Answer:

Explanation:
Hello,
In this case, for the given reaction, the equilibrium constant turns out:
![Keq=\frac{[B]}{[A]}=\frac{0.5M}{1.5M} =1/3](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D%3D%5Cfrac%7B0.5M%7D%7B1.5M%7D%20%3D1%2F3)
Nonetheless, we are asked for the reverse equilibrium constant that is:

Which is greater than one.
In such a way, the Gibbs free energy turns out:

Now, since the reverse equilibrium constant is greater than zero its natural logarithm is positive, therefore with the initial minus, the Gibbs free energy is less than zero, that is, negative.
Okay so the answer to this one is very simple 91
Answer:
Mass of NH3 produced = 1217 g or 1.217*10^3 g
Explanation:
<u>Given:</u>
Mass of N2 = 1.003*10^3 g
Mass of H2 = 5.003*10^2 g
<u>To determine:</u>
Maximum mass of NH3 that can be produced when the given amounts of N2 and H2 combine
<u>Calculation:</u>
The chemical reaction corresponding to the production of ammonia is:

Based on the reaction stoichiometry:
1 mole of N2 combines with 3 moles of H2 to form 2 moles of NH3


Since the moles of N2 is less than that of H2, the limiting reagent will be N2 which would in turn determine the amount of NH3 formed.
Based on the reaction stoichiometry the N2 : NH3 ratio = 1:2
Therefore,

Answer:
The answer to your question is the letter d. S
Explanation:
Data
Change of +4 in the oxidation number
Chemical reaction
K₂Cr₂O₇ + H₂O + S ⇒ KOH + Cr₂O₃ + SO₂
Process
1.- Calculate the oxidation numbers following the rules.
Some rules
H = +1 O = -2 Alkali metals = + 1 Alkali earth metals = +2
K₂⁺¹Cr₂⁺⁶O₇⁻² + H₂⁺¹O⁻² + S⁰ ⇒ K⁺¹O⁻²H⁺¹ + Cr₂⁺³O₃⁻² + S⁺⁴O₂⁻²
Elements that changed their oxidation numbers
Cr₂⁺⁶ ---------------- Cr₂⁺³
S⁰ --------------- S⁺⁴
Tin(II) fluoride or stannous fluoride