If P=0.998, V=1.28*10^3, and t=304K how many moles of helium do you have?

Answer 1

Mole is abbreviated to mol and given the symbol n
1 mole contains the same number of particles as there are in 12g of carbon-12 atoms by definition.
This number is called Avogadro's number or Avogadro's constant (NA) and is equal to 6.022 x 1023 particles.
1 mole of a pure substance has a mass in grams equal to its molecular mass (MM) [also known as molecular weight (MW) or formula mass (FM) or formula weight (FW)].
This is often referred to as the molar mass.
1 mole of an ideal gas has a volume of
22.4 litres (22.4L) at S.T.P. [Standard Temperature and Pressure, 0oC (273K) and 101.3kPa (1 atm)]
24.47 litres (24.47L) at S.L.C [Standard Laboratory Conditions, 25oC (298K) and 101.3kPa (1atm)]

Answer 2

use this equation
PV=nRT, n is number of moles R is the gas constant which is .0821.
This equation is the most use to solve for one of the unknown such as pressure, volutme, and temperature.
P is in atm, V is in liter, and T is in K you have to make sure of the unit or you will get the wrong answer.

.998 x 1.28x10^3 = n(.0821)(304)
n=.998 x 1.28x10^3 /(.0821 x 304)
n=51.18 mol

Answer 3

PV=nRT
n=PV/RT
=(0.998atm)(1.28e3L)/(0.0821atm mol/l K)(304K)

0.0281 is the constant R for pressure of atm and voume of L.

this is the ideal gas law.

Answer 4

n= 51.18 Grams, not mols.

One mol helium= 4.002602 grams, so divide that into 51.18 grams.

Call it 12.8 mols