Answer: CoBr3 < K2SO4 < NH4 Cl
Justification:
1) The depression of the freezing point of a solution is a colligative property, which means that it depends on the number of particles of solute dissolved.
2) The formula for the depression of freezing point is:
ΔTf = i * Kf * m
Where i is the van't Hoof factor which accounts for the dissociation of the solute.
Kf is the freezing molal constant and only depends on the solvent
m is the molality (molal concentration).
3) Since, you are assuming equal concentrations and complete dissociation of the given solutes, the solute with more ions in the molecular formula will result in the solution with higher depression of the freezing point (lower freezing point).
4) These are the dissociations of the given solutes:
a) NH4 Cl (s) --> NH4(+)(aq) + Cl(-) (aq) => 1 mol --> 2 moles
b) Co Br3 (s) --> Co(3+) (aq) + 3Br(-)(aq) => 1 mol --> 4 moles
c) K2SO4 (s) --> 2K(+) (aq) + SO4 (2-) (aq) => 1 mol --> 3 moles
5) So, the rank of solutions by their freezing points is:
CoBr3 < K2SO4 < NH4 Cl
Hello! I can help you with this. First, convert them into it’s written out standard form. 10^4 is 10,000. 10,00 * 1.26 is 12,600. 10,000 * 2.5 is 25,000. 12,600 + 25,000 = 37,600 or 3.76 * 10^4 in scientific notation. The answer in scientific notation is 3.76 * 10^4.
The bromide concentration in this solution of calcium bromide dissolved in enough water to give 469.1 mL is 1.75 × 10-⁵M.
<h3>How to calculate concentration?</h3>
The concentration of a solution can be calculated by dividing the number of moles of the substance by its volume.
No of moles of calcium bromide is calculated as follows:
moles = 1.642 ÷ 199.89 = 8.215 × 10-³moles
Molarity = 8.215 × 10-³moles ÷ 469.1mL = 1.75 × 10-⁵M
Therefore, the bromide concentration in this solution of calcium bromide dissolved in enough water to give 469.1 mL is 1.75 × 10-⁵M.
Learn more about concentration at: brainly.com/question/10725862
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Answer:
Electrons will flow from left to right through the wire.
Pb^2+ ions will be reduccd to Pb metal.
The concentration of Sn2+ ions in the left compartment will increase.
Explanation:
Looking at the relative electrode potentials of the two metals
Sn= -0.14
Pb=-0.13
Tin is expected to function as the anode (left hand half cell) and lead as the anode (right hand half cell) tin oxidizes to sn^2+ hence its concentration increases on the left compartment while lead is reduced to ordinary lead metal on the right hand half cell . since oxidation occurs on the left hand side, electrons flow from left to right.
