Ch6 Homework solutions

Ch6 Homework solutions - 3' i2gcr*r- # = 2.0 mote crH, 2.0...

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Ch" 6 Suppi;z*ar.,.'{*e Q--dU"* -- SaU+afls I rc FW= 2(23.0)+32.t +4{16.0) =142.1 2(23.0)1142.1 x 100 = 32.4% Na 32.11142.1x 100 = 22.60/o S 4(16.0)1142.1 x 100 = 45.0o/o O Q-U, molecular formula, C.HrOr: empirical formula CaH3O Lb. 4o.o% c,6.7vo H, 53.3% o MW = 180 40'0gc =3.33morec u'1n,t, =6.7moleH ?::n'o, =3'33moleo tZO gimote 1.0 g/mole 16'0 g/mole *=1.ooc 6'7 =z.oH H=1.ooo 3.33 3.33 The empirical formula is GH2O. EFW = 12.0 + 2(1.0) + 16'0 = 30'0 MWEFW = 180/30.0 = 6 The molecular formula is CuH,rOu'
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Unformatted text preview: 3' i2gcr*r- # = 2.0 mote crH, 2.0 more crH -302 kcal z x ll@ = -302 kcal 1 mole BaCl, 1 mole BaSOo 233 g BaSO4 ^ -4.16gBaClrx ^_ --x x7a;-=4.66gBaSO. " 208 g BaCl, 1 mole BaCl, 1 mole BaSuo 1?? n 3":?' x 1oo% = 87 1% yierc! 4.66 g BaSOo 'l mole AgNOt 3 " 16.ee s AsNo' x 16e6ffi; = 0.1000 mole AgNo' 1 mole NaCl 2.92 g Nacl x ,r* , _rCI = 0.0500 mole Nacl Therefore Nacl is the limiting reagent. 00s00moreNacr ' 1'"1"1n!1 - 1otInln=:' =716sAgcr 1 mole NaCl 1 mole AgCl...
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