Answer:
1.930 * 10⁻⁹ mg of Mn⁺² are left unprecipitated.
Explanation:
The reaction that takes place is:
Mn⁺² + S⁻² ⇄ MnS(s)
ksp = [Mn⁺²] [S⁻²]
If the pksp of MnS is 13.500, then the ksp is:
From the problem we know that [S⁻²] = 0.0900 M
We use the ksp to calculate [Mn⁺²]:
3.1623*10⁻¹⁴= [Mn⁺²] * 0.0900 M
[Mn⁺²] = 3.514 * 10⁻¹³ M.
Now we can calculate the mass of Mn⁺², using the volume, concentration and atomic weight. Thus the mass of Mn⁺² left unprecipitated is:
3.514 * 10⁻¹³ M * 0.1 L * 54.94 g/mol = 1.930 * 10⁻¹² g = 1.930 * 10⁻⁹ mg.
Reason1: electrons on farther layers become free easyer
2nd reason: volume of atoms grows (from Helium to Xeon) so instead of boyle-mariot law equation (PV=vRT) is more accurate to use van der walls equations that adds to the boyle-mariot equation the volume occupied by the atoms of the gas to the volume of the space between the atoms P(Vm-b)=vRT
Answer:
i think it's B sorry if i'm wrong
Long wavelengths and low frequencies
RbOH is a strong base that dissociates completely and HCl is a strong acid that too dissociates completely. the complete reaction between the acid and base is;
RbOH + HCl ---> RbCl + H₂O
stoichiometry of acid to base is 1:1
At neutralisation point
H⁺ mol = OH⁻ mol
mol = molarity x volume
if Ma - molarity of acid and Va - volume of acid reacted
Mb - molarity of base and Vb - volume of base reacted
Ma x Va = Mb x Vb
0.5 M x 52.8 mL = Mb x 60.0 mL
Mb = 0.44 M
molarity of base - 0.44 M
