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3 years ago

What is the rate of a reaction if the value of kis 0.1, [A] is 1 M, and [B] is 2 M?

Chemistry

1 answer:

mel-nik [20]3 years ago

8 0

Answer:

D. 0.4 (mol/L)/S

Explanation:

You simply have to plug in the given values into the rate law.

Rate = k[A][B]

Rate = (0.1)(1)²(2)²

Rate = (0.1)(1)²(4)²

Rate = 0.4

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.Find the pH if the [H3O+1] is 6.2  10-10 M

pishuonlain [190]

Answer:

The pH is 9,21.

Explanation:

The pH indicates the acidity or basicity of a substance. PH values between 0 and less than 7 indicate acidic solutions, 7 neutral and higher than 7 to 14 basic. It is calculated as

pH = -log (H 30+)

pH= -log (6,2 x 10^-10)

pH=9,21

3 0

2 years ago

In this experiment, you were asked to calculate your percent yield of alum assuming that you started with pure aluminum. Actuall

neonofarm [45]

Answer:

Percent yield of alum is 38.6%

Explanation:

Find attached calculations for detailed explanation.

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3 years ago

Draw the structural formula of the major product of the reaction of (S)-2,2,3-trimethyloxirane with MeOH, H . Use the wedge/hash

Katarina [22]

Answer:

(S)-3-methoxy-3-methylbutan-2-ol

Explanation:

In this case, we have an epoxide opening in acid medium. The first step then is the protonation of the oxygen. Then the epoxide is broken to generate the most stable carbocation. The nucleophile ( $CH_3OH$ ) will attack the carbocation generating a new bond. Finally, the oxygen is deprotonated to obtain an ether functional group and we will obtain the molecule (S)-3-methoxy-3-methylbutan-2-ol.

See figure 1

I hope it helps!

8 0

3 years ago

Which best describes the trends in electronegativity on the periodic table?

Wittaler [7]

Answer:

My bad i didnt mean to put that carry on.

Explanation:

7 0

3 years ago

Read 2 more answers

Draw the most stable resonance structure for the intermediate in the electrophilic aromatic bromination of aniline, anisole, and

ASHA 777 [7]

Answer:

Here's what I get

Explanation:

(a) Intermediates

The three structures below represent one contributor to the resonance-stabilized intermediate, in which the lone pair electrons on the heteroatom are participating (the + charge on the heteroatoms do not show up very well).

(b) Relative Stabilities

The relative stabilities decrease in the order shown.

N is more basic than O, so NH₂ is the best electron donating group (EDG) and will best stabilize the positive charge in the ring. However, the lone pair electrons on the N in acetanilide are also involved in resonance with the carbonyl group, so they are not as available for stabilization of the ring.

(c) Relative reactivities

The relative reactivities would be

C₆H₅-NH₂ > C₆H₅-OCH₃ > C₆H₅-NHCOCH₃

4 0

2 years ago