Answer:
Explanation:
When the amount of H2O2 is doubled while KI is kept constant, the rate of reaction doubles.
When the amount of KI is doubled and the amount of H2O2 is halved, the rate stays nearly constant.
2H2O2 (aq) → O2(g) + 2H2O (l) ------------- first order kinetics reaction.
Catalysts are KI, FeCl3 only, KCl is not a catalyst. Order: KI < MnO2 < Pb < FeCl3.
H2O2 + I– -> IO– + H2O (Step 1)
H2O2 + IO– -> I– + H2O + O2 (Step 2)
It can be seen that the iodine ion (provided by the KI solution) is a product as well as a reactant.
02(g)2Fe? (aq) + 2 H(a) 2 H 2 Fe3 (aq) H2O2(aq) + 2 Fe,Taq) H02(aq) 2 Fe (aq) 2 H (aq)
Answer:
I am looking for this as well. Can someone please help!
Explanation:
Answer:
Universal indicator can show us how strongly acidic or alkaline a solution is, not just that the solution is acidic or alkaline. This is measured using the pH scale , which runs from pH 0 to pH 14.
Explanation:
~Hope this helps
The balanced chemical reaction is:
<span>C4H8O2 + 2H2--> 2C2H6O
</span>A. How many grams of ethyl alcohol are produced by reaction of 2.7 mol of ethyl acetate with H2? 2.7 mol C4H8O2 ( 2 mol C2H6O / 1 mol C4H8O2) (46.07 g / 1 mol) = 248.78 g ethyl alcohol
B. How many grams of ethyl alcohol are produced by reaction of 13.0g of ethyl acetate with H2?
13.0 g C4H8O2 ( 1 mol / 88.11 g) ( 2 mol C2H6O / 1 mol C4H8O2) (<span>46.07 g / 1 mol) = 13.59 g ethyl alcohol</span>
C. How many grams of H2 are needed to react with 13.0g of ethyl acetate?
13.0 g C4H8O2 ( 1 mol / 88.11 g) ( 2 mol H2 / 1 mol C4H8O2) (2.02 g / 1 mol) = 0.5961 g H2
PbCl₂(s) ⇄ Pb²⁺(aq) + 2Cl⁻(aq)
Pb(CH₃COO)₂(s) → Pb²⁺(aq) + 2CH₃COO⁻(aq)
At lead acetate disolution, concentration of lead cations will increase. According to Le Chatelie's principle equilibrium will be displaced towards formation of solid lead chloride.
