The molarity of the solution is 1.0 M. This is calculated by dividing the number of moles of solute (0.5 mol Cu(NO3)2) by the volume of the solution in liters (0.5 L).
This reaction doesn't exist.
The answer is 3 moles of Cu)NO3)2..
The chemical name is copper nitrate
NaBr is a salt because it has Na in it CuNo32 may be a nonelectrolyte NH3 is a base KOH is a base CaCl2 may be a nonelectrolyte hope that somewhat helps
An aqueous solution of Cu(NO3)2 is neutral because the nitrate ion (NO3-) does not significantly contribute to acidity or basicity in water. However, copper ions (Cu2+) from Cu(NO3)2 can slightly hydrolyze in water to form Cu(OH)+ ions, which can make the solution slightly acidic.
The reaction between copper(II) nitrate (Cu(NO3)2) and sodium carbonate (Na2CO3) will form solid copper(II) carbonate (CuCO3) and soluble sodium nitrate (NaNO3) in solution. The balanced chemical equation for the reaction is: Cu(NO3)2 + Na2CO3 -> CuCO3 + 2NaNO3.
This reaction involves the reaction of copper (Cu) with silver nitrate (AgNO3) to form copper(II) nitrate (Cu(NO3)2) and silver (Ag). It is a chemical reaction that was likely conducted in a laboratory setting or for academic purposes.
The molecular equation for Cu(NO3)2 and Zn is Cu(NO3)2 + Zn -> Zn(NO3)2 + Cu. The total ionic equation is Cu^2+ + 2NO3- + Zn -> Zn^2+ + 2NO3- + Cu. The net ionic equation is Cu^2+ + Zn -> Zn^2+ + Cu.
If something being reduced, it loses an oxygen, and if something is being oxidized, it gains an oxygen. The element being reduced is called the oxidizer; the element being oxidized is called the reducer. (just remember, it's called the opposite of what happens to it) In the context of your equation, which i assume to be CuCl2+KNO3-->Cu(NO3)2+2KCl, potassium (or the K) is losing oxygen, which therefore makes that the oxidizer, the one being reduced.
No. In concentrated nitric acid, the reaction produces copper (II) nitrate, nitrogen dioxide (a toxic, brownish colored gas) and water. The reaction is as follows: Cu(s) + 4HNO3(aq) --> Cu(NO3)2(aq) + 2NO(g) + 2H2O(l) In diluted nitric acid, the reaction produces copper (II) nitrate, nitric oxide (NO), and water. The reaction is as follows: 3Cu(s) + 8HNO3(aq) --> 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l) (Source: www.angelo.edu)