Question

1. Identify each of the following reactions as a precipitation reaction, an acid-base reaction, or as a redox reaction + a. Pb(NO3)2 (aq) + 2 HCl(aq) PbCl (s) + 2 HNO, (aq) b. 2 Hxe + O2 + 2 H2O c. HNO, (aq) + KOH(aq) KNO, (aq) + H20 (1) d. Zna + 2HCl(a) - ZnClaraq) + Halle e. Ba?'(aq) + 2 Br(aq) + 2 Na(aq) + SO/"(aq) BaSO. (s) + 2 Na'(aq) + 2 Br" (aq) f. H' (aq) + OH(aq) HO (1) 2. For the following word reaction, create the balanced molecular equation, the complete ionic equation, and the net ionic equation (refer to the Crash Course video to help you). An aqueous solution of silver nitrate when combined with an aqueous solution of potassium bromide to produce a precipitate of solid silver bromide. 3. Strong acids completely disassociate in water to form H' ions (actually, H,0") and the parent anion. Strong hydroxide bases also completely disassociate in water to form OH ions and the parent metal cation. Provide the balanced chemical equation for the disassociation in water for each of the strong acids and bases. For example, HCl(aq) - H(aq) + Cl(aq); NaOH(aq) Na (aq) + OH(aq) a. HB b. HI c. HNO, d. HCIO e. H2SO4 f. KOH 8. LiOH h. Ba(OH)2 4. In aqueous solution, barium hydroxide reacts with sulfuric acid to produce solid barium sulfate and water according to the equation below: Ba(OH)2(aq) + H2SO4(aq) - BaSO4(s) + 2H2O(l) You will recognize this as a special type of precipitation reaction - an acid-base reaction. Use your knowledge of stoichiometry and unit conversion to calculate how much grams of barium sulfate will be produced from 13.23 g of barium hydroxide and an excess of sulfuric acid.

please write legible

Answer 1

1) a) precipitation reaction

B) redox reaction

C) acid-base reaction

D) redox reaction,

E) precipitation reaction

F) acid base reaction

2) AgNO3(aq) + KBr(aq) --> AgBr(S) + KNO3(aq)

Net ionic : Ag+(aq) + Br-(aq) --> AgBr(S)

3) Strong acid : HBr(aq) --> H+(aq) + Br-(aq)

Strong acid HI(aq) ---> H+(aq) + I-(aq)

Strong acid: HNO3(aq) ---> H+(aq) + NO3-(aq)

Strong acid: HClO4(aq) ---> H+(aq) + ClO4-(aq)

Strong acid : H2SO4(aq) ---> 2H+(aq) + SO₄^2-(aq)

Strong base : KOH(aq) ---> k+(aq) + OH-(aq)

Strong base : LiOH(aq) ---> Li+(aq) + OH-(aq)

Strong base : Ba(OH)2(aq) --> Ba2+(aq) + 2 OH-(aq)