Answer:
The equilibrium pressure of NO2 is 0.084 atm
Explanation:
Step 1: Data given
A reaction mixture initially contains 0.86 atm NO and 0.86 atm SO3.
Kp = 0.0118
Step 2: The balanced equation
NO( g) + SO3( g) ⇌ NO2( g) + SO2( g)
Step 3: The initial pressures
p(NO) = 0.86 atm
p(SO3) = 0.86 atm
p(NO2) = 0 atm
p(SO2) = 0 atm
Step 4: The pressure at the equilibrium
For 1 mol NO we need 1 mol SO3 to produce 1 mol NO2 and 1 mol SO2
p(NO) = 0.86 -x atm
p(SO3) = 0.86 -xatm
p(NO2) = x atm
p(SO2) = x atm
Step 5: Define Kp
Kp = ((pNO2)*(pSO2)) / ((pNO)*(pSO3))
Kp = 0.0118 = x²/(0.86 - x)²
X = 0.08427
p(NO) = 0.86 -0.08427 = 0.77573 atm
p(SO3) = 0.86 -0.08427 = 0.77573 atm
p(NO2) = 0.08427 atm
p(SO2) = 0.08427 atm
The equilibrium pressure of NO2 is 0.08427 atm ≈ 0.084 atm
NH₃:
N = 8*10²²
NA = 6.02*10²³
n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7*10²²
NA = 6.02*10²³
n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol
4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant
N₂:
n = 0.0665mol
M = 28g/mol
m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>
Answer
:
Flammable substances
Explanation
:
<em>Flammable substances</em> will catch fire and continue to burn when they contact an ignition source like a spark or a flame.
For example, <em>methanol</em> is a flammable liquid.
A flammable solid may also catch fire through friction. <em>Matches</em> are flammable solids.
Answer:
0.189 g.
Explanation:
- This problem is an application on <em>Henry's law.</em>
- Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
- Solubility of the gas ∝ partial pressure
- If we have different solubility at different pressures, we can express Henry's law as:
<em>S₁/P₁ = S₂/P₂,</em>
S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm
S₂ = ??? g/L and P₂ = 5.73 atm
- So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.
<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>
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