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
1 mol ------------ 6,02×10²³
0,17 mol ------- X
X =(0,17×6,02×10²³)/1
X = 1,0234×10²³ molecules H₂O
:•)
You can predict it by measuring the weight of the spherical object and measuring the liquid (mL) and subtracting finding the volume
