PERHITUNGAN PENCAMPURAN BAHAN BAHAN 1 Menghitung massa bahan LTO MENGHITUNG SPECIFIC CAPACITY TiO MENGUKUR KETEBALAN SAMPEL Li

LAMPIRAN A 1. PERHITUNGAN PENCAMPURAN BAHAN BAHAN 1.1 Menghitung massa bahan LTO 4LiOH.H 2 O + 5TiO 2 Li 4 Ti 5 O 12 + 3H 2 O Massa atom relatif unsur : Li = 7 Ti = 47,867 O = 15,9994 Mr LiOH.H 2 O = 41,9988 TiO 2 = 79,8658 Li 4 Ti 5 O 12 = 459,3278 • massa LiOH .H 2 O 1 x Mr LiOH .H 2 O = 20 gr Mr Li 4 Ti 5 O 12 Massa LiOH.H 2 O = 1 x Mr LiOH .H 2 O Mr Li 4 Ti 5 O 12 = 1 �41,9988 x 20 gr 459,3278 = 1,828707 gr • massa Ti O 2 1 x Mr Ti O 2 = 20 gr Mr Li 4 Ti 5 O 12 Massa TiO 2 = 1 x Mr Ti O 2 x 20 gr Mr Li 4 Ti 5 O 12 = 1 x 79,8658 x 20 gr 459,3278 = 3,47750 gr 1.2.Menghitung massa bahan NaLiTi 3 O 7 LiOH.H 2 O + 3 TiO 2 + ½ Na 2 CO 3 NaLiTi 3 O 7 + ½ CO 2 + ½ H 2 O + H 2 Massa atom relatif unsur : Universitas Sumatera Utara Na = 22,989 Li = 7 Ti = 47,867 C = 12,0107 O = 15,9994 Mr LiOH.H 2 O = 41,9988 TiO 2 = 79,8658 Na 2 CO 3 = 105,9707 NaLiTi 3 O 7 = 285,548 • massa LiOH .H 2 O 1 x Mr LiOH .H 2 O = 10 gr Mr NaLi Ti 3 O 7 Massa LiOH.H 2 O = 1 x Mr LiOH .H 2 O x 10 gr Mr NaLi Ti 3 O 7 = 1 �41,9988 x 10 gr 285,548 = 1,470814 gr • massa Ti O 2 1 x Mr Ti O 2 = 10 gr Mr NaLi Ti 3 O 7 Massa TiO 2 = 1 x Mr Ti O 2 x 10 gr Mr NaLi Ti 3 O 7 = 1 x 79,8658 x 10 gr 285,548 = 8,3907924 gr • massa Na 2 CO 3 1 x Mr Na 2 O 3 = 10 gr Mr NaLi Ti 3 O 7 Massa Na 2 CO 3 = 1 x Mr Na 2 O 3 x 10 gr Mr NaLi Ti 3 O 7 = 1 x 105,9707 x 10 gr 285,548 = 1,470814 gr Universitas Sumatera Utara

2. MENGHITUNG SPECIFIC CAPACITY TiO

2 Li + TiO 2 + e - Li x TiO 2 Berat atom Ti = 47,867 Berat atom O = 15,999 Berat atom TiO 2 = 47,867 + 215,999 = 79,856 x 10 -3 Specific capacity = N x F berat atom Dimana : N= valensi atom F = Konstanta Faraday 96485 colombmol 26,801 Ahmol Specific capacity = 1 x 26,801 79,865 x 10 -3 = 335,57 mAhg

3. MENGUKUR KETEBALAN SAMPEL Li

4 Ti 5 O 12 DAN NaLiTi 3 O 12 Tebal Cu foil : 10 чm Sampel L чm L – Cu Foil Rs Ω TiO 2 60 50 36,4 Li 4 Ti 5 O 12 90 80 72,5 NaLiTi 3 O 12 70 60 21,3 Mengukur konduktivitas L = 50 x 10 -6 m L 1 = 80 x 10 -6 m L 2 = 70 x 10 -6 m R 1 = 36,4 Ω R 1 = 72, 5 Ω R 2 = 21,3 Ω A = 5,77 mm Universitas Sumatera Utara Na- 0 = R = ρ L A 36,4 Ω = ρ 50 x 10 −6 m 5,77 x 1,67 x 10 −6 m 2 36,4 Ω = ρ 50 m 5,77 x 1,67 m ρ = 7,014 Ωm = 701,4 Ωcm σ = 1 � = 1 701,4 Ωcm = 0,001425 Scm Na- 1 = R = ρ L A 72,5 Ω = ρ 80 x 10 −6 m 5,77 x 1,67 x 10 −6 m 2 72,5 Ω = ρ 80 m 5,77 x 1,67 m ρ = 8,732 Ωm = 873,2 Ωcm σ = 1 � = 1 873,2 Ωcm = 0,00114 Scm Na- 2 = R = ρ L A 21,3 Ω = ρ 70 x 10 −6 m 5,77 x 1,67 x 10 −6 m 2 21,3 Ω = ρ 70 m 5,77 x 1,67 m ρ = 0,2932 Ωm = 293,2 Ωcm σ = 1 � Universitas Sumatera Utara = 1 293,2 Ωcm = 0,00341 Scm

4. MENGHITUNG MASSA JENIS PADA SAMPEL