Answer : The partial pressure of 
at equilibrium is, 1.0 × 10⁻⁶
Explanation :
The partial pressure of 
= 
The partial pressure of = 
The partial pressure of 
=
The balanced equilibrium reaction is,
Initial pressure 1.0×10⁻² 2.0×10⁻⁴ 2.0×10⁻⁴
At eqm. (1.0×10⁻²-2p) (2.0×10⁻⁴+p) (2.0×10⁻⁴+p)
The expression of equilibrium constant 
for the reaction will be:
Now put all the values in this expression, we get :
The partial pressure of at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶
Therefore, the partial pressure of at equilibrium is, 1.0 × 10⁻⁶
I would say B because c and d would decrease competition and a would do the same, or just kill the ecosystem.
Hope this helps and don't forget to hit that heart :)
<span>The smallest unit of an element that retains all the characteristics of that element is called an A. atom.
Atoms consist of protons, electrons, and neutrons, but they are particles and don't have the characteristics of the element which is why D is incorrect. B and C are not the smallest units - atoms are smaller than them.
</span>
Answer:
D. 4.75 m/s west
Explanation:
You can immediately get the answer once you know the formula for<em> velocity.</em>
- The formula for velocity is: Velocity (V) =
The distance is 95 meters while the time is 20 seconds. All you have to do is to divide <u>95 meters by 20 seconds</u>.
Let's solve:
- V =
- V =
- V = 4.75 meters per second
Since Brenda is going west with her her skateboard, then you have to add <em>west </em>as her direction.
Brenda's velocity is 4.75 m/s west.
Answer:
Change in entropy for the reaction is
ΔS° = -268.13 J/K.mol
Explanation:
To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
From Literature.
S°(NO₂) = 240.06 J/K.mol
S°(H₂O) = 69.91 J/K.mol
S°(HNO₃) = 155.60 J/K.mol
S°(NO) = 210.76 J/K.mol
These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.
Note that
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]
= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol
Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]
= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
ΔS° = 521.96 - 790.09 = -268.13 J/K.mol
Hope this Helps!!
