Answer:
The coefficient before potassium (K) balances this chemical equation is 2.
Explanation:
_K +Cl₂ → 2KCl
K =1 ; Cl =2
K=1 × 2 = 2
Cl = 1 × 2 = 2
2 K +Cl₂ = 2 KCl
N(CH₃OH)=3,62·10²⁴/6·10²³ 1/mol = 6,033 mol
m(CH₃OH) = 6,033 mol · 32 g/mol (molar mass) = 193,06 g.
Answer:
Colourless
Explanation:
We know that Y^3+ has the electronic configuration of;
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 (the 5s and 4d levels are empty).
According to the crystal field theory, the colour of complexes result from transitions between incompletely filled d orbitals.
As a result of this, complexes with empty or completely filled d orbitals are colourless. Thus, [Y(H2O)6]3 is colourless according to the Crystal Field Theory.
Answer:
Mole fraction of solute is 0.0462
Explanation:
To solve this we use the colligative property of lowering vapor pressure.
First of all, we search for vapor pressure of pure water at 25°C = 23.8 Torr
Now, we convert the Torr to mmHg. Ratio is 1:1, so 23.8 Torr is 23.8 mmHg.
Formula for lowering vapor pressure is:
ΔP = P° . Xm
Where ΔP = P' (Vapor pressure of solution) - P° (Vapor pressure of pure solvent)
Xm = mole fraction
24.9 mmHg - 23.8 mmHg = 23mmHg . Xm
Xm = (24.9 mmHg - 23.8 mmHg) / 23mmHg
Xm = 0.0462
