Answer : The value of equilibrium constant (K) is, 424.3
Explanation : Given,
Concentration of
at equilibrium = 0.067 mol
Concentration of
at equilibrium = 0.021 mol
Concentration of
at equilibrium = 0.040 mol
The given chemical reaction is:

The expression for equilibrium constant is:
![K_c=\frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
Now put all the given values in this expression, we get:


Thus, the value of equilibrium constant (K) is, 424.3
Answer:
202 L
Explanation:
Step 1: Write the balanced equation
C₆H₁₂O₆ + 6 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(l)
Step 2: Calculate the moles corresponding to 270 g of C₆H₁₂O₆
The molar mass of C₆H₁₂O₆ is 180.16 g/mol.
270 g × 1 mol/180.16 g = 1.50 mol
Step 3: Calculate the moles of CO₂ generated from 1.50 moles of glucose
The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of CO₂ formed are 6/1 × 1.50 mol = 9.00 mol
Step 4: Calculate the volume of 9.00 moles of CO₂ at STP
The volume of 1 mole of an ideal gas at STP is 22.4 L.
9.00 mol × 22.4 L/mol = 202 L
Answer:
Explanation:
a )
energy produced per second = 500 J
Heat produced = 500 x .8 = 400 J per second.
If m be the mass of water evaporated per unit hour
m x latent heat = 400 x 60 x 60
= m x 2.42 x 10⁶ = 1.44 x 10⁶
m = .595 kg per hour
b )
volume of water = 595 mL
bottles = 595 / 750
.8 or 4/5 of bottle. per hour.