Answer:
The new molar concentration of CO at equilibrium will be :[CO]=1.16 M.
Explanation:
Equilibrium concentration of all reactant and product:
![[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M](https://tex.z-dn.net/?f=%5BCO_2%5D%20%3D%200.24%20M%2C%20%5BH_2%5D%20%3D%200.24%20M%2C%20%5BH_2O%5D%20%3D%200.48%20M%2C%20%5BCO%5D%20%3D%200.48%20M)
Equilibrium constant of the reaction :
![K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BH_2O%5D%5BCO%5D%7D%7B%5BCO_2%5D%5BH_2%5D%7D%3D%5Cfrac%7B0.48%20M%5Ctimes%200.48%20M%7D%7B0.24%20M%5Ctimes%200.24%20M%7D)
K = 4

Concentration at eq'm:
0.24 M 0.24 M 0.48 M 0.48 M
After addition of 0.34 moles per liter of
and
are added.
(0.24+0.34) M (0.24+0.34) M (0.48+x)M (0.48+x)M
Equilibrium constant of the reaction after addition of more carbon dioxide and water:


Solving for x: x = 0.68
The new molar concentration of CO at equilibrium will be:
[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M
Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).
The new volume at standard pressure of 1 atm is 21294 liters.
Explanation:
Data given:
Initial volume of the gas V1 = 338 liters
initial pressure on the gas P1 = 63 atm
standard pressure as P2 = 1 atm
Final volume at standard pressure V2 =?
The data given shows that Boyle's law equation is to used:
P1V1 = P2V2
rearranging the equation to calculate V2,
V2 = 
Putting the values in the equation:
V2 = 
= 21294 L
as the pressure on the gas is reduced to 1 atm the volume of the gas increased incredibly to 21294 litres.
Given:
Mass, m = 51.1 g
Volume, V = 6.63 cm³
By definition,
Density = Mass/Volume
= (51.1 g)/(6.63 cm³)
= 7.7074 g/cm³
In SI units,
Density = (7.7074 g/cm³)*(10⁻³ kg/g)*(10² cm/m)³
= 7707.4 kg/m³
Answer: 7.707 g/cm³ or 7707.4 kg/m³
While I am not the brainliest I can certainly answer.
This was a chemical change because the chemical components were changed, a big giveaway to this was the fizzing, however the temperature rising was also another giveaway.