How many CL are in a L

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Sergio [31]

Answer:

Bet texting you right now!

Explanation:

3 0

3 years ago

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Amphetamine (C9H13N)(C9H13N) is a weak base with a pKbpKb of 4.2. You may want to reference (Pages 710 - 713) Section 16.8 while

Soloha48 [4]

Answer:

pH = 10.38

Explanation:

C9H13N ↔ C9H20O3N+ + OH-

∴ molar mass C9H13N = 135.21 g/mol

∴ pKb = - log Kb = 4.2

⇒ Kb = 6.309 E-5 = [OH-][C9H20O3N+] / [C9H13N]

∴ <em>C</em> sln = (205 mg/L )*(g/1000 mg)*(mol/135.21 g) = 1.516 E-3 M

mass balance:

⇒ <em>C</em> sln = 1.516 E-3 = [C9H20O3N+] + [C9H13N]......(1)

charge balance:

⇒ [C9H20O3N+] + [H3O+] = [OH-]; [H3O+] is neglected, come from water

⇒ [C9H20O3N+] = [OH-].......(2)

(2) in (1):

⇒ [C9H13N] = 1.516 E-3 - [OH-]

replacing in Kb:

⇒ Kb = 6.3096 E-5 = [OH-]² / (1.516 E-3 - [OH-])

⇒ [OH-]² + 6.3096 E-5[OH] - 7.26613 E-8 = 0

⇒ [OH-] = 2.3985 E-4 M

∴ pOH = - Log [OH-]

⇒ pOH = 3.62

⇒ pH = 14 - pOH = 14 - 3.62 = 10.38

5 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

For a typical titration, 0. 010 m naoh is the titrant (in the buret). if the initial buret reading is 2. 45 ml, and the final bu

pickupchik [31]

16.25 ml NaOH was used for the titration.

<h3>Titration:</h3>

"The process of calculating the quantity of a material A by adding measured increments of substance B, the titrant, with which it reacts until exact chemical equivalency is obtained (the equivalence point)" is the definition of titration.

A titration is a method for figuring out the concentration of an unknown solution by using a solution with a known concentration. Until the reaction is finished, the titrant (the known solution) is typically added from a buret to a known volume of the analyte (the unknown solution).

0. 010 m NaOH is the titrant (in the buret).

The initial buret reading is 2. 45 ml

The final buret reading is 18. 70 ml

The volume of NaOH = 18.70ml - 2.45 ml

= 16.25 ml

Therefore, 16.25 ml NaOH was used for the titration.

Learn more about titration here:

brainly.com/question/27107265

#SPJ4

5 0

1 year ago

If 7.84 x 10^7 J of energy is released from a fusion reaction, what amount of mass in kilograms would be lost? Recall that c = 3

docker41 [41]

Use the formula
E=mc^2

plug the values,

The amount of mass in kilograms would be lost is 8.71 x 10^-10

7 0

4 years ago

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