1 / 16

I = 20 A  =93 % d = 5 x 10 -5 dm r = 0.15 m k = 0.45364 gA -1 h -1 l = 12.0 dm t = 1 . 165 h

I = 20 A  =93 % d = 5 x 10 -5 dm r = 0.15 m k = 0.45364 gA -1 h -1 l = 12.0 dm t = 1 . 165 h.  / gcm -3 = ?. Pretvaranje jedinica: l = 12.0 dm = 120 cm r = 0.15 m = 15 cm d = 5 x 10 -5 dm = 5 x 10 -4 cm.

rollin
Download Presentation

I = 20 A  =93 % d = 5 x 10 -5 dm r = 0.15 m k = 0.45364 gA -1 h -1 l = 12.0 dm t = 1 . 165 h

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. I = 20 A  =93%d = 5 x 10-5 dm r = 0.15 mk = 0.45364 gA-1h-1l= 12.0 dmt = 1.165 h  / gcm-3 = ? Pretvaranje jedinica:l= 12.0 dm = 120 cmr = 0.15 m = 15 cmd = 5 x 10-5 dm = 5 x 10-4 cm 1. Izračunaj gustoćumagnezija u gcm-3 dobivenog elektrokemijskim postupkom ako je zadano: A = 2 * r *l * π (cm *cm) = 2 * 15 *120 *π = 11309.734cm2 = 0. 00177 Acm-2 =

  2. r = 1.74gcm-3

  3. i = 0.01678 Acm-2l= 1.5 md = 3.0 mm r = 150 mmk = 0.32338 gA-1h-1 = 20 %m = 3.054 dag t / sati = ? Pretvaranje jedinica:l= 1.5 m = 150 cmr = 150mm = 15 cmd = 3mm = 3 x 10-4 cmm = 3.054 dag = 30.54 g 2. Izračunaj vrijeme u sekundama koje je potrebno da se dobije prevlaka kroma ako je zadano: V = 2 * l * r *π *d (cm * cm *cm) = 4.24115 cm3 = 2*150*120*π*5 x10-4 = 7.2gcm-3

  4. t = 7165.08 s

  5. i = 1 Adm-2 = 10.5 gcm-3t = 5.218 min  = 90 %k = 4,02454 gA-1h-1 l = 1500 mmr = 36.34 cm Ar = 107.868 gmol-1 n / mol = ? Pretvaranje jedinica:i= 1 Adm-2= 0.01 Acm-2l = 1500mm = 150cm 3. Izračunaj količinu tvari u molima dobivene prevlake srebra ako je zadano: = 3 * 10-4 cm V = 2 * l * r *π *d(cm * cm * cm) = 2*150*36.34*π*3 x10-4 = 10.2749 cm3 m = V* (cm3 * gcm-3) = 107.88645g

  6. n = 1 mol

  7. I = 7500 A  = 98 %d = 3x 10-3 dml = 18 dmt = 0.2605 h = 8.92 gcm-3k = 2,3709 gA-1h-1 r/ m = ? Pretvaranje jedinica:d= 3 * 10-3 dm=3 * 10-2 cml = 18dm = 180cm 4. Izračunaj polumjer bakrenog Ballardovog cilindra u m za duboki tisak ako je zadano: = 0.4421 Acm-2 = 16964.4877 cm2

  8. r = 0.15 m

  9. d =8 mm I = 1000 At = 520.73 s r = 200 mmk = 1,09534 gA-1h-1 = 96 % = 8,90 gcm-3 l / m = ? Pretvaranje jedinica:d= 8 mm=8 * 10-4 cmr= 200mm = 20cm 5. Izračunaj dužinu Ballardovog plašta u dm pri postupku niklanja ako je zadano: =0.04681 Acm-2 = 21362.96 cm2 = 170 cm

  10. l = 17 dm

  11. i = 20 Adm-2 = 7.14 gcm-3t = 10.98 min  = 80 %k = 1,21952 gA-1h-1m = 35.7dag A / m2 = ? Pretvaranje jedinica:i= 20 Adm-2= 0.2 Acm-2m= 35.7dag = 357cm 6. Izračunaj pocinčanu površinu tiskovne forme za plošni tisak u m2 ako je zadano: = 5.0 * 10-3 cm = 50 cm3 = 10000 cm2

  12. A= 1 m2

  13. d = 4.0 x 10-5 m  = 75 %k = 2.08395 gA-1h-1r = 2.5 dmt = 17.694 min l= 1400 mm = 7.86 gcm-3 I / A = ? Pretvaranje jedinica:d= 4 * 10-5 m= 40 * 10-4 cml= 1400 mm = 140 cmr = 2.5 dm = 25 cm 7. Izračunaj potrebnu jakost struje u A za elektrokemijsko taloženje željeza ako je zadano: = 0.06821 Acm-2 A = 2 * r * l * p (cm * cm) = 2 * 25 *140 * p = 21991.15 cm2 I = i * A (Acm-2 * cm2) = 0.06821 * 21.991,15

  14. I= 1500 A

  15. I = 200 A l= 10 dmd = 1.0 x 10-4 m r = 0.12 mt = 4957.44 s = 7.28 gcm-3k = 2.21409 gA-1h-1 Pretvaranje jedinica:d= 1.0 * 10-4 m= 100 * 10-4 cml= 10 dm = 100 cmr = 0.12 m = 12 cm h / % = ? 8. Izračunaj iskorištenje struje pri elektrokemijskom taloženju kositra ako je zadano: A = 2 * r * l * p (cm * cm) = 2 * 12 *100 * p = 7539.822 cm2 = 0.02653 Acm-2

  16. h= 90 %

More Related