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Alexeev081 [22]
3 years ago
6

Write a balanced chemical equation for the reaction that occurs when heptane, C7H16(l) , burns in air.

Chemistry
1 answer:
Pavlova-9 [17]3 years ago
8 0
Start with Unbalanced Equation and balance it, so...
C7H16+O2--->CO2+H2O
There are 7 C atoms on the left-hand side, so we need 7 C atoms on the right-hand side. Add a 7 in front of the CO2...7CO2+H2O on right side now.
We have fixed 16 H atoms on the left-hand side, so we need 16 H atoms on the right-hand side. Add an 8 in front of H2O to make 16 (8x2)...7CO2+8H2O on right side now.
There are 22 O atoms on the right-hand side: 14 from the CO2 and 8 from the H2O. Add an 11 in front of the O2 on the left side to make 22 (11x2).
Every formula now has a fixed coefficient. You should have a balanced equation of...
C7H16+11O2--->7CO2+8H2O
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Consider the addition of NaCl to the ice-water mix in Part II. Which of the following statements is TRUE?
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6 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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