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3 years ago

The envelope of air that surrounds the earth is called

Chemistry

1 answer:

svlad2 [7]3 years ago

4 0

Atmosphere is the envelope of air that surrounds the earth.

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A 1.00 cm2 blackbody surface emits 201 watts of radiant power. What is the temperature of the surface? Give to the nearest whole

erastovalidia [21]

Explanation:

Formula for black body radiation is as follows.

$\frac{P}{A} = \sigma \times T^{4} J/m^{2}s$

where, P = power emitted

A = surface area of black body

$\sigma$ = Stephen's constant = $5.6703 \times 10^{-8} watt/m^{2}.K^{-4}$

As area is given as 1.0 $cm^{2}$ . Converting it into meters as follows.

$1.00 cm^{2} \times \frac{(10^{-2})^{2} m^{2}}{1 cm^{2}}$ (as 1 m = 100 cm)

= $1 \times 10^{-4} m^{2}$

It is given that P = 201 watts. Hence,

$\frac{201 watts}{1 \times 10^{-4} m^{2}}$ = $5.6703 \times 10^{-8} watt/m^{2}.K^{-4} \times T^{4}$

$T^{4}$ = $35.45 \times 10^{12}$

T = $(35.45 \times 10^{12})^{1/4}$

= 8862.5 K

Thus, we can conclude that the temperature of the surface is 8862.5 K.

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3 years ago

What is this sign plz tell for 100 points

steposvetlana [31]

It's a hazard symbol for irritant.

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3 years ago

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Describe the process that ice on Mount Everest goes through when being heated from 10 degrees Fahrenheit to 305 degrees Kelvin.

ladessa [460]

Since the atmospheric pressure at Mt Everest is constant, the process will take place at constant pressure and such process at constant pressure is called Isobaric process. Hope this helps!

3 0

3 years ago

Calculate how many grams of the product form when 16.7 g of calcium metal completely reacts. Assume that there is more than enou

swat32

39.96 g product form when 16.7 g of calcium metal completely reacts.

<h3>What is the stoichiometric process?</h3>

Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Equation:

$Ca(s) + Cl_2(g)$ → $CaCl_2(s)$

In this case, for the undergoing reaction, we can compute the grams of the formed calcium chloride by noticing the 1:1 molar ratio between calcium and it (stoichiometric coefficients) and using their molar mass of 40 g/mol and 111 g/mol by using the following stoichiometric process:

$m_{ca_C_l_2}$ = 16.7 g Ca x $\frac{1 mol \;of \;Ca}{40g Ca}$ x $\frac{1 mol \;of \;CaCl_2}{1 mol \;Ca}$ x $\frac{111g of \;CaCl_2}{1 mol \;CaCl_2}$