Answer:
The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.
Explanation:
- To solve this problem, we use Clausius Clapeyron equation: ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂).
- The first case: P₁ = 1 atm = 760 torr and T₁ = 451.0 K.
- The second case: P₂ = <em>??? needed to be calculated</em> and T₂ = 61.5 °C = 334.5 K.
- ΔHvap = 48.8 KJ/mole = 48.8 x 10³ J/mole and R = 8.314 J/mole.K.
- Now, ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂)
- ln(760 torr /P₂) = (48.8 x 10³ J/mole / 8.314 J/mole.K) (1/451 K - 1/334.5 K)
- ln(760 torr /P₂) = (5869.62) (-7.722 x 10⁻⁴) = -4.53.
- (760 torr /P₂) = 0.01075
- Then, P₂ = (760 torr) / (0.01075) = 70691.73 torr.
So, The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.
Answer:
Kc = [Pb²⁺]³.[PO₄³⁻]²
Explanation:
Let's consider the following reaction at equilibrium.
Pb₃(PO₄)₂(s) ⇄ 3 Pb²⁺(aq) + 2 PO₄³⁻(aq)
The concentration equilibrium constant is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.
Kc = [Pb²⁺]³.[PO₄³⁻]²
This equilibrium constant is known as the solubility product of Pb₃(PO₄)₂.
Answer: The correct answer is option B.
Explanation:
- Ionic bond is formed when there is complete transfer of electrons from one element to another.
- Covalent bond is formed by the sharing of electrons between two elements.
- Metallic bond is formed when two metals combine together.
Electronic Configuration of Group VII-A : 
This group require 1 electron to attain stable electronic configuration.
Electronic configuration of Group I-A : 
This group will loose 1 electron to attain stable electronic configuration.
So, there will be complete transfer of electron from Group VII-A to Group I-A and hence, will form ionic bond between them.
Hence, the correct option is B.
Answer:
it shows that no atom have been gained or loosed in that reaction
