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Illusion [34]
3 years ago
14

Describe how to find calorie content in food

Chemistry
1 answer:
Fiesta28 [93]3 years ago
4 0

<u>To find the calorie content in food:</u>

Importantly, a calorie will not be considered as full or empty and not a thing too.  You cannot tie calories in a bottle. A calorie is a unit of measure of energy. We can also say that, it is the energy amount that is needed to rise the temperature of one ml of water by 1 degree Celsius.

Earlier, a calorimeter was used to measure the calorie content of food. That is, a certain amount of food in which water content has evaporated, and this food was put in a vessel surrounded by some water. The container was then sealed, oxygen was supplied in, and the food was ignited.

From the temperature increase of water, the calorie content of food was determined. Today, producers use the “At water indirect system” to find calorie content by adding the calories containing the energy nutrients: protein, carbohydrate, fat and alcohol.

The At water system makes use of the average values of 4 K cal/g for protein, 4 K cal/g for carbohydrate, 9 K cal/g for fat and  7 K cal/g for alcohol that were determined by burning these substances in a calorimeter.

Thus the label on a 45 gram Perk that contains 3 g of protein, 29 g of carbohydrate (22 grams of which are simple sugars) and 12 g of fat would read 230 Calories.

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You are given a solid that is a mixture of na2so4 and k2so4.
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Here we have to calculate the amount of SO_{4}^{2-} ion present in the sample.

In the sample solution 0.122g of SO_{4}^{2-} ion is present.

The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-

K₂SO₄ = 2K⁺ +  SO_{4}^{2-}

(Na)₂SO₄=2Na⁺ +  SO_{4}^{2-}

Thus, BaCl₂+  SO_{4}^{2-} = BaSO₄↓ + 2Cl⁻ .

(Na)₂SO₄ and  K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of  SO_{4}^{2-} ion is precipitated in this reaction.  

The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.

Now the molecular weight of BaSO₄ is 233.3 g/mol.

We know the molecular weight of sulfate ion (SO_{4}^{2-}) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of SO_{4}^{2-} ion is present.

Or. we may write in 233.3 g of BaSO₄ 96.06 g of SO_{4}^{2-} ion is present. So in 1 g of BaSO₄ \frac{96.06}{233.3}=0.411 g of SO_{4}^{2-} ion is present.

Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of SO_{4}^{2-} ion is present.        

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3 years ago
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