Answer:
ΔSv = 0.1075 KJ/mol.K
Explanation:
Binary solution:
∴ a: solvent
∴ b: solute
in equilibrium:
- μ*(g) = μ(l) = μ* +RTLnXa....chemical potential (μ)
⇒ Ln (1 - Xb) = ΔG/RT
∴ ΔG = ΔHv - TΔSv
⇒ Ln(1 -Xb) = ΔHv/RT - ΔSv/R
∴ Xb → 0:
⇒ Ln(1) = ΔHv/RT - ΔSv/R
∴ T = T*b....normal boiling point
⇒ 0 = ΔHv/RT*b - ΔSv/R
⇒ ΔSv = (R)(ΔHv/RT*b)
⇒ ΔSv = ΔHv/T*b
∴ T*b = 80°C ≅ 353 K
⇒ ΔSv = (38 KJ/mol)/(353 K)
⇒ ΔSv = 0.1075 KJ/mol.K
It is A. This is because the rate of reaction is defined as the speed at which the reactants are converted into the products.
Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v
There are 100 degrees between the freezing (0°) and boiling points (100°) of water on the Celsius scale and 180 degrees between the similar points (32° and 212°) on the Fahrenheit scale.
Problem: Two scientists are doing an experiment designed to identify the boiling point
Answer: 250°F is the higher temperature by 2°F
<span>If the concentration of H⁺ ions will decrease then the concentration of OH⁺ ions will increase.</span>
