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3 years ago

HELP PLEASE Chemical name for As2S3

Chemistry

2 answers:

sattari [20]3 years ago

7 0

Answer:

Arsenic Sulfide

Explanation:

irina1246 [14]3 years ago

4 0

Answer:

Chemical name for As2S3 is Arsenic sulfide

Explanation:

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Z is a covalent substance. In an experiment, a sample of pure solid Z was continually heated for 11 minutes

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When a pure solid Z sample it a covalent compound is heated continually for 11 minutes it will undergo disintegration leaving the bonds intact.

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The anions formed from the atoms of the elements in family VIA should carry a +6 charge.

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The answer would be true

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4 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!

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3 years ago

50 pts

ki77a [65]

Answer:

The paper no longer has smooth edges

The paper has not produced gas when torn

The paper has not changed colours

The paper is still paper with the same chemical makeup

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3 years ago

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Answer:

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Explanation:

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