Answer:
As solute concentration increases, vapor pressure decreases.
Step-by-step explanation:
As solute concentration increases, the number of solute particles at the surface of the solution increases, so the number of <em>solvent </em>particles at the surface <em>decreases</em>.
Since there are fewer solvent particles available to evaporate from the surface, the vapour pressure decreases.
C. and D. are <em>wrong</em>. The vapour pressure depends <em>only</em> on the number of particles. It does not depend on the nature of the particles.
Answer:
569K
Explanation:
Q = 3.5kJ = 3500J
mass = 28.2g
∅1 = 20°C = 20 + 273 = 293K
∅2 = x
c = 0.449
Q = mc∆∅
3500 = 28.2×0.449×∆∅
3500 = 12.6618×∆∅
∆∅ = 3500/12.6618
∆∅ = 276.4220
∅2 - ∅1 = 276.4220
∅2 = 276.4220 + ∅1
∅2 = 276.4220 + 293
∅2 = 569.4220K
∅2 = 569K
Answer:
d. 12.3 grams of Al2O3
Explanation:
Based on the reaction:
4Al + 3O2 → 2Al2O3
<em>Where 4 moles of Al reacts in excess of oxygen to produce 2 moles of aluminium oxide.</em>
<em />
To solve this question we must find the moles of Aluminium. With these moles we can find the moles of aluminium oxide using the reaction:
<em>Moles Al -Molar mass: 26.9815g/mol-</em>
6.50g * (1mol / 26.9815g) = 0.241 moles Al
<em>Mass Al₂O₃ -Molar mass: 101.96g/mol-</em>
0.241 moles Al * (2 mol Al2O3 / 4 mol Al) = 0.120 moles Al2O3
0.120 moles Al2O3 * (101.96g / mol) =
12.3g of Al2O3 are produced.
Right answer is:
<h3>d. 12.3 grams of Al2O3
</h3>
Answer:
0.08 mol L-1
Explanation:
Sulfuric acid Formula: H2SO4
Ammonia Formula: NH3
Ammonium sulfate Formula: (NH₄)₂SO₄
H2SO4 + 2NH3 = 2NH4+ + SO4 2-
H2SO4 + 2NH3 = (NH₄)₂SO₄
H2SO4 = (1/2)x (32.8 x 10^-3 L x 0.116 mol L-1)/25 x 10^-3 L
= 0.08 mol L-1
B.
Explanation:
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