Boiling-point elevation is a colligative property.
That means, the the boiling-point elevation depends on the molar content (fraction) of solute.
The dependency is ΔTb = Kb*m
Where ΔTb is the elevation in the boiling point, kb is the boiling constant, and m is the molality.
A solution of 6.00 g of Ca(NO3) in 30.0 g of water has 4 times the molal concentration of a solution of 3.00 g of Ca(NO3)2 in 60.0 g of water.:
(6.00g/molar mass) / 0.030kg = 200 /molar mass
(3.00g/molar mass) / 0.060kg = 50/molar mass
=> 200 / 50 = 4.
Then, given the direct proportion of the elevation of the boiling point with the molal concentration, the solution of 6.00 g of CaNO3 in 30 g of water will exhibit a greater boiling point elevation.
Or, what is the same, the solution with higher molality will have the higher boiling point.
Answer:first D. 88L
Second A 2*10^24
Explanation:
At stp 1 mole = 22.4L
mw Cl2= 70.9
280 g =280/70.9 moles, about 4
4*22.4 = about 88
aw Sr 87.6 —> 6.02214076*10^23 atoms = 1 mole
Ductility - a materials ability to stretch, ie if you pull it apart does it stretch to a wire.
density - ratio of volume to mass
conductivity - materials ability to conduct a current.
hopefully with these definitions you can figure out the answer.
Chromium , silver, zinc...
Answer:
2.2×10^8
Explanation:
Cu(OH)2(s)<---------> Cu^2+(aq) + 2OH^-(aq) Ksp=2.2 x 10 ^-20
2H3O^+(aq) + 2OH^-(aq) <-------> 4H2O(l). Kw= 1×10^14
Cu^2+(aq) + 4H2O(l) <--------> [Cu(H2O)4]^2+(aq)
Overall ionic reaction:
Cu(OH)2(s) +2H3O^+(aq) <---------> [Cu(H20)4]^2+(aq)
Equilibrium constant for the reaction: Ksp×Kw= 2.2 x 10 ^-20 × (1/(1×10^-14))^2
Keq= 2.2×10^8
Kw= ion dissociation constant of water