A balloon originally has 0.100 moles of helium and has a volume of 0.500 L. If 0.590 grams of He are added to the balloon, what will the new volume be, in L? (Assume temperature and pressure do not change.)

1.24 L

Explanation:

Avogadro's Law

Convert 0.590 g of He to moles.

0.590 g x (1 mol/4.00 g) = 0.1475 mol He

Add 0.1475 mol to 0.100 mol since the initial moles (n1) increases by this amount.

0.100 mol + 0.1475 mol = 0.2475 mol He

This will represent the final moles (n2)

The initial volume is given, 0.500 L. Plug in the values.

"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"