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Question:a) 2AgNO3 (aq) + Na2SO4 (aq) ---> How would i write the ionic and net ionic equation for this without knowing the charge for Ag (silver)? I'm really lost =[ can someone give me an explaination as well? that would be really helpful!

Answers:Ag is +1 and Na is also +1 (remember HNO3 where H is +1)

Question:Write a balanced formula equation, complete ionic equation, and net ionic equation for the reaction between: (Be sure to include phases.) a. an alkaline earth salt and sulfuric acid, being sure to identify the precipitate. b a halogen with a less active halide, being sure to identify which is oxidized and reduced.

Answers:(A) Among the alkaline earth metals you can chose the soluble salts of calcium, strontium or barium as one of the reactants, because the sulfates of these salts are slightly soluble. The least soluble one is barium sulfate and suppose we choose it as the product. All nitrates of metals are soluble, therefore we can choose barium nitrate as the reactant. Formula equation; Ba(NO3)2(aq) + H2SO4(aq) -------> BaSO4(s) + 2HNO3(aq) Ionic equation: Ba^2+(aq) + 2NO3^-(aq) + 2H^+(aq) + SO4^2-(aq) -------> BaSO4(s) + 2H^+(aq) + 2NO3^-(aq) Net ionic equation: (obtained by eliminating the spectator ions from both sides) Ba^2+(aq) + SO4^2-(aq) -------> BaSO4(aq) (B) Activity of halogens decreases from top to bottom within the group ( F > Cl > Br > I ) In the elemental state all halogens are diatomic molecules. F2 and Cl2 are gases, Br2 is liquid and I2 is solid. F2 replaces all other halogens. Cl2 replaces Br2 and I2. Br2 can only replace I2. Since I2 is the least active one it cannot replace any halogen. Formula equation; Cl2(g) + 2NaBr(aq) -------> 2NaCl(aq) + Br2(l) (note: all salts of sodium, potassium and ammonium are soluble) Ionic equation: Cl2(g) + 2Na^+(aq) + 2Br^-(aq) ------> 2Na^+(aq) + 2Cl^- (aq) + Br2(l) Net ionic equation: Cl2(g) + 2Br^-(aq) ------>2Cl^- (aq) + Br2(l) As it is clearly seen from the net ionic equation, Cl2 is reduced from 0 to -1 and Br^- is oxidized from -1 to 0.

Question:Write overall, total ionic and net ionic equations when two beakers containing the following solutions are mixed. Include phases for each species. A. Acid Base Reaction: hydrochloric acid and potassium hydroxide. B. Acid-Base Reaction: ammonia and nitric acid. C. Activity Series: Mn(s) and hydrochloric acid(aq). I'm pretty clueless as to where to begin with these... can you explain to me? Thank you!

Answers:A. Hydrochloric acid is a strong acid and potassium hydroxide is a strong base.. In a solution; - A strong acid completely ionizes. HCl(aq) + H2O(l) -----> H3O+(aq) + Cl-(aq) or HCl(aq) -----> H+(aq) + Cl-(aq) - A strong base completely dissociates. KOH(aq) --------> K+(aq) + OH-(aq) Strong acid - strong base reactions produces a slightly ionizable water. Overall: HCl(aq) + KOH(aq) -------> KCl(aq) + H2O(l) Total ionic: H+(aq) + Cl-(aq) + K+(aq) + OH-(aq) ----> K+(aq) + Cl-(aq) + H2O(l) Net ionic: H+(aq) + OH-(aq) ---->H2O(l) B. Ammonia is a weak base and nitric acid is a strong acid. In a solution; - A strong acid completely ionizes. HNO3(aq) -----> H+(aq) + NO3-(aq) - A weak base cannot ionize completely. NH3(aq) + H2O(l) <-----> NH4+(aq) + OH-(aq) Although this equation looks similar to the ionization of HCl, the remarkable difference is the shapes of the arrows representing the equations. -----> represents the complete ionization <----> represents the partial ionization (equilibrium reaction) The other difference is not visible, but it is a fact that the extent of the ionization cannot exceed 5%. Therefore, OH-(aq) cannot represent a weak base. Overall: NH3(aq) + HNO3(aq) ----> NH4NO3(aq) + H2O(l) Total ionic: NH3(aq) + H+(aq) + NO3-(aq) ---> NH4+(aq) + NO3-(aq) + H2O(l) Net ionic: NH3(aq) + H+(aq) + ---> NH4+(aq) + H2O(l) C. To complete such oxidation - reduction reactions, the activities of metals should be known. Manganese (Mn) is more active than H2. In other words, Mn(s) can reduce H+ to H2(g) (oxidation number = 0) or H+ can oxidize Mn(s) (oxidation number = 0) to Mn^2+(aq). Overall: Mn(s) + 2HCl(aq) -------> MnCl2(aq) + H2(g) Total ionic: Mn(s) + 2H^+(aq) + 2Cl^-(aq) -------> Mn^2+(aq) + 2Cl^-(aq) + H2(g) Net ionic: Mn(s) + 2H^+(aq) -------> Mn^2+(aq) + H2(g) I hope I am not late this time.

Question:Write the total and net ionic equations for the reaction in which the antacid Al(OH)3 neutralizes the stomach acid HCl. This reaction is a double-displacement reaction. A. Identify the spectator ions in this reaction. B. What would be the advantages of using Al(OH)3 as an antacid rather than NaHCO3 which undergoes the following reaction with stomach acid? NaHCO3 + HCl ---> NaCl + H20 + CO2 I apologize for posting such a long question, and I normally don't. I just truely need help with this and my book has not come in handy at all. I appreciate anyone's help. Thank you so much in advance.

Answers:A. Molecular Equation Al(OH)3 + 3HCl ---> AlCl3 + 3H2O Total Ionic Al3+ + 3OH- + 3H+ + 3Cl- ----> Al3+ + 3Cl- + 3H2O Net Ionic 3OH- + 3H+ ----> 3H2O The spectator ions are Al3+ and Cl-. B. Notice that the NaHCO3 releases CO2 gas, so a person might have gas pains or feel bloated. There is no gas released when using Al(OH)3.

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Net Ionic Equation :Free Science Help at Brightstorm! brightstorm.com How to write a net ionic equation.

Net Ionic Equations :You can see copies of the math at www.coolstudyguides.com We got a double replacement- We got the equation, it says find the net ionic of the precipitate, son. First we convert the whole thing into our ions. This way of writing, it's called complete ionic equation. Next we find the precipitate, we look on our 'lil sheet. It says that this one's insoluble, even in heat. We get rid of all the ions watching it occur, not participating or even precipitating, just being spectators. You've got it now. Word.