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djverab [1.8K]
2 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]2 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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6 0
3 years ago
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Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has
Ymorist [56]

Answer:

\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

Explanation:

Given that:

The Half-life of ^{235}U = 7.13 \times 10^8 \ years is less than that of ^{238} U = 4.51 \times 10^9 \ years

Although we are not given any value about the present weight of ^{235}U.

So, consider the present weight in the percentage of ^{235}U to be  y%

Then, the time elapsed to get the present weight of ^{235}U = t_1

Therefore;

N_1 = N_o e^{-\lambda \ t_1}

here;

N_1 = Number of radioactive atoms relating to the weight of y of ^{235}U

Thus:

In( \dfrac{N_1}{N_o}) = - \lambda t_1

In( \dfrac{N_o}{N_1}) =  \lambda t_1 --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of ^{235}U to be = t_2

Then:

In( \dfrac{N_o}{N_2}) =  \lambda t_2  ---- (2)

here;

N_2 =  Number of radioactive atoms of ^{235}U relating to 3.0 a/o weight

Now, equating  equation (1) and (2) together, we have:

In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) =  \lambda( t_1-t_2)

replacing the half-life of ^{235}U = 7.13 \times 10^8 \ years

In( \dfrac{N_2}{N_1})  = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)      ( since \lambda = \dfrac{In 2}{t_{1/2}} )

∴

\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is  \mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

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