11 × 44 = 484 g
mol × molar mass
The balanced equation for the above reaction is
2K₃PO₄ + 3NiCl₂ ---> 6KCl + Ni₃(PO₄)₂
stoichiometry of K₃PO₄ to NiCl₂ is 2:3
the number of NiCl₂ moles reacted - 0.0110 mol/L x 0.154 L = 1.69 x 10⁻³ mol
if 3 mol of NiCl₂ reacts with - 2 mol of K₃PO₄
then 1.69 x 10⁻³ mol of NiCl₂ reacts with - 2/3 x 1.69 x 10⁻³ = 1.13 x 10⁻³ mol of K₃PO₄
molarity of K₃PO₄ solution given - 0.205 M
there are 0.205 mol in 1 L
therefore 1.13 x 10⁻³ mol are in - 1.13 x 10⁻³ mol / 0.205 mol/L = 5.51 mL
volume of K₃PO₄ required - 5.51 mL
SrSo4 = Sr(2+) + SO4(2-)
Let’s say that the initial concentration of SrSo4 was 1. ( or we have 1 mole of this reagent).
When The reaction occurs part of SrSo4is dissociated. And we get X mole Sr(2+) and So4(2-).
Ksp=[Sr(2+)]*[SO4(2-)]
X^2=3.2*10^-7
X=5.6*10^-4
Answer:
Work = 90.65 j
Explanation:
Given data:
Mass = 0.500 Kg
Distance = 18.5 m
Work done = ?
Solution:
Work = force . distance
Force = mg
Work = mg.distance
Work = mgh
Work = 0.500 Kg × 9.8 m/s²× 18.5 m
Work = 90.65 Kg .m²/s²
Kg .m²/s² = j
Work = 90.65 j
