What is the volume of 6.9 mol oxygen (o2) gas at 233 k and a pressure of 4.0 atm? (the universal gas constant is 0.0821 l•atm/mol•k.)

Volume of Oxygen O2 would be 33 L.

Solution : Given information - Mole (n) = 6.9 mol

Pressure (P) = 4.0 atm

Temperature (T) = 233 K

Universal gas constant (R) = 0.0821 L.atm/mol.K

Volume(V) is unknown , which we need to find.

Volume can be calculated by using the Ideal gas equation which is :

PV = nRT  (Ideal gas equation)

On rearranging the above formula we will get :

Where , P = pressure , V = volume , n = mole , R = universal gas constant , T = temperature.

We will plugin the values of P , n , R and T in the ideal gas equation and then calculate Volume (V)

V = 32.998 L or 33 L

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

metric time was an attempt to create a standard unit of time based off of 10, but it never took off an is not included in the metric system (si)