What is the ∆g for the following reaction under standard conditions (t = 298 k) for the formation of nh4no3(s)?
2nh3(g) + 2o2(g) nh4no3(s) + h2o(l)

given:
nh4no3(s): ∆hf = -365.56 kj ∆sf = 151.08 j/k.
nh3(g): ∆hf = -46.11 kj ∆sf = 192.45 j/k.
h2o(l): ∆hf = -285.830 kj ∆sf = 69.91 j/k.
o2(g): ∆hf = 0.00 kj ∆sf = 205 j/k.

a.) 186.6 kj
b.) 6.9 kj
c.) -10.4 kj
d.) -126.3 kj
e.) -382 kj

Use the formula for ΔHrxn= (sum product) - (sum reactants)
and formula for ΔSrxn= (sum product) - (sum reactants)
and ΔG= ΔH-TΔS

there are different ways to solve this question, but I think the simplest is to find the ΔG of each one of the substance and solve for ΔG as (sum product) - (sum reactants). also because the answers are in Kj, you need to convert the ΔS to Kj by simply dividing the value by 1000. 

ΔG= ΔH-TΔS

ΔGNH4NO3= (-365.56) - (298 X 151.08x10-3)= -410.58
ΔG NH3= (-46.11) - (298 X 192.45x10-3)= -103.46
ΔG H2O= (-285.830) - (298 X 69.91x10-3)= -306.66
ΔG O2= (0) - (298 X 205x10-3)= -61.09

ΔGrxn= product- reactants

ΔGrxn= [∆Gf(NH4NO3) + ∆Gf(H2O)] - [2 ∆Gf(NH3) + 2 ∆Gf(O2)] 
ΔGrxn= [ -410.58 + -306.66] - [ (2 x -103.46) + (2 x -61.09)]= -388.14 kj

the answers is closest to E

I’d have to say no because without oxygen in the water the fish would all be dead. so if you want to fish for dead fish, pick the ponds with the lowest oxygen concentrations.

If a hydrogen atom is neutral, it must also have  one electron. an oxygen atom (o) has an atomic number of eight (8). this means a neutral oxygen atom has eight protons and eight electrons. the element actium (ac) has an atomic number of 89, so it has 89 protons and 89 electrons in a neutral atom