Answer:
Total pressure increased
Explanation:
When gas C is added in the vessel then number of mole increases and number of collision depends on the number of molecules present in the vessel and on adding gas C ,mole also increases hence number of collision increases therefore pressure also increases because number of collision increases.
Total pressure increases.
The highest electronegativity is in the elements in the top left corner of the periodic table, and the lowest in the bottom right corner. Therefore, traveling up or to the left across the periodic table will increase the electronegativity
Answer:
5 moles of electrons
Explanation:
The balance equation is as follow,
<span> 5 Ag</span>⁺ + Mn⁺²<span> + 4 H</span>₂O →<span> 5 Ag + MnO</span>₄⁻<span> + 8 H</span>⁺
Reduction of Ag:
Ag⁺ + 1 e⁻ → Ag
Or,
5 Ag⁺ + 5 e⁻ → 5 Ag
Oxidation of Mn:
Mn⁺² → MnO₄⁻ + 5 e⁻
Result:
Hence 5 moles of Ag⁺ accepts 5 electrons from 1 mole of Mn⁺².
Answer:
Ksp = 2.74 x 10⁻⁵
Explanation:
The solubility equilibrium for Ca(OH)₂ is the following:
Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2 OH⁻(aq)
I 0 0
C + s + 2s
E s 2s
According to the ICE table, the expression for the solubility product constant (Kps) is:
Ksp = [Ca²⁺] x ([OH⁻])² = s x (2s)² = 4s³
Then, we calculate Ksp from the solubility value (s):
s = 0.019 M
⇒ Ksp = 4s³ = 4 x (0.019)³ = 2.74 x 10⁻⁵
Answer:
The van't hoff factor of 0.500m K₂SO₄ will be highest.
Explanation:
Van't Hoff factor was introduced for better understanding of colligative property of a solution.
By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.
a) For NaCl the van't Hoff factor is 2
b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]
Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.
c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.
