Answer:
The answer to this is
Unsaturated solution
Explanation:
An unsaturated solution has the property of having a solute concentration lower than the the solubility at equilibrium at a given temperature hence it has the capacity to dissolve more solutes. Is is a solution containing a lower amount of solute than a saturated solution
The two processes that occur on dissolving a solute in a solvent are dissolution and crystallization and in an unsaturated solution the rate of dissolution is greater than the rate of crystallization
Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108
A change of 1 Kelvin is exactly the same as a change of 1 degree Celsius.
Answer:
B) 7.7
Explanation:
For the reaction Ag2CO3(s) + CrO42‒(aq) → Ag2CrO4(s) + CO32‒(aq)
Kc = (CO₃²⁻) / (CrO₄²⁻)
and the Ksp given are
Ag₂CO₃ ⇒ 2 Ag⁺(aq) + CO₃²⁻(aq) Ksp₁ = (Ag⁺)²(CO₃²⁻)
Ag₂CrO₄ ⇒ 2 Ag⁺(aq)+ CrO₄²⁻(aq) Ksp₂ = (Ag⁺)²(CrO₄²⁻)
Where (...) indicate concentrations M
Notice if we divide the expressions for Ksp we get:
Ksp₁/Ksp₂ = (CO₃²⁻) / (CrO₄²⁻) = 8.5 x 10⁻¹² / 1.1 x 10⁻¹² = 7.7
which is the desired answer.
Even though Hydrogen is originally in group 1, based on this property, we can say it is in group 6.
Because:
Group 6 would mean that it only needs 2 more valence electrons till the octet (8 valence electrons). This would make it reactive, yet, in normal conditions, unlike group 7.
