Assuming the partial pressure of oxygen in air (0.20 atm) and nitrogen in air (0.80 atm). Calculate the mole fractions of oxygen and nitrogen in water at 298 K. FOR THIS QUESTION report the mole fraction of OXYGEN
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Ответ:
oxygen = 4.7 * 10^-6
Nitrogen = = 9.7 * 10^6
Explanation:
partial pressure of oxygen = 0.20 atm
partial pressure of Nitrogen = 0.80 atm
calculate the mole fractions of oxygen and Nitrogen in water
Temp = 298k
applying henry's law
molar conc of oxygen in water ( Coxygen )
= Kp = 1.3 * 10^-3 Mol/L.atm * 0.20 atm = 2.6 * 10^-4 Mol
molar conc of Nitrogen in water ( Cnitrogen )
= Kp = 6.8 * 10^-4 Mol/L.atm * 0.80 atm = 5.4 * 10^-4
next Given that the number of moles in 1 liter of water = 55.5 mol
therefore the mole fraction of oxygen
= 2.6 * 10^-4 / 55.5
= 4.7 * 10^-6
mole fraction of Nitrogen
= 5.4 * 10^-4 / 55.5
= 9.7 * 10^6
Ответ:
dependent
The variable that you change during an experiment is the independent variable, also called the manipulated variable. The variable that is observed during the experiment is the dependent variable, also called the responding variable.
Explanation: