There are about 6.022*10^23 molecules in 1 mole of carbon
Lets say that the required number of moles will be x
1 mole —6.022*10^23
x moles—7.87*10^7
Cross multiply
x=7.87*10^7/6.022*10^23
Answer x=1.3*10^-16
Answer:
V = 1.79 L
Explanation:
Given data:
Volume of gas = ?
Number of moles of gas = 0.080 mol
Temperature = standard = 273.15 K
Pressure = standard = 1 atm
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
1 atm × V = 0.080 mol × 0.0821 atm.L/ mol.K × 273.15 K
V = 1.79 atm.L / 1 atm
V = 1.79 L
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