Answer:
The precipitated are option a and d.
Explanation:
2 LiI(aq) +Hg2(NO3)2(aq) → Hg2I2(s) ↓ + 2 LiNO3(aq)
Cation Hg2+ 2 in the presence of iodide, a precipitated is formed.
Zn(s) + 2AgNO3(aq) → 2 Ag(s) ↓ +Zn(NO3)2(aq)
Zinc starts to get rid, and some white particles also stick to it. Afterwards the solution becomes cloudy and a precipitate appears, which is the solid silver
6.02 x10^23 atom
3.5g x 1mol/63.55g Cu x 6.02 x 10^23/ 1mol=
3.32 x 10^22 atoms
Answer:
zero
Explanation:
I I think one should be so accurate with measurements and experiments
