Answer:
a=28600J; b=90.6 J/K; c=402 torr
Explanation:
(a) considering the data given
Vapour pressure P1 =0 at Temperature T1 = 42.43˚C,
Vapour pressure P2 = 273.15 at Temperature T2= 315.58 K)
Using the Clausius-Clapeyron Equation
ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
In 760/140 = ΔH/8.314 J/mol/K × (1/315.58K -- 1/273.15K)
ΔH vap= +28.6 kJ/mol or 28600J
(b) using the Equation ΔG°=ΔH° - TΔS to solve forΔS.
Since ΔG at boiling point is zero,
ΔS =(ΔH°vap/Τb)
ΔS = 28600 J/315.58 K
= 90.6 J/K
(c) using ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
ln P298 K/1 atm = 28600 J/8.314 J/mol/K × (1/298.15K - 1/315.58K)
P298 K = 0.529 atm
= 402 torr

Actually Welcome to the Concept of the Lens and Ray Optics.
We must understand the Concept that,
1.) When a beam of parallel light pass through a Convex lens, all the rays get deviated and converges to a special point at the axis that is a Focus.
2.) When a beam of light passes through the Concave lens,the light beams diverge from one another forming a virtual focus in the primary side of lens,
Hence the above given diagram beam rays are diverging, it is a CONCAVE LENS
answer is C.) Concave lens
Answer:
e- 7.25 x 10³.
Explanation:
∵ ΔG = -RTlnK,
where, ΔG is the free energy change.
R is the general gas constant (R = 8.324 J/mol.K).
K is the equilibrium constant of the reaction.
- For the reaction: <em>N₂(g) + 3H₂(g) → 2NH₃(g),</em>
K = (PNH₃)²/(PN₂)(PH₂)³ = (0.65)²/(1.9)(1.6)³ = 5.43 x 10⁻².
∵ ΔG = -RTlnK.
∴ ΔG = -(8.314 J/mol.K)(298 K) ln(5.43 x 10⁻²) = 7.218 x 10³ J/mol.
A rod of a neutron-absorbing substance used to vary the output power of a nuclear reactor.
The relative mass of the sample : 18.94
<h3>Further explanation
</h3>
The elements in nature have several types of isotopes
Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.
So Isotopes are elements that have the same Atomic Number (Proton)
Atomic mass is the average atomic mass of all its isotopes
Mass atom X = mass isotope 1 . % + mass isotope 2.%
F-19 = 94% of the sample
F-18 = 100%-94%=6% of the sample
The relative mass :
