The Keq for the reaction N₂ + 3H2 = 2NH3 if the equilibrium concentrations are Keq = 1.5. The correct option is D.
<h3>What is Keq?</h3>
Keq is the ratio of the concentration of reactant to the concentration of the product.
The balanced equation is
N₂ + 3H₂ = 2NH₃
The equilibrium constant is ![\rm \dfrac{[NH_3]^2}{[N_2]\; [H_2]^3}](https://tex.z-dn.net/?f=%5Crm%20%5Cdfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5C%3B%20%5BH_2%5D%5E3%7D)
The given concentrations of the compounds have been:
Ammonia = 3 M
Nitrogen = 1 M
Hydrogen = 2 M

Thus, the correct option is D. Keq = 1.5.
Learn more about Keq
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I don’t think you can get it for
Answer:
Rate of formation of SO₃
= 7.28 x 10⁻³ M/s
Explanation:
According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)
Rate of disappearance of reactants = rate of appearance of products
⇒
-----------------------------(1)
Given that the rate of disappearance of oxygen =
= 3.64 x 10⁻³ M/s
So the rate of formation of SO₃
= ?
from equation (1) we can write
![\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D%20%3D%202%20%5B-%5Cfrac%7Bd%5BO_%7B2%7D%5D%20%7D%7Bdt%7D%20%5D)
⇒
= 2 x 3.64 x 10⁻³ M/s
⇒
= 7.28 x 10⁻³ M/s
∴ So the rate of formation of SO₃
= 7.28 x 10⁻³ M/s
3.0 × 10¹¹ RBC's (or) 3E11 RBC's
Solution:
Step 1: Convert mm³ into L;
As,
1 mm³ = 1.0 × 10⁻⁶ Liters
So,
0.1 mm³ = X Liters
Solving for X,
X = (0.1 mm³ × 1.0 × 10⁻⁶ Liters) ÷ 1 mm³
X = 1.0 × 10⁻⁷ Liters
Step 2: Calculate No. of RBC's in 5 Liter Blood:
As given
1.0 × 10⁻⁷ Liters Blood contains = 6000 RBC's
So,
5.0 Liters of Blood will contain = X RBC's
Solving for X,
X = (5.0 Liters × 6000 RBC's) ÷ 1.0 × 10⁻⁷ Liters
X = 3.0 × 10¹¹ RBC's
Or,
X = 3E11 RBC's
The steel rods will enable the concrete to form without any bumps and it will add shape to the cement and strength, so no odd massive lumps are formed.
That is what I think anyways :)