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xeze [42]
3 years ago
11

The rate constant of an SN1 reaction depends on the nucleophile b. The rate constant of an SN2 reaction does not depend on the n

ucleophile c. SN1 reactions proceed via carbocation intermediates d. The SN2 mechanism does not involve an intermediate
Chemistry
1 answer:
Sergio039 [100]3 years ago
7 0

Answer:

The rate constant of an SN1 reaction depends on the nucleophile

The rate constant of an SN2 reaction does not depend on the nucleophile

Explanation:

Let us recall that in an SN1 reaction, the rate determining step involves only the alkyl halide substrate and not the nucleophile. Hence;

Rate = k[RX]

Therefore;

k= Rate/[RX]

For an SN2 reaction, the rate determining step involves both the nucleophile and the alkyl halide substrate.

Hence;

Rate = k[Nu-] [RX]

k= Rate/[Nu-] [RX]

Note that;

[Nu-] = concentration of the nucleophile

[RX] =concentration of alkyl halide substrate

k= rate constant

We can see from the above derivations that;

1) The rate constant of an SN1 reaction does not depend on the nucleophile

2) The rate constant of an SN2 reaction depends on the nucleophile

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Answer:

ane, al, keto

ol, al, keto

ol, al, one

ol, ane, one.

Explanation:

The suffix –ol is used in organic chemistry principally to form names of organic compounds containing the hydroxyl (–OH) group, mainly alcohols (also phenol). The suffix was extracted from the word alcohol. The suffix also appears in some trivial names with reference to oils (from Latin oleum, oil).

Functional group is a ketone, therefore suffix = -one

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2 years ago
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A chromate ion consist of four oxygen atom bonded to be a chromium atom. It has two extra electrons. the formula for this
natali 33 [55]

Answer:

Cr04^2-

Explanation:

 

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3 years ago
What is another common name for the lanthanoids on the periodic table?
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Lanthanide Lanthanoid, also called Lanthanide
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What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C
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Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1= initial pressure at 26.5^oC = ?

P_2 = final pressure at -5.1^oC = 100 mm Hg

= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol

R = gas constant = 8.314 J/mole.K

T_1= initial temperature = 26.5^oC=273+26.5=299.5K

T_2 = final temperature =-5.1^oC=273+(-5.1)=267.9K

Now put all the given values in this formula, we get

\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]

\log  (\frac{P_1}{100})=-0.576

\frac{P_1}{100}=0.265

P_1=26.5mmHg

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3 years ago
Is this statement true or false?
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