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bija089 [108]
3 years ago
14

In pressure duntuwingfrost shattering​

Chemistry
1 answer:
Cerrena [4.2K]3 years ago
7 0

Answer:

i dont get it???????

Explanation:

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As the [h+] in a solution decreases, what happens to the [OH^-]?
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as the Hydrogen (acid) ion decreases there is a corresponding increase in the Hydroxide (basic) ion.

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Which two models would be most useful in predicting weather​
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A 20.0 mL sample of glycerol has a mass of 25.2 grams. What is the mass of a 57 mL sample of glycerol?
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Read 2 more answers
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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