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chubhunter [2.5K]
2 years ago
8

Radiation Project: Solar Cooker

Chemistry
1 answer:
Andrew [12]2 years ago
4 0

In the solar cooker, reflective material like aluminum is used to collect and trap the heat from the sunlight while the interior is painted to reduce heat loss.

<h3>What is a solar cooker?</h3>

A solar cooker is a cooker which uses the energy of sunlight to cook food.

In designing a solar cooker, a solar collector such as reflective material like aluminum is used to collect and trap the heat from the sunlight.

The interior of the cooker is painted black to avoid heat loss by radiation.

Cardboard boxes are used as insulating materials.

In conclusion, a solar cooker is designed to maximize solar energy collected and minimize heat loss.

Learn more about solar cooker at: brainly.com/question/727221

#SPJ1

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Calculate the oxidation number of the iodine (I) in each compound: HIO4 = I2 = NaI = HIO3 =
Makovka662 [10]
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What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom
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Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

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So, 5.5 g of lead contains \frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22} atoms

and,

As, 118.71 g of lead contains 6.022\times 10^{23} atoms

So, 94.5 g of lead contains \frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23} atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%

and,

\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

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