1) Zn(CH₃COO)₂(s) + 2KOH(aq) = Zn(OH)₂(s) + 2CH₃COOK(aq)
Ksp{Zn(OH)₂}=1.2*10⁻¹⁷
2) Zn(CH₃COO)₂(s) + 2NaCN(aq) = Zn(CN)₂(s) + 2CH₃COONa(aq)
Ksp{Zn(CN)₂}=2.6*10⁻¹³
Ksp{Zn(OH)₂}<Ksp{Zn(CN)₂}
Zn(OH)₂ precipitates first
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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