<u>Answer:</u>
"Boyle's Law" is based on the graph that is shown below.
<u>Explanation:</u>
Boyle's law or Boyle – Mariotte law or Mariotte's law, is an experimental gas law that discusses how a gas's pressure tends to rise as the container volume start declining. This shows the relationship between pressure and volume for a fixed mass at a constant temperature, i.e., number of a gas molecules.This rule visualizes the actions of gas molecules in a confined space. This law can be understood from following equation:
p₁V₁ = p₂V₂
Above the product of the initial volume and pressure is equal to the product of the volume and pressure after a change.
Answer:
B)−6,942 J
/mol
Explanation:
At constant temperature and pressure, you cand define the change in Gibbs free energy, ΔG, as:
ΔG = ΔH - TΔS
Where ΔH is enthalpy, T absolute temperature and ΔS change in entropy.
Replacing (25°C = 273 + 25 = 298K; 25.45kJ/mol = 25450J/mol):
ΔG = ΔH - TΔS
ΔG = 25450J/mol - 298K×108.7J/molK
ΔG = -6942.6J/mol
Right solution is:
<h3>B)−6,942 J
/mol</h3>
Answer:
c. contraction
Explanation:
Heating will cause substances to expand, or change their state (like solid to liquid) or it may be a chemical reaction.
CH4 is <u>not</u> soluble in water
whereas CH3OH <u>is</u> soluble in water.
Answer:
Kc = 3.90
Explanation:
CO reacts with
to form
and
. balanced reaction is:

No. of moles of CO = 0.800 mol
No. of moles of
= 2.40 mol
Volume = 8.00 L
Concentration = 
Concentration of CO = 
Concentration of
= 

Initial 0.100 0.300 0 0
equi. 0.100 -x 0.300 - 3x x x
It is given that,
at equilibrium
= 0.309/8.00 = 0.0386 M
So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M
At equilibrium
= 0.300 - 0.0386 × 3 = 0.184 M
At equilibrium
= 0.0386 M
![Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2O%5D%5BCH_4%5D%7D%7B%5BCO%5D%5BH_2%5D%5E3%7D)
