Answer:
volume = 0.288 L
Explanation:
To find the volume, you need to (1) convert grams LiBr to moles LiBr (via molar mass) and then (2) calculate volume (via molarity equation). It is important to arrange the ratios in a way tat allows for the cancellation of units (desired units in the numerator).
<u>(Step 1)</u>
Molar Mass (LiBr): 6.9410 g/mol + 79.904 g/mol
Molar Mass (LiBr): 86.845 g/mol
100 grams LiBr 1 mole
----------------------- x ------------------ = 1.15 moles LiBr
86.845 g
<u>(Step 2)</u>
Molarity (M) = moles / volume (L)
4 M = 1.15 moles / volume
(4 M) x volume = 1.15 moles
volume = 1.15 moles / 4 M
volume = 0.288 L
This is a question about the colligative property known as freezing point depression. Freezing point depression (the amount the normal freezing point of the solvent is decreased) can be calculated with this equation:
ΔT = i Kf<span> m
</span>
Where i (the van't Hoff factor) is the degree of dissociation of the solute, Kf is the freezing point depression constant, and m is the molality of the solution.
Here i = 2 (KCl dissociates into 2 ions, K+ and Cl-), Kf = 1.86 C/m (for water), and m = 0.743m).
ΔT = 2 x 1.86 C/m x 0.743m = <span>2.764C
</span>
That means the freezing point of the solution is 2.764C less than the pure solvent (water), making it 0C - 2.764C = -2.764C.
There is an increase in the frequency of particle collisions which leads to the increased rate of the reaction.
<u>Explanation:</u>
Hydrochloric acid reacts faster with the powdered zinc than with the equal mass of zinc strips.
Since the surface area of the Zinc particle increases when it is powdered and has more active spots when compared to less number of active sites in Zinc strips and so the number of collision between the Zinc particles and the particles of Hydrochloric acid also increases and consequently the rate of the reaction also enhances.
So there is an increase in the frequency of particle collisions.
2C5H12 + 16O2 —> 10CO2 + 12H2O
2•5=10 C 10=1•10
2•12=24 H 24=2•12
16•2=32 O 32=(10•2)+(1•12)
=20+12