<h2>Acetic Acid</h2>
Explanation:
- Initial alcohol that is formed from methane is methanol that can be converted to acetic acid.
- Methanol in presence of
and hydrogen gas gets oxidized to acetic acid with the release of water.
Hence, compound A will be Acetic acid.
→ 
- The acetic acid formed can be transformed into -
→ 
The product formed is methane and carbon monoxide.
2. → 
The product formed is formaldehyde and water.
3. 
→ 
The product formed is sodium acetate, carbon dioxide, and water.
Answer:
2.52 g NaCl
Explanation:
(Step 1)
To find the mass, you first need to find the moles NaCl. This value can be found using the molarity ratio:
Molarity = moles / volume (L)
After you convert mL to L, you can plug the given values into the equation and simplify to find moles.
136.9 mL / 1,000 = 0.1369 L
Molarity = moles / volume
0.315 M = moles / 0.1369 L
0.0431 = moles
(Step 2)
Now, you can use the molar mass to convert moles to grams.
Molar Mass (NaCl): 22.990 g/mol + 35.453 g/mol
Molar Mass (NaCl): 58.443 g/mol
0.0431 moles NaCl 58.443 g
------------------------------ x ------------------- = 2.52 g NaCl
1 mole
Answer:
C more collision between molecules
Explanation:
increase in heat causes increase in kinetic energy of the particles
Answer:
Rate of formation of SO₃ ![[\frac{d[SO_{3}] }{dt}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D%5D)
= 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
⇒ ![-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BSO_%7B2%7D%20%5D%7D%7Bdt%7D%20%3D%20-%5Cfrac%7Bd%5BO_%7B2%7D%20%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BSO_%7B3%7D%20%5D%7D%7Bdt%7D)
-----------------------------(1)
Given that the rate of disappearance of oxygen = ![-\frac{d[O_{2} ]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BO_%7B2%7D%20%5D%7D%7Bdt%7D)
= 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)
⇒ ![\frac{d[SO_{3}] }{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D)
= 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
