Answer:
see below
Explanation:
The rate constant is missing in question, but use C(final) = C(initial)e^-kt = 0.200M(e^-k·10). Fill in k and compute => remaining concentration of reactant
Answer:
1. 48 mols
2. 0.2 M
5. 1.25 L
Explanation:
Molarity= mols divided by liters
Hope this helps not sure about 3 and 4
Answer:
Cu(NO3)2(aq)+Pb(s) ⇌ Pb(NO3)2(aq)+Cu(s)
Explanation:
If we look at the both reactions closely, we will quickly discover that the reaction CuSO4(aq)+Pb(s) ⇌ PbSO4(s)+Cu(s) involves PbSO4.
The compound PbSO4 is insoluble in water and sinks to the bottom of the reaction vessel. When this occurs, the concentration of Pb^2+ becomes low. This will bring about a low voltage in the cell.
On the other hand, Pb(NO3)2 is soluble in water hence the cell voltage in this case is higher than the former.
Answer:
Rubidium
Rubidium is the first element placed in period 5.
1. The third option is the least soluble in water because it is the chain with the most number of hydrocarbons. Next is the second option while the first one is the most soluble.
2. Statements 1 and 2 are true. The third option is not true all the time because it depends on the structure of the compound.
