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densk [106]
3 years ago
5

A sample of oxygen gas occupies 3.60 liters at a pressure of 1.00 atm. If temperature is held constant, what will be the volume

of the oxygen gas at a pressure of 2.50 atm?
Chemistry
1 answer:
Soloha48 [4]3 years ago
3 0

Answer: The volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L

At constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts, then

PV = c

Thus, if the pressure increases, the volume decreases, and if the pressure decreases, the volume increases.

It is not necessary to know the exact value of the constant c to be able to use this law since for a fixed amount of gas at constant temperature, it is satisfied that,

P₁V₁ = P₂V₂

Where P₁ and P₂ as well as V₁ and V₂ correspond to pressures and volumes for two different states of the gas in question.

In this case the first oxygen gas state corresponds to P₁ = 1.00 atm and V₁ = 3.60 L while the second state would be P₂ = 2.50 atm and V₂ = y. Substituting in the previous equation,

1.00 atm x 3.60 L = 2.50 atm x y

We cleared y to find V₂,

V₂ = y = \frac{1.00 atm x 3.60 L}{2.50 atm} = 1.44 L

Then, <u>the volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L</u>

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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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