The octet rule does not apply to transition metals.
The stock name for Cu2+ is copper(II).
Cu2+ is a cation as are all positive ions. One way to remember is to think of the "t" in cation as a plus sign.
To find the molarity of Cu2+ ions, first calculate the moles of Cu2+ from the given mass of copper. Next, use the total volume of the solution to calculate the molarity. The molarity of Cu2+ ions in the solution is 0.377 M.
In the cell, the half-reaction for silver will be Ag+ (aq) + e- -> Ag (s) with a standard reduction potential of +0.80 V. The half-reaction for copper will be Cu2+ (aq) + 2e- -> Cu (s) with a standard reduction potential of +0.34 V. The silver half-reaction will occur at the cathode, while the copper half-reaction will occur at the anode in the cell.
The name of Cu^2+ or (Cu2+) is called: "The name of Cu2+ is _________ ion or ________ ion" 1st blank is: copper(II) 2nd blank is: cupric It is correct. i am a teacher.
The stock name for Cu2+ is copper(II).
Cu2+ + I- --> Cu2I The compound created is Copper(I) Iodide
Cu(s) | Cu2+(aq) Au+(aq) | Au(s)
Cu + Mg2 --------> Cu2 + Mg Cu --------------> Cu2 + 2e Mg2 + 2e --------> Mg Cu --------------> Cu2 + 2e (E = +0.35) Mg2 + 2e --------> Mg (E = -2.36V) +0.35 + (-2.36) = -2.01V --------------------------------------… Mg + Cu2 --------> Mg2 + Cu Mg --------------> Mg2 + 2e Cu2 + 2e --------> Cu Mg --------------> Mg2 + 2e (E = +2.36V) Cu2 + 2e --------> Mg (E = -0.35V) +2.36 + (-0.35) = +2.01V
Cu2+ + O2- + H+ + Cl- = Cu2+ + 2 Cl- + H2O
E2xoy4=cu8
Cu2+ is a cation as are all positive ions. One way to remember is to think of the "t" in cation as a plus sign.
To find the molarity of Cu2+ ions, first calculate the moles of Cu2+ from the given mass of copper. Next, use the total volume of the solution to calculate the molarity. The molarity of Cu2+ ions in the solution is 0.377 M.
29 protons and 27electrons are present in Cu2+ ion.
Cu2+ has 29 electrons (since copper has 29 protons) and 27 protons, after losing 2 electrons to become a positively charged ion.
In the biuret test, the complex formed between Cu2+ and protein is a coordination complex where the peptide bonds in the protein act as ligands to chelate with the Cu2+ ions. This complexation causes a color change, typically from blue to violet or purple, indicating the presence of proteins in the sample.
This is an acid-base reaction or proton exchange, but there are many more reaction possibillities depending of Cu2+, OH- and NH3 concentration