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
Neon I think. Go to the periodic table and see which one is the 11th
Answer:- 3. 
and 
Explanations:- An empirical formula is the simplest whole number ratio of atoms of each element present in the molecule/compound.
For example, the molecular formula of benzene is 
. The ratio of C to H in it is 6:6 that could be simplified to 1:1. So, an empirical formula of benzene is CH.
In the first pair, the ratio of C to H in first molecule is 2:4 that could be simplified to 1:2 and the empirical formula is 
. In second molecule the ratio of C to H is 6:6 and it could be simplified to 1:1. and the empirical formula is CH. Empirical formulas are different for both the molecules of first pair and so it is not the right choice.
In second pair, C to H ratio in first molecule is 1:2, so the empirical formula is . The C to H ratio for second molecule is 1:4, so the empirical formula is 
. Here also, the empirical formulas are not same and hence it is also not the right choice.
In third pair, C to H ratio in first molecule is 1:3, so the empirical formula is . In second molecule the C to H ratio is 2:6 and it is simplified to 1:3. So, the empirical formula for this one is also . Hence. this is the correct choice.
In fourth pair, first molecule empirical formula is CH. Second molecule has 2:4 that is 1:2 mole ratio of C to H and so its empirical formula is . As the empirical formulas are different, it is not the right choice.
So, the only and only correct pair is the third one. 3. and
1) Radioactive decay is the spontaneous decomposition of the unstable nucleus of an atom.
2) The emission of a particle or a photon.
For example, alpha decay is radioactive decay in which an atomic nucleus emits an alpha particle (helium nucleus).
3) The result is usually more stable element with smaller atomic number.
For example, in alpha decay atom transforms into an atom with an atomic number that is reduced by two and mass number that is reduced by four.
For example nuclear fission is radioactive decay process in which the nucleus of an atom splits into smaller parts and huge amount of energy is released.
Answer:
1552.83J Released
Explanation:
1. mass/m=225
Initial temp:86C, final:32.5C
Changed Temp: 32.5-86= -53.5C
s=0.129 J/gC
Formula: q= m times s times changed Temp.
q=(225)(0.129)(-53.5)
q= -1552.83 J
q=1552.83 J Released
