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NARA [144]
2 years ago
12

When optically active (R)-2-methylcyclohexanone is treated with either aqueous base or acid, racemization occurs. Explain.

Chemistry
1 answer:
IrinaK [193]2 years ago
3 0

Answer:

See explanation

Explanation:

Racemization is said to occur when a 1:1 ratio of (+) and (-) enantiomers of a compound are produced in a reaction.

The reaction of optically active (R)-2-methylcyclohexanone with either aqueous base or acid leads to the formation of a planar enol species for reaction with acid and a planar enolate species for reaction with base.

Both reactions involves the formation of achiral species which reverts back to the chiral product with equal chances of the formation of both enantiomers of the product during the process. This leads to racemization of the product in both cases.

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aleksklad [387]
The gravitational pull
3 0
3 years ago
A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from
sweet-ann [11.9K]

Answer:

T_i~=163.1 ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

<u>Initial temperature of gold= Unknow</u>

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the <u>heat equation</u>:

Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT

Q_A_u=m_A_u*Cp_A_u*deltaT

We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT

Now we can <u>put the values into the equation</u>:

60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)

Now we can <u>solve for the initial temperature of gold</u>, so:

T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20

T_i~=163.1 ºC

I hope it helps!

5 0
3 years ago
Nitric acid is a strong acid, sodium hydroxide is a strong base, and sodium nitrate is a soluble salt. Which of the following is
Evgesh-ka [11]
Can’t help with this but good luck
4 0
3 years ago
Save the ________ used to extract metals.
MrRa [10]

Answer:

Save the *minerals* used to extract metals.

6 0
2 years ago
In the following reaction, how many grams of NaBr will react with 311 grams of Pb(NO3)2?
Mashutka [201]
Molar mass :

NaBr = 103 g/mol

Pb(NO3)2 = 331.20 g/mol

<span><span /><span>Balanced chemical equation :

</span></span>2 NaBr + 1 Pb(NO3)2 = 2 NaNO3 + 1 PbBr<span>2
</span><span>
2*103 g NaBr ------------> 1 * 331.20 g Pb(NO3)2
      g NaBr  -------------------> 311 g Pb(NO3)2

331.20  g  =   2*103*311

331.20 g = 64066

mass ( NaBr ) =  64066 / 331.20

mass ( naBr)  = 193,43 g of NaBr

hope this helps!.


</span>
5 0
3 years ago
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