4HCl + MnO₂ → MnCl₂ + 2H₂O + Cl₂
2Cl⁻ → Cl₂ + 2e⁻
MnO₂ + 4H⁺ + 2e⁻ → Mn²⁺ + 2H₂O
2Cl⁻ + MnO₂ + 4H⁺ → Cl₂ + Mn²⁺ + 2H₂O
Answer:
cis-hex-2-ene as produced as major product.
Explanation:
- In presence of (Lindlar catalyst) and , alkynes are reduced to alkene.
- molecules first adsorbed onto metal surface through chemisorption.
- When alkyne molecule comes in close proximity to the metal surface, H atoms add onto triple bond to reduce it to alkene.
- Two H atoms add onto alkyne from same face of metal. So, (Z) or cis-isomer of alkene is produced as a major product.
- Reaction mechanism has been shown below.
Answer:
I believe the answer would be D
Explanation:
The reason why is that I divided 5.75 with 5 which results in 1.15
Answer:
gas at s.t.p when 5g of zinc trioxocarbonate is heated strongly.
Answer:
pH = 12.6
Explanation:
The HF reacts with LiOH as follows:
HF + LiOH → LiF + H₂O
To solve this question we need to find the moles of each reactant:
<em>Moles HF:</em>
0.1500L * (0.20mol / L) = 0.030 moles HF
<em>Moles LiOH:</em>
0.600L * (0.10mol / L) = 0.060 moles LiOH
That means there is an amount of LiOH in excess, that is:
0.060 mol - 0.030 mol = 0.030 moles LiOH
In 600.0mL + 150.0mL = 750.0mL = 0.750L
The molarity of LiOH is:
0.030 moles LiOH / 0.750L =
0.040M LiOH = [OH⁻]
As:
Kw = 1x10⁻¹⁴ = [H⁺] [OH⁻]
1x10⁻¹⁴ = [H⁺] [0.040M]
2.5x10⁻¹³M = [H⁺]
As pH = -log [H⁺]
<h3>pH = 12.6</h3>