Answer:
(a) 1.77*10^15 Xe atoms
(b) 2.10*10^16 Ne atoms
(c) 2.10*10^16 He atoms
Explanation:
Total V = 0.900 mm * 0.300mm * 10.0mm
Total V = 2.7mm cubed
Converting cubed to litres:
2.7mm cubed = 2.7 * 10^-6 Liters
Converting pressure from torr to atm,
(500 torr) * (1 atm ÷ 760 torr) = 0.6578947368 atm
Room temperature is assumed to be 21 degrees celcius, or 294.15 Kelvin
Applying Ideal Gas Equation:
PV = nRT
Therefore, n = PV ÷ RT,
n = [(2.7*10^-6 Liters)(.6578947368 atm)] ÷ [(0.08206)(294.15)]
n = 7.359017079 * 10^-8
Therefore, the total moles present in plasma is 7.359017079 * 10^-8.
(a) To find Xe atoms,
Multiply the total moles present by 4 percent. Therefore,
(7.359017079 * 10^-8)(.04) = 2.943606832*10^-9,
this value gives you Xe moles.
Then multiply by Avogadro's constant:
[(2.943606832*10^-9)(6.022*10^23)] = 1.77*10^15 Xe atoms
(b) and (c) are the same.
Using the total moles number we found in the beginning and minus the number of moles for Xe.
We have:
7.359017079 * 10^-8 - 2.943606832*10^-9 = 7.064656396*10^-8
And Since Ne and He are in a 1:1 ratio, divide this number by two.
We have:
7.064656396*10^-8 ÷ 2 = 3.532328198*10-8 moles
By multiplying this number by Avogadro's number we get atoms:
[(3.532328198*10-8 moles)(6.022*10^23)] = 2.10*10^16
Therefore, the answer for both (b) and (c) is 2.10*10^16.
