Mass/volume = density. We know the volume (5 cm cubed) and the density (8.96). We plug these into the equation to find the mass:
M/5 = 8.96
M = 44.8 grams
This uses the concept of freezing point depression. When faced with this issue, we use the following equation:
ΔT = i·Kf·m
which translates in english to:
Change in freezing point = vant hoff factor * molal freezing point depression constant * molality of solution
Because the freezing point depression is a colligative property, it does not depend on the identity of the molecules, just the number of them.
Now, we know that molality will be constant, and Kf will be constant, so our only unknown is "i", or the van't hoff factor.
The van't hoff factor is the number of atoms that dissociate from each individual molecule. The higher the van't hoff factor, the more depressed the freezing point will be.
NaCl will dissociate into Na+ and Cl-, so it has i = 2
CaCl2 will dissociate into Ca2+ and 2 Cl-, so it has i = 3
AlBr3 will dissociate into Al3+ and 3 Br-, so it has i = 4
Therefore, AlBr3 will lower the freezing point of water the most.
Answer:
About 547 grams.
Explanation:
We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.
To do so, we can use the initial value and convert it to grams using the molar mass.
Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:
Dimensional Analysis:
In conclusion, about 547 grams of copper (II) bicarbonate is produced.
