Answer: 0.4533mol/L
Explanation:
Molar Mass of CaCO3 = 40+12+(16x3) = 40+12+48 = 100g/mol
68g of CaCO3 dissolves in 1.5L of solution.
Xg of CaCO3 will dissolve in 1L i.e
Xg of CaCO3 = 68/1.5 = 45.33g/L
Molarity = Mass conc.(g/L) / molar Mass
Molarity = 45.33/100 = 0.4533mol/L
The trough and the hill part
The volume of a gas that is required yo react with 4.03 g mg at STP is 1856 ml
calculation/
- calculate the moles of Mg used
moles=mass/molar mass
moles of Mg is therefore=4.03 g/ 24.3 g/mol=0.1658 moles
- by use of mole ratio of Mg:O2 from the equation which is 2:1
the moles 02=0.1679 x1/20.0829 moles
- at STP 1 mole of a gas= 22.4 l
0.0895 moles=? L
- =0.0895 moles x22.4 l/ 1 mole=1.8570 L
into Ml = 1.8570 x1000=1856 ml approximately to 1860
Here try this presentation, it should help :)
https://prezi.com/6ofdo3fadl9a/exothermic-and-endothermic-reactions-in-everyday-life/
