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3 years ago

15

Como descargo el codm version china enlace porfavor xxxxddddddd

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pantera1 [17]3 years ago

5 0

Answer:

ano po yan? hind ko po maintindihan

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A sample of argon gas is at a pressure of 1.725x10^4 kPa and a temperature of 24°C in a rigid 20 L tank. How many moles of argon

Paraphin [41]

To find how many moles of the gas you can use ideal gas formula. Remember to change the temperature unit to Kelvin. Since the pressure using kpa, the constant used would be 8.314 kpa*L / mol*K
PV=nRT
n= PV/RT
n= 17,250 kpa * 20 / 8.314 * (24+273.15)K
n=139.64moles

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4 years ago

How many grams are in 3.4 moles of C2H5OH molecules? (use molar mass and round answer to the correct number of sig figs)

Salsk061 [2.6K]

Molar mads of C2H5OH:
12*2+1*5+16+1
24+5+17
46
no of moles = mass in grams/molar mass
3.4=m/46
3.4*46=m
m=156.4
m=156g

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3 years ago

A wave is known to have a frequency of 20 Hz, wavelength of 0.5 m, and amplitude of 1.2 m. If the amplitude of the wave is doubl

Komok [63]

The wave will have a frequency that is larger

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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

Which of the following is a direct result of hydrogen bonding?

prisoha [69]

Cohesion and adhesion

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3 years ago