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NARA [144]
3 years ago
12

1. How did the change of stress (adding or removing reactants or products) cause a shift in the equilibrium system of your solut

ions? Use data to support your answer. Make sure you discuss all four stress changes: a. Adding a reactant b. Adding a product c. Removing a reactant d. Removing a product
Chemistry
2 answers:
tatyana61 [14]3 years ago
8 0

Answer:

<em><u>When additional reactant is added, the equilibrium shifts to reduce this stress: it makes more product. When additional product is added, the equilibrium shifts to reactants to reduce the stress. 2. Concentration is just one stress that can affect the equilibrium position.</u></em>

Explanation:

<h2>HOPE IT WILL HELP YOU✌✌✌✌✌</h2>
Charra [1.4K]3 years ago
7 0

Answer:

If reactant or product is removed, the equilibrium shifts to make more reactant or product, respectively, to make up for the loss.

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Which statement best describes chemical properties of matter? Chemical properties, such as density, must be observed when a subs
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Answer:

Chemical properties, such as combustibility, are generally observed as the identity of a substance changes and one or more new substances form.

Explanation:

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It takes a slug 20 minutes to travel from the grass to the trash can, a trip of 15 meters. How far could the slug travel in 60 m
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45 meters

Explanation:

If it takes the slug 20 minutes to travel 15 meters, you multiply the 15 by three to get the distance for 60 minutes.

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What is the condition of the outside air at a certain time and place?
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How can the conductivity of a salt solution be decreased?
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Answer:

B. Add more water to decrease the concentration.

Explanation:

If we add more water to the solution, the concentration is decreased and like wise the conductivity of the solution.

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  • Adding more water increases the amount of solvent and the concentration reduces.
4 0
3 years ago
Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral molecules, but
alukav5142 [94]

Answer:

a) For nicotine, the protonated form is the present in stomach.

b) For caffeine, the neutral base form is the present in stomach.

c) For Strychinene, the protonated form is the present in stomach.

d) For quinine, the protonated form is the present in stomach.

Explanation:

In a basic dissociation for molecules with basic nitrogen, the equilibrium is:

A + H₂O ⇄ AH⁺ + OH⁻

<em>Where A is neutral base and AH⁺ is protonated form</em>

The basic dissociation constant, kb, is:

K_{b} = \frac{[AH^+][OH^-]}{[A]}

As pH in stomach is 2,5:

[OH] =10^{-[14-pH]}

[OH] = 3,16x10⁻¹² M

Thus:

\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]} <em>(1)</em>

Using (1) it is possible to know if you have the neutral base or the protonated form, thus:

(a) nicotine Kb = 7x10^-7

\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

\frac{7x10^{-7}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

221359 = \frac{[AH^+]}{[A]}

[AH⁺}>>>>[A]

For nicotine, the protonated form is the present in stomach

(b) caffeine,Kb= 4x10^-14

\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

\frac{4x10^{-14}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

0,0127 = \frac{[AH^+]}{[A]}

[AH⁺}<<<<[A]

For caffeine, the neutral base form is the present in stomach

(c) strychnine Kb= 1x10^-6

\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

\frac{1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

314456 = \frac{[AH^+]}{[A]}

[AH⁺}>>>>[A]

For Strychinene, the protonated form is the present in stomach

(d) quinine, Kb= 1.1x10^-6

\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

\frac{1,1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

347851= \frac{[AH^+]}{[A]}

[AH⁺}>>>>[A]

For quinine, the protonated form is the present in stomach.

I hope it helps!

7 0
3 years ago
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