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Marrrta [24]
3 years ago
11

According to Le Châtelier's principle, how will an increase in concentration of a reactant affect the equilibrium system?

Chemistry
2 answers:
beks73 [17]3 years ago
8 0

Answer:

The answer is Shift it toward the products

Explanation:

In a system that is in equilibrium, an increase in the concentration of a substance will cause the system to move so that the substance that was added is partially used. On the other hand, if the concentration of a substance in a system that is in equilibrium is decreased, it will cause the system to move in the sense that it partially replaces the substance that was removed.

In the case of exercise, if the concentration of a reagent is increased, the direct sense of the reaction, that is, towards the products, will be favored.

Svetach [21]3 years ago
3 0

Answer:

Shift it toward the products

Explanation:

Imagine this equilibrium:

2A  +  B   ⇄  A₂B

Let's think the expression for Kc:

Kc = [A₂B] / [A]² . [B]

Now, we have this situation:

2A ↑  +  B   ⇄  A₂B

As we are not in equilibrium, we have to think that Kc would decrease.

↓Kc = [A₂B] / [A]² ↑ . [B]

so we have to use Qc (reaction quotient). For this case

Qc < Kc, so the reaction → Product side

We have to make more product to counteract the reagent increase and thus return the system to equilibriumr

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motikmotik

Answer:

The resultant molarity is 0,7 M.

Explanation:

A dilution consists of the decrease of concentration of a substance in a solution (the higher the volume of the solvent, the lower the concentration).

We use the formula for dilutions:

C1 x V1 = C2 x V2

3,5 M x 20 ml= C2 x 100ml

C2= (3,5 M x 20 ml)/100ml

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3 years ago
A compound is 52.14% C, 13.13% H, and 34.73% O. What is the empirical formula of the compound?
alina1380 [7]
% H = 100 - ( 52.14 + 34.73 )=13.13 % 

<span>assume 100 g of this compound </span>
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<span>moles O = 34.73 g/ 15.999 g/mol=2 </span>

<span>the empirical formula is C4H13O2</span>
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