Question

3 A compound of sodium and sulfur is analyzed to contain following mass of each elements Calculate is the empirical formula. Show all calculation steps by using conversion factor. give calculator value (c.v.) and Final value (f.v) to get full credit. Show steps (1.23) (Na-5.891 g and S-4.109 g) (6 pt. Step 1: Step 2: Step 3: Empirical Formula is: In the following chemical reaction. Balance the molecular equation by adding coefficient. Write the net ionic equation by adding phases. Do not write ionic Eq. (4 pts) 4 Molecular Eq.: HNO, (aq) + Al(OH), (aq) Al (NO) (aq) + Но () Net ionic Eq. In the following chemical reaction. Balance the molecular equation by adding coefficient. Write the net ionic equation by adding phases. Do not write ionic Eq (4 pts) KNO, (aq) KS (aq) Al:Ss (s) + Al (NOs)s (aq) Molecular Eq.: + Net ionic Eq.: Determine the oxidation number of the element in the following ions or compounds. (1 pt each) e) PO P b) CIO4 CI= SnO2 Sn a) Determine if the following compounds are soluble or insoluble in aqueous solutions.(I pt each) 7 Ca(PO4) b) KCO (A) a) AgBr

Answer 1

mass Na= 5.891 g Molar mass = 22.99 g/mol moles mass = 5.891 fxmol Molar mass = 0.2562 mol 22.998 -- mass, s = 4.109 g Molar mass= 32.065 gol moles = mol = 0.1281 mol 4.109 of 32.065 g = - Divide byth! both by smaller unit, Na = 0.2562 = 2 S=0.1281 = 1 0.1281 0.1281 so, there are 2 Na atom and I s atom Epperical Formula is- Na₂S O 3 HNO3(aq) + Al(OH), (2) All NO3)3 (2) + 3 H20 (1) Net ionic equation - → 3H+ + Al(OH) → Al + 3H₂O Al253 (s) + 6 KNO₃(aq) ① 2 Al (NO3)₂ (aq) + 3 K₂5 (aq) → Net ionic equation 2 Al 3 (4) + 35² (aq) - Al,S3 (5)

la Sho, (6) ao Sn=74 a=+7 P=+3