a) Cl2 + F2 ClF(g) 3-hard  (g) Cl2 + F2 2ClF(g). (g)  b) HBr(aq)+ NaOH(aq) NaBr(aq) + H2O(l)  Already balanced. Acid-base reaction. 17) Would gold (III) chloride and sodium hydroxide (based on their chemical composition) form a precipitation Yes, the net ionic reaction? How do you know? 3-hard equation for gold (III) chloride and sodium hydroxide shows that they would form a precipitate. Au(OH)3(s), Au +Cl3 + 3Na + OH Au(OH)3(s)  3+(aq) -1(aq) +1(aq) -1(aq) + 3Na+(aq) + 3Cl-(aq) 18) Balance the following decomposition reaction and then identify the products: 2-medium BaCO3 BaO(s) + CO2(g) (s)  2BaCO3 2BaO(s) + 2CO2(g). Barium (s)  oxide and carbon dioxide are the two products that BaCO3 decomposed into. 65/10.5=6.19
a) Cl2 (g) + F2  ClF(g) 3-hard Cl2 (g) + F2  2ClF(g). b) HBr(aq)+ NaOH(aq)  NaBr(aq) + H2O(l) Already balanced. Acid-base reaction. 17) Would gold (III) chloride and sodium hydroxide (based on their chemical composition) form a precipitation reaction? How do you know? 3-hard Yes, the net ionic equation for gold (III) chloride and sodium hydroxide shows that they would form Au(OH)3(s), a precipitate. Au 3+(aq) +Cl3 -1(aq) + 3Na +1(aq) + OH -1(aq)  Au(OH)3(s) + 3Na+(aq) + 3Cl-(aq) 18) Balance the following decomposition reaction and then identify the products: 2-medium BaCO3 (s)  BaO(s) + CO2(g) 2BaCO3 (s)  2BaO(s) + 2CO2(g). Barium oxide and carbon dioxide are the two products that BaCO3 decomposed into. 65/10.5=6.19
SAS_HC_CR_Chap_8_PracProb_Groner_Edit_Green.docx