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The unknown of this problem is the experimental percent of water in the compound in order to remove the water of hydrogen, given the following:
Mass of crucible, cover and contents before heating 23.54 g
Mass of empty crucible and cover 18.82 g
Mass of crucible, cover, and contents after heating to constant mass 20.94 g
In order to get the answer, determine the following:
Mass of hydrated salt used = 23.54 g – 18.82 g = 4.72 g
Mass of dehydrated salt after heating = 20.94 g – 18.82 g = 2.12 g
Mass of water liberated from salt = 4.72 g – 2.12 g = 2.60 g
Then solve the percent of water in the hydrated salt by:
% water = (mass of water / mass of hydrated salt) x 100
% water = 2.60 g / 4.72 g x 100
% water = 55.08 % in the compound
To answer this question, we will use the general gas law which states that:
PV = nRT where:
P is the pressure of the gas = <span>10130.0 kPa
</span>V is the volume of the gas = 50 liters
n is the number of moles that we want to calculate
R is the gas constant = <span>8.314 L∙kPa/K∙mol
T is the temperature = 300+273 = 573 degree kelvin
Substitute with the givens in the equation to get the number of moles as follows:
</span><span>10130 * 50 = n * 8.314 * 573
506500 = 4763.922 n
n = </span>506500 / 4763.922
n = 106.3199 moles
<u>Answer:</u> The moles of oxygen and carbon dioxide in air is 
and 
respectively
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of atmosphere = 
Average molar mass of atmosphere = 28.96 g/mol
Putting values in above equation, we get:
We know that:
Percent of oxygen in air = 21 %
Percent of carbon dioxide in air = 0.0415 %
Moles of oxygen in air = 
Moles of carbon dioxide in air = 
Hence, the moles of oxygen and carbon dioxide in air is and respectively
