CHM 201-003 Answer the questions in order in the exam book. H, 1.0; 0 16.0; N, 1

Question

CHM 201-003 Answer the questions in order in the exam book. H, 1.0; 0 16.0; N, 14.0; Zn 65.3,S-32.0; CI, 35.5 1.A compound zinc and chlorine only was found to contain 47.9 % zinc. what is the empirical formula of Lab Exam Time Allowed: 1 hr the compound? 2. Balance the equation below then calculate how many grams of carbon dioxide can be obtained by burning 6.00 g of C,Hs with 6.00 g of oxygen: GM (g) + O2(g) co,(g) + H20(1) 3. Complete and balance each of the following equations: a) Na,cols) + HCl(aq) b) Ba(NO3)(aq) + K,SO.(aq) c) A1(OH),(s) + HSO4(aq) d) FeSO4(aq) + NaOH(aq) above. 5. Write a balanced equation for the reaction of sulfamic acid (H2NSO3H) and sodium hydroxide then calculate the molarity of a NaOH solution if 0.5026 g of the acid required 20.55 ml of a NaOH solution for titration to a phenolphthalein endpoint. 6. Use the data below to calculate the enthalpy of reaction for: NaOH(s) + H (aq) + Cl (aq)-H20(1) +Na"(aq) + cr(aq) (1) H--33.44kJ/mol NaOH(s) Na+(aq) + OH(aq) . A:-33.44ki/mol H'(aq) + CI (aq) + Na"(aq) + OH(aq) NaTaq) + ci (aq) + H20(. (2) Is the reaction exothermic or endothermic? Briefly explain your answer.

Explanation / Answer

1. Consider 100 g. of sample.

Element:

Zn 47.9 g.

Cl 100 - 47.9 = 52.1 g.

Moles:

Zn 47.9 / 65.3 = 0.734 mol

Cl       52.1 / 35.5 = 1.47 mol

division by small number:

Zn        0.734 / 0.734 = 1

Cl         1.47 / 0.734 = 2

Therefore,

Emperical formula is ZnCl2

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