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The percentage of the total energy contributed by fat = 51.59% of fat
<h3>Calculation of total energy in fat</h3>The quantity of fat in the hamburger = 36grams
But 1 gram of fat = 9 kcal( standard energy value per gram of fat).
That is , 9kcal = 1 gram
X kcal = 36 grams
cross multiply,
X kcal = 9 × 36
= 324kcal
To calculate the percentage of 324kcal of fat in 628 kcal,
% = 51.59% of fat
The percentage of the total energy contributed by fat = 51.59%
Learn more about fats here:
Answer:
93,32 % (w/w)
Explanation:
The Ca²⁺ of CaCO₃ was completely precipitate to CaC₂O₄ that results in H₂C₂O₄. The titration of this one is:
3H₂C₂O₄ + 2H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O + 6CO₂
As you required 35,62mL of 0,1092M MnO₄⁻, the moles of H₂C₂O₄ are:
0,03562L× =
3,890x10⁻³mol of MnO₄⁻× = <em>0,01167 mol of H₂C₂O₄</em>
The number of moles of CaCO₃ are the same number of moles of H₂C₂O₄ because every Ca²⁺ was converted in CaC₂O₄ that was converted in H₂C₂O₄. That means: <em>0,01167 mol of CaCO₃</em>
0,01167 mol of CaCO₃ are:
0,01167 mol of CaCO₃× = <em>1,168 g of CaCO₃</em>
As the mass of the initial mixture is 1,2516 g, the percentage by weight of CaCO₃ is:
×100 = <em>93,32 % (w/w)</em>
I hope it helps!