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
787.57 grams GIVE ME BRAINLIEST
0.781 moles
Explanation:
We begin by balancing the chemical equation;
O₂ (g) + 2H₂ (g) → 2H₂O (g)
21.8 Liters = 21.8 Kgs
To find how many moles are in 28.1 Kg H₂O;
Molar mass of H₂O = 18 g/mol
28.1/18
= 1.56 moles
The mole ratio between water vapor and oxygen is;
1 : 2
x : 1.56
2x = 1.56
x = 1.56 / 2
x = 0.781
0.781 moles
Answer:
12.09 L
Explanation:
Step 1: Convert 826.1 mmHg to atm
We will use the conversion factor 760 mmHg = 1 atm.
826.1 mmHg × 1 atm/760 mmHg = 1.087 atm
Step 2: Convert 427.8 J to L.atm
We will use the conversion factor 101.3 J = 1 L.atm.
427.8 J × 1 L.atm/101.3 J = 4.223 L.atm
Step 3: Calculate the change in the volume
Assuming the work done (w) is 4.223 L.atm against a pressure (P) of 1.087 atm, the change in the volume is:
w = P × ΔV
ΔV = w/P
ΔV = 4.223 L.atm/1.087 atm = 3.885 L
Step 4: Calculate the final volume
V₂ = V₁ + ΔV
V₂ = 8.20 L + 3.885 L = 12.09 L
Answer:
0.719M AgNO₃
Explanation:
Based on the reaction:
MgBr₂ + 2AgNO₃ ⇄ 2AgBr + Mg(NO₃)₂
<em>1 mole of magnesium bromide reacts completely with 2 moles of AgNO₃</em>
<em />
To find molarity of AgNO₃ solution we need to determine moles of AgNO₃ and, as molarity is the ratio of moles over liter (13.9mL = 0.0139L). Now, to determine moles of AgNO₃ we need to use the reaction, thus:
<em>Moles AgNO₃:</em>
<em />
Moles of MgBr₂ are:
50.0mL = 0.050L * (0.100mol / L) = 0.00500 moles of MgBr₂.
As the silver nitrate reacts completely and 2 moles of AgNO₃ reacts per mole of MgBr₂:
0.00500 moles MgBr₂ * (2 moles AgNO₃ / 1 mole MgBr₂) =
0.0100 moles of AgNO₃ are in the solution.
And molarity is:
0.0100 moles AgNO₃ / 0.0139L =
<h3>0.719M AgNO₃</h3>
