Answer:
0.33 mol/kg NH₃
Explanation:
Data:
b(NH₃) = 0.33 mol/kg
b(Na₂SO₄) = 0.10 mol/ kg
Calculations:
The formula for the boiling point elevation ΔTb is
i is the van’t Hoff factor — the number of moles of particles you get from a solute.
(a) For NH₃,
The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.
1 mol NH₃ ⟶ 1 mol particles
i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water
(b) For Na₂SO₄,
Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)
1 mol Na₂SO₄ ⟶ 3 mol particles
i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water
The NH₃ has more moles of particles, so it has the higher boiling point.
Step 1 - Since 3.3mm^3 = 0.0033cm^3, convert that to 3.3x10^-3 cm^3.
Step 2 - Since 1cm^3 = 1x10^-6 m^3, times 3.3 by that. (3.3 x 1 x 10^-6) = 0.0000033
Step 3 - 0.0000033 = 3.3 x 10 ^-6 m^3, which is your answer choice A.
*if you do not understand, message me and I will go through it more thoroughly!!*
Have a great day!! :)
The balanced reaction is:<span>
Mg (s) + 2HCl (aq)
--> MgCl 2 (aq) + H 2 (g)
We
are given the amount hydrochloric acid to be used for the reaction. This will
be the starting point of the calculation.
40.0 g
HCl ( 1 mol HCl / 36.46 g HCl)
(1 mol H2 / 2 mol HCl) (2.02 g H2 / 1 mol H2) = 1.11 g H2</span>
Answer:
The answer is no. C. Rate =[A] [B]
