When Lead (II) acetate and Hydrogen sulfide react, they form Lead sulfide and Acetic acid. The reaction is a reduction-oxidation (redox) reaction.
The balanced chemical reaction is this:
Pb(C2H3O2)2 + H2S --> PbS + 2C2H4O2
And the net ionic reaction is this:
Pb2+ + S2- --> PbS
Answer:
KO is the limiting reactant.
0.11 mol O₂ will be produced.
Explanation:
4 KO₂ + 2 H₂O ⇒ 4 KOH + 3 O₂
Find the limiting reagent by dividing the moles of the reactant by the coefficient in the equation.
(0.15 mol KO₂)/4 = 0.0375
(0.10 mol H₂O)/2 = 0.05
KO₂ is the limiting reagent.
The amount of product produced depends on the limiting reagent. To find how much is produced, take moles of limiting reagent and multiply it by the ratio of reagent to product. You can find the ratio by looking at the equation. For every 4 moles of KO₂, 3 moles of O₂ are produced.
0.15 mol KO₂ (3 mol O₂)/(4 mol KO₂) = 0.1125 mol O₂
0.11 mol O are produced.
Answer:
Products will be K2SO4 and Cu(NO3)2. A precipitate will not form.
Explanation:
Double replacement reactions are the switching of cations between two compounds. So in this case, the K+ and Cu 2+ cations will switch places.
KNO3 + CuSO4 -> K2SO4 + Cu(NO3)2
According to solubility rules, any substances containing NO3 or a Group 1 ion (K+ in this case) will be definitely soluble. There are a few exceptions to the Group 1 rule but it does not apply here. Potassium sulfate (K2SO4) contains a Group 1 ion so it will be soluble and not be a precipitate. And copper nitrate (Cu(NO3)2) contains nitrate so it will also be soluble and not be a precipitate.
Answer:
<h3>The answer is 5.0 g/mL</h3>
Explanation:
The density of a substance can be found by using the formula

From the question
mass = 100 g
volume = 20 mL
So we have

We have the final answer as
<h3>5.0 g/mL</h3>
Hope this helps you