A4.5-liter sample of a gas at 2.0 atm and 300.0 k has 0.80 mole of the gas. if 0.35 mole of the gas is added to the sample at the same temperature and pressure, what is the final volume of the gas? \

The final volume = 6.469 L.

Explanation:

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

(V₁n₂) = (V₂n₁)

V₁ = 4.5 L, n₁ = 0.8 mol.

V₂ = ??? L, n₂ = 0.8 mol + 0.35 mol = 1.15 mol.

∴ V₂ = (V₁n₂)/n₁ = (4.5 L)(1.15 mol)/(0.8 mol) = 6.469 L.

Also 6.5

Explanation:

"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"