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denpristay [2]
3 years ago
11

Food dye allergies have increased in recent years. One study calculated that each 47.9 g serving of M&M's candy contains 240

calories 10 g of fat, 34 g of carbohydrates , 2 g of proteinand 29.5 mg of food dyes. What is the mass percent of food dyes in the M &M's candy ?
A ) 0.0616%
B) 4.18%
C)20.9
D) 71.5 %
Chemistry
1 answer:
77julia77 [94]3 years ago
3 0

Answer:

Mass percent of food dyes  = 0.0616%

Explanation:

Given data:

Mass of candy = 47.9 g

Calories = 240

Mass of fat = 10 g

Mass of carbohydrate = 34 g

Mass of protein = 2 g

Mass of food dyes = 29.5 mg

Mass percent of food dyes = ?

Solution:

First of all we will convert the mg into g.

Mass of food dyes = 29.5 mg × 1g /1000 mg = 0.0295 g

Mass percent of food dyes  = mass of food dyes / total mass× 100

Now we will put the values.

Mass percent of food dyes  = 0.0295 g / 47.9 g × 100

Mass percent of food dyes  =  0.000616 × 100

Mass percent of food dyes  = 0.0616%

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<h3>Further explanation</h3>

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A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of
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c) pH = 5.503

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a) 0.0 mL KOH:

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⇒ <em>C</em> CH3COOH = [CH3COOH] + [CH3COO-] = 0.100 M

charge balance:

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⇒ 1.75 E-5 = [H3O+]²/(0.100 - [H3O+])

⇒ [H3O+]² + 1.75 E-5[H3O+] - 1.75 E-6 = 0

⇒ [H3O+] = 1.314 E.3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.88

b) 5.0 mL KOH:

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∴ <em>C </em>CH3COOH = ((0.025)(0.100) - (5 E-3)(0.200))/(0.025+5 E-3)

⇒ <em>C</em> CH3COOH = 0.05 M

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⇒ <em>C</em> CH3COOH + <em>C</em> KOH = [CH3COOH] + [CH3COO-] = 0.05 + 0.033 = 0.083 M

charge balance:

⇒ [H3O+] + [K+] = [CH3COO-]

⇒ [CH3COO-] = [H3O+] + 0.033

⇒ 1.75 E-5 = ([H3O+]*([H3O+] + 0.033))/(0.083 - ([H3O+] + 0.033))

⇒ 1.75 E-3 = ([H3O+]² + 0.033[H3O+])/(0.05 - [H3O+])

⇒ 8.75 E-7 - 1.75 E-5[H3O+] = [H3O+]² + 0.033[H3O+]

⇒ [H3O+]² +0.03302[H3O+] - 8.75 E-7 = 0

⇒ [H3O+] = 2.523 E-5 M

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∴ Kw/Ka = 1 E-14/1.75 E-5 = 5.714 E-10 = [CH3COOH]*[OH-]/[CH3COO-]

∴ [CH3COO-] = (0.025)(0.100))/(0.025+0.0125) = 0.066 M

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⇒ 0.066 = [CH3COOH] + [CH3COO-]..........(1)

charge balance:

⇒ [K+] = [OH-] + [CH3COO-] = 0.066 M.........(2)

∴ [K+] = <em>C</em> CH3COO- = 0.066 M

(1) = (2):

⇒ [OH-] = [CH3COOH].......(3)

⇒ 5.714 E-10 = [OH-]² / (0.066 - [OH-])

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⇒ [OH-] = 6.1408 e-6 m

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⇒ [OH-] ≅ <em>C</em> KOH = 0.0125 M

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