Compute the boiling point of this solution:
21.6 g nis04 in 1.00 • 102 g h20
(assume 100% ionization)

The boiling point of the solution is 101,42ºC

Explanation:

Formula for boiling point elevation is:

ΔT = Kb · molality . i

Water boils at 100ºC, at 1 atm.

NiSO4 > Ni2+  +  SO4-2  (assume 100% ionization, the value of i, is 2)

Tº boiling point solution - Tº boiling point = Ebulloscopic constant . molality . Van't Hoff factor.

Tº bp solution - 100ºC = 0,512 ºC/molal . molality . 2

Molality means moles of solute in 1kg of solvent (1000g)

Molar mass NiSO4 = 154,75 g/m

Mass/ Molar mass = Moles --> 21,6 g / 154,75 g/m = 0,139 moles

These moles are in 100 g of H2O, so we have to make a rule of three to find, the moles in 1000g

100 g 0,139 moles

1000 g 1,39 moles (molality)

Tº bp solution - 100ºC = 0,512 ºC/molal . 1,39molal . 2

Tº bp solution = (0,512 ºC/molal . 1,39molal . 2) + 100ºC

Tº bp solution = 101,42ºC

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

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.