The boiling point elevation of an aqueous sucrose solution is found to be 0.39°c.

what mass of sucrose (molar mass= 342.30 g/mol) would be needed to dissolve in 500.0 g of water?

kb (water) = 0.512°c/m.

130.4 grams of sucrose, would be needed to dissolve in 500 g of water.

Explanation:

Colligative property of boiling point elevation:

ΔT = Kb . m . i

In this case, i = 1 (sucrose is non electrolytic)

ΔT = Kb . m

0.39°C = 0.512°C/m . m

0.39°C /0.512 m/°C = m

0.762 m (molality means that this moles, are in 1kg of solvent)

If in 1kg of solvent, we have 0.712 moles of sucrose, in 500 g, which is the half, we should have, the hallf of moles, 0.381 moles

Molar mass sucrose = 342.30 g/m

Molar mass . moles = mass

342.30 g/m . 0.381 m = 130.4 g

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer