Answer:
![K=K_1*K_2\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}](https://tex.z-dn.net/?f=K%3DK_1%2AK_2%5C%5C%5C%5CK%3D%5Cfrac%7B%5BH_2%5D%5E3%5BCO_2%5D%5BH_2%5D%7D%7B%5BCH_4%5D%5BH_2O%5D%5BH_2O%5D%7D)
Explanation:
Hello there!
In this case, for the given chemical reaction, it turns out firstly necessary to write the equilibrium expression for both reactions 1 and 2:
![K_1=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \\\\K_2=\frac{[CO_2][H_2]}{[CO][H_2O]}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%20%5C%5C%5C%5CK_2%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D)
Now, when we combine them to get the overall expression, we infer these two are multiplied to get:
![K=K_1*K_2\\\\K=\frac{[CO][H_2]^3}{[CH_4][H_2O]} *\frac{[CO_2][H_2]}{[CO][H_2O]}\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}](https://tex.z-dn.net/?f=K%3DK_1%2AK_2%5C%5C%5C%5CK%3D%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%20%2A%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D%5C%5C%5C%5CK%3D%5Cfrac%7B%5BH_2%5D%5E3%5BCO_2%5D%5BH_2%5D%7D%7B%5BCH_4%5D%5BH_2O%5D%5BH_2O%5D%7D)
Regards!
Answer:
Below.
Explanation:
This is the combustion of the hydrocarbon propane.
C3H8 + 5O2 --> 3CO2 + 4H2O
They should report the density as 1.11 g/L
.
Density = mass/volume = 2.260g/2.04 mL
My calculator says the density <em>1.107 843 137 g/mL</em>
However, the answer can have <em>no more</em> significant figures than are in the number with the <em>fewest </em>significant figures.
The volume measurement has only three significant figures, so we must round off the density to three significant figures.
We drop all the digits after the zero.
The digit to be dropped is 7, so we <em>round up </em>the last significant figure of the answer.
1.10<u>7 843 137</u> → 1.11
Answer:
The solution is not ideal.
The relative strengths of the solute-solvent interactions are greater compared to the solute-solute and solvent-solvent interactions
Explanation:
The total vapor pressure is the sum of the partial pressures of water and methanol, and they are calculated by the Raoult´s law equation:
Pₐ = Xₐ Pºₐ, where Pₐ is the partial pressure of component A
Xₐ is the molar fraction of A
P⁰ₐ is the pressure of pure A
So lets calculate the partial pressures of methanol and water and compare them with the given total vapor pressure of solution:
X H2O = 0.312 ⇒ X CH3OH = 1 - 0.312 = 0.688
PH2O = 0.312 x 55.3 torr = 17.3 torr
PCH3OH = 0.688 x 256 torr = 176.1 torr
Ptotal = PH2O + PCH3OH = 17.3 torr + 176.1 torr = 193.4 torr
This pressure is less than the experimental value of 211 torr. So the solution is not ideal. The relative strength of the solute-solvent interactions are greater than the solute-solute and solvent-solvent interactions.
The reason for this is the presence of hydrogen bonding between methanol and water.
