Answer: The volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L
Explanation:
According to ideal gas equation:
P = pressure of gas = 1 atm (at STP)
V = Volume of gas = ?
n = number of moles = 0.684
R = gas constant =
T =temperature = (at STP)
Thus the volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L
Answer:
0.008945 atm
Explanation:
In the reaction:
2H2S(g) ⇌ 2 H2(g) + S2(g)
Kp is defined as:
<em>Where P is the pressure of each compound in equilibrium.</em>
If initial pressure of H2S is 3.00atm, concentrations in equilibrium are:
H2S = 3.00 atm - 2X
H2 = 2X
S2: = X
Replacing:
0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶
Solving for X:
X = 0.008945 atm
As in equilibrium, pressure of S2 is X, <em>pressure is 0.008945 atm</em>