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AVprozaik [17]
3 years ago
8

The degree of unsaturation, or index of hydrogen deficiency, is the number of pi bonds plus rings in a molecule. Specify the deg

ree of unsaturation (index of hydrogen deficiency) of the following formulas:
(a) C5H6
(b) C10H6Cl2
(c) C4H3NO
Chemistry
1 answer:
yaroslaw [1]3 years ago
5 0

Explanation:

The degree of unsaturation is given as;

Degree of unsaturation = (2C + 2 + N - H - X) / 2

(a) C5H6

Using the formular above;

Degree of unsaturation = [ 2(5) + 2 - 6] / 2

Degree of unsaturation = [ 12 - 6 ] / 2 = 3

(b) C10H6Cl2

Using the formular above;

Degree of unsaturation = [ 2(10) + 2 - 6 -2 ] / 2

Degree of unsaturation = [ 22 - 8 ] / 2 = 7

(c) C4H3NO

Using the formular above;

Degree of unsaturation = [ 2(4) + 2 - 3 -1 ] / 2

Degree of unsaturation = [ 10 - 4 ] / 2 = 3

Note: Oxygen is ignored because its presence has no effect on the degree of unsaturation.

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Explanation:

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How many grams of precipitate will be formed when 20.5 mL of 0.800 M
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There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

<u>Step 1</u>: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

<u>Step 2:</u> Data given

Volume of 0.800 M  CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

<u>Step 3:</u> Calculate moles of CO(NO3)2

Moles CO(NO3)2  = Molarity * volume

Moles CO(NO3)2  = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step  6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

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