Explanation:
1.
Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)
2.
Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)
A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.
3.
Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)
2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)
Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)
2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)
4.
The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.
Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.
Fluorine is a nonmetal and as such it would need to take in or obtain an electron from a metal to have a stable octet, or full number of 8 electrons in its valence shell. Thus due to the indifference of electrons and protons it becomes an anion, a negatively charged ion.
Answer: one molecule of O2.
Explanation: sweet i just took a guess but I believe that if 3 o2 molecules - 2 h2 molecules I think that its just basic maths and it is C because 3-2 = 1 and its o2 remaining, sorry if I’m wrong.
Answer:
No, there is not because it would form H2 instead of methane if hydrogen bonded with itself.
Explanation:
from the shape of methane which is tetrahedral it's evident there's no hydrogen bond only C-H bond.
