Answer:
71.1
Explanation:
1 mol Fe = 10.4 g/55.85 g/mol = 0.186
1 mol AgNo3 = 28.4 g/169.87 g/mol = 0.178 mol AgNo3
then since Ag:Fe is 1:3, AgNo3 is the limiting reactant
So now
0.178 moles * 1/3 * 241.83 g/mol Fe(NO3)3 = 14.35 g Fe(NO3)3
Excess reactant: 0.178 moes AgNO3 * 1/3 = 0.059
0.186 - 0.059 = 0.127 moles Fe * 55.85 g/mol Fe = 7.1 g Fe excess
They can use carbon dating or lithium argon dating.
In order to find the answer, use an ICE chart:
Ca(IO3)2...Ca2+......IO3-
<span>some.......0..........0 </span>
<span>less.......+x......+2x </span>
<span>less........x.........2x
</span>
<span>Ca(IO₃)₂ ⇄ Ca⁺² + 2 IO⁻³
</span>
K sp = [Ca⁺²][IO₃⁻]²
K sp = (x) (2 x)² = 4 x³
7.1 x 10⁻⁷ = 4 x³
<span>x = molar solubility = 5.6 x 10</span>⁻³ M
The answer is 5.6 x 10 ^ 3 M. (molar solubility)
