Mass of SO2 = mass of sulfur+ mass of oxygen
⇒ 64 g= mass of sulfur+ 32 g
⇒ mass of sulfur= 64 g-32 g
⇒ mass of sulfur= 32 g.
There is 32 g of sulfur in 64 g of SO2~
As the volume of a gas increases <em>at constant temperature</em>, the number of particle impacts per unit area decreases.
There is the same number of impacts, but they are spread over a larger surface area.
Thus, the number of impacts per unit area decreases.
The ionic eqn is as follow:
1 Al(OH)3(s) + 3 H+(aq) + 3 NO3(-1) --> 1 Al(3+)(aq) + 3 NO3(-)(aq) + 3 H2O(l)
3moles of No3- ion on both sides cancels out to give the net ionic eqn:
1 Al(OH)3(s) + 3 H+(aq) --> 1 Al(3+)(aq) + 3 H2O(l)
Answer:
The equation: (NH₄)₂SO₄ = 2NH4(+) + SO4(-2)
The number of moles = 5 g / 132.14 g/mol = 0.038 mol
The number of molecules = 0.038 X 6.022x10^23 = 2.29x10^23
the number of positive ions present in the ammonium sulphate solution:
2 positive ions for every 1 molecule of (NH₄)₂SO₄
so 2 x 2.29x10^23 = 4.58x10^23
the number of negative ions present in the ammonium sulphate solution
1 negative ion for every 1 molecule of (NH₄)₂SO₄
so 1 x 2.29x10^23 = 2.29x10^23
the total number of ions present in the ammonium sulphate solution
4.58x10^23 + 2.29x10^23 = 6.87x10^23
