Answer:
(a) O = Valance Electrons (6), Inner electrons (8)
(b) Sn = Valance Electrons (2), Inner electrons (36)
(c) Ca = Valance Electrons (2), Inner electrons (20)
(d) Fe = Valance Electrons (2), Inner electrons (26)
(e) Se = Valance Electrons (6), Inner electrons (34)
Answer:
Option a.
0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i
Explanation:
To determine which of the solute is going to have a greatest effect on colligative properties we have to consider the Van't Hoff factor (i)
These are the colligative properties:
ΔP = P° . Xm . i → Lowering vapor pressure
ΔT = Kb . m . i → Boiling point elevation
ΔT = Kf . m . i → Freezing point depression
π = M . R . T → Osmotic pressure
Van't Hoff factor are the numbers of ions dissolved in the solution. For nonelectrolytes, the i values 1.
CaCl₂ and KNO₃ are two ionic solutes. They dissociate as this:
CaCl₂ → Ca²⁺ + 2Cl⁻
We have 1 mol of Ca²⁺ and 2 chlorides, so 3 moles of ions → i = 3
KNO₃ → K⁺ + NO₃⁻
We have 1 mol of K⁺ and 1 mol of nitrate, so 2 moles of ions → i = 2
Option a, is the best.
Answer
MnO₄ + 2H⁺ +3NO₂⁻ →3NO₃⁻ + Mn²⁺ +H₂O
Explanation
This is a redox reaction (oxidation-reduction reaction) which involves the transfer of electrons between two species. i.e
Mn + 6e⁻→Mn²⁺ (reduction)
3N³⁺- 6e⁻→3Mn⁵⁺(oxidation)
Answer:
Roughly C100 H140 N3 O
Explanation:
Gilsonite is a bituminous product that resembles shiny black obsidian.
It contains more than 100 elements.
Its mass composition varies but is approximately 84 % C, 10 % H, 3 % N, and 1 % O.
Its empirical formula is roughly C100 H140 N3 O.
Atoms of sulfur = 9.60⋅g32.06⋅g⋅mol−1×6.022×1023⋅mol−1 . Because the units all cancel out, the answer is clearly a number, ≅2×1023 as required.
