The concentration is 5 g/L.
Concentration = mass/volume = 10 g/2 L = 5 g/L
Answer:
1.089%
Explanation:
From;
ν =1/2πc(k/meff)^1/2
Where;
ν = wave number
meff = reduced mass or effective mass
k = force constant
c= speed of light
Let
ν =1/2πc (k/meff)^1/2 vibrational wave number for 23Na35 Cl
ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl
The between the two is obtained from;
ν' - ν /ν = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2
Therefore;
ν' - ν /ν = [meff/m'eff]^1/2 - 1
Substituting values, we have;
ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2 -1
ν' - ν /ν = -0.01089
percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;
ν' - ν /ν * 100
|(-0.01089)| × 100 = 1.089%
When The balanced equation is:
2Al + 3CuCl2 ⇒3 Cu + 2AlCl3
So, we want to find the limiting reactant:
1- no. of moles of 2Al = MV/n = (Wt * V )/ (M.Wt*n*V) = Wt / (M.Wt *n)
where M= molarity, V= volume per liter and n = number of moles in the balanced equation.
by substitute:
∴ no. of moles of 2Al = 0.2 / (26.98 * 2)= 0.003706 moles.
2- no.of moles of 3CuCl2= M*v / n = (0.5*(15/1000)) / 3= 0.0025 moles.
So, CuCl2 is determining the no.of moles of the products.
∴The no. of moles of 3Cu = 0.0025 moles.
∴The no.of moles of Cu= 3*0.0025= 0.0075 moles.
and ∵ amount of weight (g)= no.of moles * M.Wt = 0.0075 * M.wt of Cu
= 0.0075 * 63.546 =0.477 g
