So the answer is: no; 3Cu+8h++ 2 NO3-= 3c U2 ++ 2NO ↑+ 4H2O； The gas in the flask changed from colorless to red; Balloon inflation;
(2) A. Adding distilled water will slow down the reaction speed of nitric acid dilution, so A is wrong;
B. Sodium acetate will react with nitric acid to generate acetic acid, and the acid will change from strong acid to weak acid, and the reaction speed will be slow, so B is wrong;
C. The primary battery formed by graphite, metallic copper and nitric acid takes metallic copper as the negative electrode, which accelerates the reaction speed, so C is correct;
D. Adding AgNO3 _ 3 solution (a small amount), metallic copper will replace metallic silver. In the formed primary battery, metallic copper is the negative electrode, which speeds up the reaction, so D is correct;
So choose CD.
(3) Nitric oxide and nitrogen dioxide can be absorbed by water in a certain proportion to obtain nitric acid solution, so the gas is gradually reduced, and nitrogen oxides are converted into nitric acid to generate 4NO2+O2+2H2O ═ 4NO3, or 4NO+3O2+2H2O ═ 4NO3, so the answer is: NO2+O2+2H2O ═ 4NO3 (or 4NO
(4) ③ n (Cu) = 0.384g64g/mol = 0.006mol, from 3cu+8h+2no3-= 3cu2+2no = 4h2o, it can be known that the Cu oxidized by nitric acid is 0.01l×1.2mol/l×/kloc-.
According to 2Na2O2+2CO2═2Na2CO3+O2, 4NO+3O2+2H2O=4HNO3, 3cu+8h+2no3-= 3cu2+2no↑+4h2o, The total reaction is 2cu+4hno3+O2 ═ 2cu (NO3) 2+2h2o.n (O2) = 0.003mol× 22.4l/mol = 0.0672l = 67.2ml, where A contains air under standard conditions (oxygen accounts for one fifth) and the flask is 250mL, so oxygen accounts for (250ml).
Therefore, the answer is: 2na2o3+2co2 ═ 2n2co3+O2; 2cu+4hno3+o2═2cu(no3)2+2h2o； 38.4mL。