What mass of precipitate (in g) is formed when 250.0 mL of 0.150 M CuCl₂ is mixed with excess KOH in the following chemical reaction? CuCl₂(aq) + 2 KOH(aq) → Cu(OH)₂(s) + 2 KCl(aq)

3.6487g

CuCl2 moles reacted = (0.15×250)/1000

according to balanced chemical equation

precipitated Cu(OH)2 moles = Reacted CuCl2 moles

molar mass of Cu(OH)2 = 63.5+ (17+1)×2 = 97.5

mass of precipitate = (97.5 × 0.15×250)/1000

= 3.648g

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be