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Iteru [2.4K]
2 years ago
8

Calculate the energy provided by a slice of bread containing 2 g of protein, 1 g of fat, and 20 g of carbohydrate. record your a

nswer using whole number. kcal
Chemistry
1 answer:
vichka [17]2 years ago
8 0
 the energy gained by proteins and carbohydrates differs from the energy gained by fats.
proteins and carbohydrates both give 4 kcal per gram 
fats give 9 kcal per gram
mass of proteins - 2 g
energy given by proteins - 2 g x 4 kcal/g = 8 cal
mass of carbohydrates - 20 g 
energy given by carbohydrates - 20 g x 4 kcal/g = 80 cal
mass of fat - 1 g 
energy given by fat - 1 g x 9 kcal/g = 9 cal
total energy = 8 + 80 + 9 = 97 kcal
energy = 97 kcal
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A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce
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Explanation:

The given data is as follows.

       Volume of lake = 15 \times 10^{6} m^{3} = 15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}

        Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = 5.6 \times 15 \times 10^{9} mg

                                                                    = 84 \times 10^{9} mg

                                                                    = 84 \times 10^{3} kg

Flow rate of river is 50 m^{3} sec^{-1}

Volume of water in 1 day = 50 \times 10^{3} \times 86400 liter

                                          = 432 \times 10^{7} liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are 2.9792 \times 10^{10} mg or 2.9792 \times 10^{4} kg

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Volume of sewage water in 1 day = 6048 \times 10^{4} liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = 1.8144 \times 10^{10} mg or 1.8144 \times 10^{4}kg

Therefore, total concentration of lake after 1 day = \frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l

                                        = 6.8078 mg/l

                 k_{D} = 0.2 per day

       L_{o} = 6.8078

Hence, L_{liquid} = L_{o}(1 - e^{-k_{D}t}

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Hence, the remaining concentration = (6.8078 - 1.234) mg/l

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Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

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