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
The number of years required for 1/4 cobalt-60 to remain after decay is calculated as follows
after one half life 1/2 of the original mass isotope remains
after another half life 1/4 mass of original mass remains
therefore if one half life is 5.3 years then the years required
= 2 x 5.3years = 10.6 years
When the concentration is expressed in mass percentage, that means there is 3 g of solvent H₂O₂ in 100 grams of the solution. Then, that means the remaining amount of solute is 97 g. We use the value of molarity (moles/liters) to determine the amount of solution in liters, denoted as V. The solution is as follows:
0.02 mol KMnO4/L solution * 158.034 g KMnO4/mol * V = 97 g KMnO4
Solving for V,
V = 30.69 L
Answer:
1/45 or 0.0222 g/s
Explanation:
rate of reaction = quantity of reactant used/time taken
= (10-8)/90
= 1/45 or 0.0222 g/s
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Answer:
Heat required = 1.23×10⁵J
Explanation:
Given:
Mass (m) = 500 gram
Specific heat = 6,090 J/g
heat of fusion = 247 J/g.
Find:
Heat required
Computation:
Heat required = 247 J/g× 500 g
Heat required = 1.23×10⁵J
