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3 years ago

12

Buttery popcorn contained in a large 1bowl has a mass of about 50 and about 650 calories.

2 answers:

Fudgin [204]3 years ago

6 0

Answer:

Buttery popcorn contained in a large 1 <u>liter</u> bowl has a mass of about 50 <u>grams</u> of fat and about 650 calories

Explanation:

The question is about using correct units. The capacity of a bowl must be measured in a volume unit; the options were milliliter and liter. 1 milliliter is a very small amount (about 0.002 pint), so 1 liter is the correct option.

For the second blank, the options (mass units) are: microgram, milligram, gram, kilogram. Milligram is used to measure the mass of minerals, microgram is even smaller than milligram, the buttered popcorn weights much less than 50 kilograms. So, the correct option is 50 grams.

Rasek [7]3 years ago

4 0

Answer:

Buttery popcorn contained in a large 1 liter bowl has a mass of about 50 mg and about 650 calories.

Explanation:

Liter is the most appropriate unit to measure a bowl. Usually 1 liter of liquid has a mass of 1000 gram. Since popcorn is something lightweight and only a few can fill the bowl quickly so 50 mg makes perfect sense with 1 liter of bowl and 650 calories in buttery popcorn.

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3 years ago

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

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Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

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$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

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Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

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Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

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