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8090 [49]
3 years ago
8

A particular kind of Spanish orange is completely fat free. If one of the oranges has 40 Calories total and a test shows that it

contains 1gram of protein, how many grams of carbohydrates does it have
Chemistry
1 answer:
LenaWriter [7]3 years ago
5 0

Answer:

                      9 grams of Carbohydrates

Explanation:

                     The energies contained by each class of these compounds are as;

(i)  Fats:

            1 grams of Fat contains 9 Calories

(ii)  Proteins:

            1 grams of Proteins have 4 Calories

(iii)  Carbohydrates:

            1 grams of Carbohydrates contains 4 Calories

(iv)  Vitamins and Minerals:

            Vitamins and Minerals have no Calories

Solution:

As 1 gram of Protein Contains 4 Calories so, we will subtract 4 Calories from the total Calories contained by the Juice as,

Calories by Carbohydrates  =  Total Calories - Calories by Protein

Calories by Carbohydrates  =  40 - 4

Calories by Carbohydrates  =  36 Calories.

Hence,

As,

           4 Calories were contained by  =  1 gram of Carbohydrate

So,

       36 Calories will be contained by  =  X grams of Carbohydrates

Solving for X,

                      X =  36 Calories × 1 gram / 4 Calories

                       X  =  9 grams of Carbohydrates

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4 years ago
How much heat is required to convert 20.0 g of ice at 50.0⁰C to liquid water at 0.0⁰C? The specific heat of ice is 2.06 J/(g∙⁰C)
Alex777 [14]

Answer:

8740 joules are required to convert 20 grams of ice to liquid water.

Explanation:

The amount of heat required (Q), measured in joules, to convert ice at -50.0 ºC to liquid water at 0.0 ºC is the sum of sensible heat associated with ice and latent heat of fussion. That is:

Q = m\cdot [c\cdot (T_{f}-T_{o})+L_{f}] (1)

Where:

m - Mass, measured in grams.

c - Specific heat of ice, measured in joules per gram-degree Celsius.

T_{o}, T_{f} - Temperature, measured in degrees Celsius.

L_{f} - Latent heat of fussion, measured in joules per gram.

If we know that m = 20\,g, c = 2.06\,\frac{J}{g\cdot ^{\circ}C}, T_{f} = 0\,^{\circ}C, T_{o} = -50\,^{\circ}C and L_{f} = 334\,\frac{J}{g }, then the amount of heat is:

Q = (20\,g)\cdot \left\{\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot [0\,^{\circ}C-(-50\,^{\circ}C)]+334\,\frac{J}{g} \right\}

Q = 8740\,J

8740 joules are required to convert 20 grams of ice to liquid water.

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3 years ago
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