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2 years ago

In a reaction where A + B AB, what does an equilibrium position of A= 14%, B = 14%, and AB= 72% indicate about the reaction?

Chemistry

2 answers:

MissTica2 years ago

5 0

The answer is c I just answered mine .

fomenos2 years ago

3 0

A.) Because A and B are equal, the reaction has reached equilibrium.
B.) There is more A and B than AB at equilibrium, and the reaction favors the reactants.
C.) There is more AB than A and B at equilibrium, and the reaction favors the products.
D.) There is more AB than A and B at equilibrium, and the reaction favors the reactants.
E.) The reaction is not at equilibrium.
These are the answer choices pls help

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OverLord2011 [107]

Answer:

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3 years ago

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Katen [24]

Answer:

Invasive species are an organism that causes ecological or economic harm in a new environment where it's not native.

Explanation:

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3 years ago

For a certain experiment, a student requires 100 milliliters of a solution that is 8% HCl (hydrochloric acid). The storeroom has

exis [7]

Answer:

70 mL of 5% HCl and 30 mL of 15% HCl

Explanation:

We will designate x to be the fraction of the final solution that is composed of 5% HCl, and y to be the fraction of the final solution that is composed of 15% HCl. Since the percentage of the final solution is 8%, we can write the following expression:

5x + 15y = 8

Since x and y are fractions of a total, they must equal one:

x + y = 1

This is a system of two equations with two unknowns. We will proceed to solve for x. First, an expression for y is found:

y = 1 - x

This expression is substituted into the first equation and we solve for x.

5x + 15(1 - x) = 8

5x+ 15 - 15x = 8

-10x = -7

x = 7/10 = 0.7

We then calculate the value of y:

y = 1 - x = 1 - 0.7 = 0.3

Thus 0.7 of the 100 mL will be the 5% HCl solution, so the volume of 5% HCl we need is:

(100 mL)(0.7) = 70 mL

Similarly, the volume of 15% HCl we need is:

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3 years ago

Why pie bond are not perticipate in hybridaization

DedPeter [7]

The first bond between two atoms is always a sigma bond and the other bonds are always pi bonds and a hybridized orbital cannot be involved in a pi bond. Thus we need to leave one electron (in case of Carbon double bond) to let the Carbon have the second bond as a pi bond.

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2 years ago