In the given situation, the reaction is-
NO + H2 ↔ Products
The rate of the reaction can be expressed (in terms of the decrease in the concentration of the reactants) as-
Rate = -dΔ[NO]/dt = -dΔ[H2]/dt
Now, if the concentration of NO is decreased there will be fewer molecules of the reactant NO which would decrease the its collision with H2. As a result the rate of the forward reaction would also decrease.
Ans) A decrease in the concentration of nitrogen monoxide decreases the collisions between NO and H2 molecules. the rate of the forward reaction then decreases.
Answer:
6
Explanation:
Formula: Al2O3
If we require 2Al2O3
We divide 2 by 3
Sorry for the shadow of phone and fingers.
Option C: Sulfur Dioxide is the answer
Hope this helps
Answer:
pH = 5.54
Explanation:
The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:
- pH = pKa + log
![\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
For acetic acid, pKa = 4.75.
We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:
- CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
- CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH
The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.
Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:
- pH = 4.75 + log
= 5.54