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3 years ago

You are on an alien planet where the names for substances and the units of measures are very unfamiliar.

Chemistry

1 answer:

enyata [817]3 years ago

8 0

Given that

1 skvarnick = 45 quibs

3 quibs = 7 sleps

Now if we have 45 quibs it means we have you have one skvarnick

three quibs = 7 sleps

so one quib = 7/3 sleps

so 45 quibs = 7 X 45 / 3 = 105 sleps

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3 years ago

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3 years ago

Suppose a 500.mL flask is filled with 1.9mol of NO3 and 1.6mol of NO. The following reaction becomes possible: NO3gNOg 2NO2g The

never [62]

There is an error in the first sentence of the question; the right format is:

Suppose a 500.mL flask is filled with 1.9mol of NO3 and 1.6mol of NO2.

It should be NO2 and not NO.

Answer:

The equilibrium molarity of NO = 0.21695 m

Explanation:

Given that :

the volume = 500 mL = 0.500 m

number of moles of $NO_3 = 1.9 \ mol$

number of moles of $NO_2 = 1.6 \ mol$

Then we can calculate for their respectively concentrations as :

$[NO_3] = \frac{number \ of \ moles}{volume}$

$[NO_3] = \frac{1.9}{0.500}$

$[NO_3] = 3.8 \ M$

$[NO_2] = \frac{number \ of \ moles}{volume}$

$[NO_2] = \frac{}{} \frac{1.6}{0.500}$

$[NO_2] = 3.2 \ M$

The chemical reaction can be written as:

$NO_3_{(g)} + NO_{(g)} \to 2NO_2_{(g)}$

The ICE table is as follows;

$NO_3_{(g)} + NO_{(g)} \to 2NO_2_{(g)}$

Initial 3.8 - 3.2

Change +x x -2x

Equilibrium 3.8+x +x 3.2 - 2x

$K_c=\frac{[NO_2]^2}{[NO_3][NO]}$

$where \ K_c = 8.33$

$8.33 = \frac{(3.2-2x)^2}{(3.8+2x)x} \\ \\ 8.33 = \frac{(3.2-2x)^2}{(3.8x+2x^2)}$

$8.33(3.8x + 2x^2) = (3.2-2x)^2 \\ \\ 31.654x + 16.66x^2 = (3.2-2x)(3.2-2x) \\ \\ 31.654x + 16.66x^2 = 10.24 - 12.8x +4x^2 \\ \\ 10.24 - 44.454x -12.66 x^2 = 0 \\ \\ 12.66x^2 +44.454x -10.24 = 0$