fe³⁺ + e⁻ ↔ Fe²⁺, E ° = 0.77 V

 que es más alta que la del cobre presente, por lo tanto, el cobre se oxidará en el ánodo

Por lo tanto, en el ánodo, tendremos

Cu → Cu²⁺ + 2e⁻ (E ° = -0.34 V)

En el cátodo

2Ag + 2e⁻ → 2Ag (E ° = 0.80 V)

E_{cell} = E_c + E_a = -0.34 + 0.8 = +0.46 \, V

El potencial celular estándar, E_{cell} = +0.46 V

4 0
3 years ago
HELP PLEASE I HAVE A TEST TODAY AND I DON'T UNDERSTAND ANY OF THIS...
myrzilka [38]

Answer:

About 67 grams or 67.39 grams

Explanation:

First you would have to remember a few things:

 enthalpy to melt ice is called enthalpy of fusion.  this value is 6.02kJ/mol

  of ice  

 it takes 4.18 joules to raise 1 gram of liquid water 1 degree C

 water boils at 100 degrees C and water melts above 0 degrees C

 1 kilojoules is 1000 joules

  water's enthalpy of vaporization (steam) is 40.68 kJ/mol

  a mole of water is 18.02 grams

  we also have to assume the ice is at 0 degrees C

Step 1

Now start with your ice.  The enthalpy of fusion for ice is calculated with this formula:

q = n x ΔH    q= energy, n = moles of water, ΔH=enthalpy of fusion

Calculate how many moles of ice you have:

150g x (1 mol / 18.02 g) = 8.32 moles

Put that into the equation:

q = 8.32 mol x 6.02 = 50.09 kJ of energy to melt 150g of ice

Step 2

To raise 1 gram of water to the boiling point, it would take 4.18 joules times 100 (degrees C)  or 418 joules.

So if it takes 418 joules for just 1 gram of water, it would take 150 times that amount to raise 150g to 100 degrees C.  418 x 150 = 62,700 joules or 62.7 kilojoules.

So far you have already used 50.09 kJ to melt the ice and another 62.7 kJ to bring the water to boiling.  That's a total of 112.79 kJ.

Step 3

The final step is to see how much energy is left to vaporize the water.

Subtract the energy you used so far from what you were told you have.

265 kJ - 112.79 kJ = 152.21 kJ

Again q = mol x ΔH (vaporization)

You know you only have 152.21 kJ left so find out how many moles that will vaporize.

152.21 kJ = mol x 40.68  or   mol = 152.21 / 40.68  = 3.74 moles

This tells you that you have vaporized 3.74 moles with the energy you have left.

Convert that back to grams.

3.74 mol   x  ( 18.02 g / 1 mol ) = 67.39 grams

5 0
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