Answer:
C. The intensity of 20F signal would be higher, but the lifetime would be unchanged
Explanation:
The half live value is constant, life term of an exponentially decaying quantity, and it a property of the exponential decay equation. Half life does not decay on the initial concentration. Hence, there won't be any change in observation.
C. The intensity of 20F signal would be higher, but the lifetime would be unchanged
The answer is osha ensure that employers have right to succeed in what they do best receive information
Answer is C.) Hope I could help
Answer:
(a) 151.84 kJ
(b) 2.922 kJ/K
Explanation:
(a) The parameters given are;
Mass of hydrogen gas, H₂ = 0.1 kg = 100 g
Molar mass of H₂ = 2.016 g/mol
Number of moles of H₂ = 100/2.016 = 49.6 moles
V₁ = mRT/P = 0.1×4.12×300/1000 = 0.1236 m³
P₁/P₂ = (V₂/V₁)^k
V₂ = (P₁/P₂)^(1/k)×V₁ =0.1236 × (1000/500)^(1/1.4) = 0.3262 m³
Boundary work done = (V₂ - V₁)(P₂ + P₁)/2 = (0.3262 - 0.1236)*(500 + 1000)/2 = 151.84 kJ
(b) Entropy generated ΔS = Cv · ㏑(T₂/T₁) + R ·㏑(v₂/v₁)
=10.18 × ㏑(270/300) + 4.12 ·㏑(0.3262/0.1236) = 2.922 kJ/K.