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djverab [1.8K]
3 years ago
5

The nutritional energy content (in Calories) present in 86g of broccoli with 6g of carbohydrates, 2.6g of protein, and 0.3g of f

at is __________.
Chemistry
1 answer:
asambeis [7]3 years ago
5 0

Answer:

37.1 calories.

Approximately, 37.1 = 40 calories.

Explanation:

So, without mincing words let's dive straight into the solution to the question above.

We are given the following parameters which are going to help in solving this particular Question.

The mass of broccoli = 86g of broccoli, mass of carbohydrates present = 6g of carbohydrates, the mass of protein present = 2.6g of protein and the mass of fat present = 0.3g of fat.

Therefore, the nutritional energy content (in Calories) = (6 × 4) + (2.6 × 4) + (0.3 × 9) = 10.4 + 24 + 2.7 = 37.1

Hence, the nutritional energy content (in Calories) = 37.1 calories.

Approximately, 37.1 = 40 calories.

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A chemistry student in lab needs to fill a temperature-control tank with water. The tank measures 24.0 cm long by 21.0 cm wide b
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Answer:

The required volume of water the student needs is 4.9 litres of water

Explanation:

From the diagram related to the question, we have;

The dimensions of the tank are;

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Width of tank = 21.0 cm = 0.21 m

Depth of tank = 13.0 cm. = 0.13 m

Allowance provided between the top of the tank and the top of the water = 2.0 cm

Diameter of the round bottom flask, D = 10.5 cm = 0.105 m

Therefore, the radius of the round bottom flask, r = 0.105/2 = 0.0525 m

Therefore we have;

Depth of water in the tank = Depth of tank - Allowance provided between the top of the tank and the top of the water

∴ Depth of water in the tank = 13.0 - 2.0 = 11.0 cm = 0.11 m

Given that the flask is immersed in the water contained in the tank to raise the tank water level, we have;

Volume of water + Volume of flask in the tank = Length of tank × Width of tank × Depth of water in the tank

Volume of water + Volume of flask in the tank =  0.24 × 0.21 × 0.11 = 0.005544 m³ = 0.005544 m³× 1000 l/m³ = 5.544 l

The volume of the spherical flask = 4/3·π·r³ = 4/3·π·0.0525³ = 6.06×10⁻⁴ m³ = 6.06×10⁻⁴ m³ × 1000 l/m³ = 0.606 l

The required volume of water the student needs , V = Volume of water + Volume of flask in the tank - The volume of the spherical flask = 5.544 l - 0.606 l = 4.9 l.

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