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elixir [45]
3 years ago
12

Help please it’d be greatly appreciated lol

Chemistry
2 answers:
Reil [10]3 years ago
5 0

6. c: electrons

7. a : Na and Cu

omeli [17]3 years ago
3 0

6. c: electrons

7. a: Na and Cu

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Draw structural formulas for the alkoxide ion and the alkyl(aryl)bromide that may be used in a williamson synthesis of the ether
Juliette [100K]
Williamson synthesis is the most common way for obtaining ethers, called after its developer Alexander Williamson. It is an organic reaction of forming ethers from an organohalide and an alkoxide. The reaction is carried out according to the SN2 mechanism.

On the attached picture it is shown required alkoxide ion, <span>alkyl(aryl)bromide and the ether that forms from the reactants. </span>

3 0
3 years ago
The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism
MatroZZZ [7]

Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

<em>Where K' = K1 * K2</em>

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just <em>mechanism A and B are consistent with observed rate law</em>

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just <em>mechanism A is consistent with the observation of J. H. Sullivan</em>

5 0
3 years ago
The N2O4−NO2 reversible reaction is found to have the following equilibrium partial pressures at 100∘C. Calculate Kp for the rea
timofeeve [1]

Answer:

K_{p} for the reaction is 18.05

Explanation:

Equilibrium constant in terms of partial pressure (K_{p}) for this reaction can be written as-

                K_{p}=\frac{P_{NO_{2}}^{2}}{P_{N_{2}O_{4}}}

where P_{NO_{2}} and P_{N_{2}O_{4}} are equilibrium partial pressure of NO_{2} and N_{2}O_{4} respectively

Hence K_{p}=\frac{(0.095)^{2}}{(0.0005)} = 18.05

So, K_{p} for the reaction is 18.05

3 0
3 years ago
Which are balanced and which are unbalanced?
erica [24]

A. is balanced

B. is not balanced

C. is not balanced

3 0
3 years ago
What happens to an object's temperature if the kinetic energy of its particles decrease
ser-zykov [4K]
The temperature decreases
8 0
3 years ago
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