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nika2105 [10]
2 years ago
8

What is the composition and percentage of the gases in earths atmosphere?

Chemistry
2 answers:
ryzh [129]2 years ago
7 0

Nitrogen accounts for 78% of the atmosphere, oxygen 21% and argon 0.9%. Gases like carbon dioxide, nitrous oxides, methane, and ozone are trace gases that account for about a tenth of one percent of the atmosphere.

olasank [31]2 years ago
3 0

Nitrogen accounts for 78% of the atmosphere, oxygen 21% and argon 0.9%. Gases like carbon dioxide, nitrous oxides, methane, and ozone are trace gases that account for about a tenth of one percent of the atmosphere.

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Water vapor enters a compressor at 35 kpa and 160°c and leaves at 300 kpa with the same specific entropy as at the inlet. What a
Annette [7]
<h3><em><u>solution</u></em><em><u>:</u></em></h3>

<em><u>The initial entropy is obtained from the initial pressure and temperature with data from A-6 using interpolation:</u></em>

<em><u>s</u></em><em><u>=</u></em><em><u> </u></em><em><u>8</u></em><em><u>.</u></em><em><u>26</u></em><em><u>5</u></em><em><u>2</u></em><em><u> </u></em><em><u>kJ</u></em><em><u>/</u></em><em><u>kgK</u></em>

<em><u>The final temperature is determined from the entropy and the final pressure with data from A-6 using interpolation:</u></em>

<em><u>T₂ = T₁+</u></em><em><u> </u></em><em><u>T₂ - </u></em><em><u>T₁</u></em><em><u>/</u></em><em><u> </u></em><em><u>8</u></em><em><u>₂</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>8</u></em><em><u>₁</u></em><em><u> </u></em><em><u>(</u></em><em><u> </u></em><em><u>s</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>s</u></em><em><u>₁</u></em><em><u>)</u></em>

<em><u>= </u></em><em><u>(</u></em><em><u>400 +</u></em><em><u> </u></em><em><u>500 - 400</u></em><em><u>/</u></em><em><u>8.3271</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>8.0347</u></em><em><u> </u></em><em><u>(8.2652 - 8</u></em><em><u>)</u></em><em><u>)</u></em>

<em><u>= 478.83°C</u></em>

<em><u>The final enthalpy is determined in the same way:</u></em>

<em><u>h₂= h₁</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>h₂</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>h₁</u></em><em><u>/</u></em><em><u>s</u></em><em><u>₂</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>s</u></em><em><u>₁</u></em><em><u> </u></em><em><u>( s - s₁)</u></em>

<em><u>= (</u></em><em><u> </u></em><em><u>3275.5</u></em><em><u>+</u></em><em><u> </u></em><em><u>3486.6 </u></em><em><u>-</u></em><em><u> </u></em><em><u>3275.5</u></em><em><u>/</u></em><em><u> </u></em><em><u>8.3271</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>8.0347</u></em><em><u>)</u></em><em><u> </u></em><em><u>(8.265</u></em><em><u>)</u></em>

<em><u>=</u></em><em><u> </u></em><em><u>3441.91 </u></em><em><u>kJ</u></em><em><u>/</u></em><em><u>kg</u></em>

6 0
2 years ago
Which of the following statements about Bohr's model of the atom is true? Bohr's model explains the chemical behavior of all ato
dsp73

I believe the answer is "Bohr's model explains the chemical behavior of all atoms."

3 0
3 years ago
Read 2 more answers
To find the atomic mass of a particular atom you need to know
Andreyy89
You just need the number of protons and number of neutrons as the mass of eelctrons is negligible
5 0
3 years ago
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A reaction rate assay measures enzyme _______________; and a colorimetric endpoint assay measures ________________.
VMariaS [17]

Answer:

The correct option is: A. activity; concentration

Explanation:

A reaction rate essay is a laboratory method to determine the activity of an enzyme. It is necessary for determining the enzyme kinetics and inhibition.

Whereas, a colorimetric analysis is a method to determine the concentration of a chemical compound in a solution. Therefore, a colorimetric endpoint assay measures the concentration.

Therefore, Enzyme activity is measured by a reaction rate assay and the concentration is measured by a colorimetric endpoint assay.

4 0
3 years ago
Read 2 more answers
Calculate the empirical formula for each natural flavor based on its elemental mass percent composition.
algol [13]
The complete question is;
Calculate the empirical formula for each of the following naturalflavors based on their elemental mass percent composition.

Q1)
methyl butyrate (component of apple taste andsmell):  C -58.80 %  H- 9.87 % 
O -31.33.%Express your answer as a chemical formula.


Q2)
vanillin (responsible for the taste and smellof vanilla):  C - 63.15%  H- 5.30 % 
O - 31.55%Express your answer as a chemical formula.

Q1)
empirical formula is the simplest ratio of whole number of elements in the compound. as the percentages have been given, lets calculate for 100 g of compound 
                                          C                         H                         O
mass                             58.80 g                  9.87 g                   31.33
molar mass                   12 g/mol                 1 g/mol                 16 g/mol
number of moles           58.80/12                9.87/1                    31.33/16
                                      = 4.9                      =9.87                     = 1.95
then divide number of moles by least number of moles - 1.95 in this case
                                      4.9/1.95 = 2.51      9.87/1.95 = 5.06    1.95/1.95= 1
next multiply by 2 to get numbers that can be rounded off to whole numbers
                                       2.51x2 = 5.02        5.06x2 = 10.12      1x2 = 2
when rounded off to the nearest whole number 
C - 5
 H - 10
 O - 2
therefore empirical formula is C₅H₁₀O₂

Q2) for this too since elemental composition has been given in percentages lets calculate for 100 g of compound 
                                          C                         H                         O
mass                               63.15 g                5.30 g                 31.55 g
molar mass                     12 g/mol              1 g/mol                16 g/mol
number of moles             63.15/12             5.30/1                  31.55/16
                                        =5.26                  =5.30                   =1.97
divide the number of moles by the least number of moles - 1.97
                                        5.26/1.97            5.30/1.97               1.97/1.97 
                                        =2.67                   = 2.69                      = 1
multiply each by 3 to get numbers that can be rounded off to whole numbers
                                        2.67x3 = 8.01     2.69x3 = 8.07         1x3 = 3
rounded off to the nearest whole numbers 
C - 8
H - 8
O - 3
empirical formula = C₈H₈O₃
7 0
3 years ago
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