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
1) As can be seen from any 1H NMR chemical shift ppm tables, hydrogens which have δ values from 2ppm to 2.3ppm are hydrogens from carbon which is bonded to a carbonyl group. From this, we can conclude that our hydrogens belong to the type, but from 2 different alkyl groups because of 2 different signals.
2) So, one alkyl group is CH3 and second one can be CH or CH2.
3) If we know that ratio between two types of hydrogens is 3:2, it can be concluded that second alkyl group is CH2.
4) Finally, we don't have any other signals and it indicates that part of the compound which continues on CH2 is exactly the same as the first part.
The ratio remains the same, 3:2 ie 6:4
Answer:
Endothermic
Explanation:
Because Endothermic is cold and exothermic is hot and if you are using an ice pack it would be Endothermic and if you were using something that was hot it would be exothermic
Answer:
The balanced equation for methanol when is burned in the air, is
CH3OH + O2 -----> 3/2 CO2 +2 H2O and as you see coefficient of oxygen is 3/2
Explanation:
When you always burn something you are doing combustion. The reactives are your compound + O2, and as products you have CO2 and H2O
