How do know what happens when #A.# #"calcium nitrate"# is added to #"sodium carbonate"#, and #B.# #"sodium chloride"# is added to #"copper nitrate"#?

Answer 1

How else do you predict but by reference to PRIOR experiment?

A chemical reaction takes place when a salt, like common salt, dissolves in water:

#NaCl(s) rarr Na^(+) + Cl^(-)#
The ionic species are the aquated ions; which we could represent as #[Na(OH_2)_6]^(+)# or #[Cl(H_2O)_(4-6)]^(-)#. Now we conceive these species to float around as discrete particles in aqueous solution. Add some silver salt to solution, say as #AgNO_3#, likewise we get #Ag^(+)# or #NO_3^(-)# (again as the aquated ions), the silver ions react irreversibly with the chloride ions to form #AgCl(s)# as a curdy white precipitate.
#Ag^(+) + Cl^(-) rarr AgCl(s)darr#
So for your reactions (#A.#):
#Ca(NO_3)_2(aq) + Na_2CO_3(aq) rarr CaCO_3(s)darr + 2NaNO_3(aq)#

Alternatively, the net ionic equation

#Ca^(2+) + CO_3^(2-) rarr CaCO_3(s)darr#

It is reasonable to say that calcium carbonate is insoluble in water and separates from solution. However, this is a matter best left to experimentation.

For #B.#, metathesis (which is only #"conceived"# to occur) gives #"potential"# salts that are both soluble. I write #"potential"# in that we don't know a priori that the salts are soluble; this is the province of experiment.
#NaCl(aq) + Cu(NO_3)_2(aq) rarr Na^(+) + Cl^(-) +Cu^(2+) + 2NO_3^(-)#

The partner exchange is conceptual rather than real. Ionization is undoubtedly occurring; that is, discrete sodium, chloride, cupric, etc. ions are in solution; however, no precipitation of any salt occurs because an insoluble salt is NOT formed.

Does this make sense? If not, I (as well as others) will be happy to provide feedback, clarifications, and/or corrections.

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Answer 2

A. When calcium nitrate is added to sodium carbonate, a double displacement reaction occurs, resulting in the formation of calcium carbonate and sodium nitrate.

B. When sodium chloride is added to copper nitrate, a single displacement reaction takes place, leading to the formation of copper chloride and sodium nitrate.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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