Answer:
A and D are true , while B and F statements are false.
Explanation:
A) True. Since the standard gibbs free energy is
ΔG = ΔG⁰ + RT*ln Q
where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R
when the system reaches equilibrium ΔG=0 and Q=Keq
0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)
therefore the first equation also can be expressed as
ΔG = RT*ln (Q/Keq)
thus the standard gibbs free energy can be determined using Keq
B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions
C) False. From the equation presented
ΔG⁰ = (-RT*ln Keq)
ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1
for example, for a reversible reaction ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)
D) True. Standard conditions refer to
T= 298 K
pH= 7
P= 1 atm
C= 1 M for all reactants
Water = 55.6 M
Physics<span> beyond the Standard </span>Model<span> (BSM) refers to the theoretical developments needed to ... Yet, the Standard </span>Model<span> does not supply any fundamental </span>particles<span> that are good dark matter candidates. .... compared to their mass—while formation of structures </span>similar<span> to the galaxies in our </span>universe<span>requires cold dark matter.</span>
Answer: option E. Evaporation to dryness
Explanation:
Answer:
-3.82ºC is the freezing point of solution
Explanation:
We work with the Freezing point depression to solve the problem
ΔT = m . Kf . i
ΔT = Freezing point of pure solvent - freezing point of solution
Let's find out m, molality (moles of solute in 1kg of solvent)
15 g / 58.45 g/mol = 0.257 moles of NaCl
NaCl(s) → Na⁺ (aq) + Cl⁻(aq)
i = 2 (Van't Hoff factor, numbers of ions dissolved)
m = mol /kg → 0.257 mol / 0.250kg = 1.03 m
Kf = Cryoscopic constant → 1.86 ºC/m (pure, for water)
0ºC - Tºf = 1.03m . 1.86ºC/m . 2
Tºf = -3.82ºC
