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Bad White [126]
2 years ago
14

What change would you expect on the rate of the SN2 reaction of 1-iodo-2-methylbutane with cyanide ion if the nucleophile concen

tration is halved and the alkyl halide concentration is doubled?
a. divided by 4
b. halved
c. no effect
d. increased 2x
e. increased 3x
f. increased 6x
g. increased 9x
Chemistry
1 answer:
n200080 [17]2 years ago
6 0

Answer:

no effect

Explanation:

In an SN2 reaction, the mechanism is bi-molecular and first order in both the alkyl halide and the nucleophile.

Hence the rate of reaction is;

Rate = k [1-iodo-2-methylbutane] [cyanide ion]

Given that it is a bimolecular reaction, if we double the concentration of 1-iodo-2-methylbutane and the concentration of the cyanide ion is halved, the rate of reaction remains the same.

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29. What is E for a system which has the following two steps:
jasenka [17]

Answer:

Zero

Explanation:

Recall that;

E = q + w

Where;

q = heat, w = work done

When heat is absorbed by the system q is positive

When heat is evolved by the system q is negative

When the system does work, w is negative

When work is done on the system w is positive

Step 1

ΔE1= 60 KJ + 40 KJ = 100KJ

Step 2

ΔE2= (-30 KJ) + (-70 KJ) = (-100) KJ

ΔE1 + ΔE2= 100KJ + (-100) KJ = 0KJ

4 0
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siniylev [52]

Answer: Yes! you're all good. Alkali metals in group 1 are the most metallic :)

6 0
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motikmotik

Answer:

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Explanation:

Zn + HCl → ZnCl2 +

The complete equation is given below:

Zn+ HCl → ZnCl2 + H2

Now we can balance the equation by doing the following:

There are 2 atoms of Cl and 2 atoms of H on the left. This can be balanced by putting 2 in front of HCl as shown below:

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