Answer:
94.2 g/mol
Explanation:
Ideal Gases Law can useful to solve this
P . V = n . R . T
We need to make some conversions
740 Torr . 1 atm/ 760 Torr = 0.974 atm
100°C + 273 = 373K
Let's replace the values
0.974 atm . 1 L = n . 0.082 L.atm/ mol.K . 373K
n will determine the number of moles
(0.974 atm . 1 L) / (0.082 L.atm/ mol.K . 373K)
n = 0.032 moles
This amount is the weigh for 3 g of gas. How many grams does 1 mol weighs?
Molecular weight → g/mol → 3 g/0.032 moles = 94.2 g/mol
The answer is d. particles move from an area of high concentration to low concentration.
Answer:
substitution is the best method or collecting like terms
Explanation:
Volume of the tank is 5.5 litres.
Explanation:
mass of the CO2 is given 8.6 grams
Pressure of the gas is 89 Kilopascal which is 0.8762 atm
Temperature of the gas is 29 degrees ( 0 degrees +273.5= K) so (29+273)
R = gas constant 0.0821 liter atmosphere per kelvin)
FROM THE IDEAL GAS LAW
PV=nRT ( P Pressure, V Volume, n is number of moles of gas, R gas constant, Temperature in Kelvin)
no of moles = mass/atomic mass
= 8.6/44
= 0.195 moles
now putting the values in equation
V=nRT/P
= 0.195*0.0821*302/ 0.8762
= 5.5 litres.
As the carbon dioxide gas occupies the volume os the tank hence volume of tank is 5.5 litres.
A stock solution is a solution of a known concentration. Stock solution has a high concentration and therefore, the known amount of stock solution is used for preparing different concentrations, by diluting the same with the known amount of the solvent being used, such as water.