BTC4404 7 Nov 2014 by mzulkhairi

2014-11-07 02:41:28 UTC

g3 q4

2014-11-07 03:30:22 UTC

8.5 - 5.1 x 300/9 = 113 mg/L

group 1

2014-11-07 03:30:27 UTC

df=300/9

=33.3

BOD = (DO final -DO initial) x df

= (8.5-5.1) x (300/9)

= 113 mg/L

G4

2014-11-07 03:30:31 UTC

(8.5-5.1Lmg/L) x 300ml÷ 9ml = 113 mg/L

G5

2014-11-07 03:30:33 UTC

(8.5 mg/l - 5.1 mg/l) x 300 ml ÷ 9 ml

= 113 mg/l

g2

2014-11-07 03:31:09 UTC

(8.5-5.1)x300

g2

2014-11-07 03:31:10 UTC

(8.5-5.1)x300/

g2

2014-11-07 03:31:11 UTC

(8.5-5.1)x300

G4. Q8

2014-11-07 03:33:14 UTC

(245-22mg/L) ×100% ÷ 245mg/L = 91%

g^

zulaikhahalim92 — 2014-11-07 03:35:46 UTC

Q4) BOD=(initial DO-final DO)xdf

= (8.5-5.1)x (300/9)

= 113mg/l

Q8) (245-22)/245 x100

= 91%

Q11) (200x1000) / 2500

= 80 mL/g

g3 q8

2014-11-07 03:36:21 UTC

245

g3 q8

2014-11-07 03:36:21 UTC

(245 - 22)/245 x 100 = 91

G4. Q11

2014-11-07 03:36:52 UTC

200ml/L x 1000mg/g ÷ 2500 mg/L = 80 mg

g3

2014-11-07 03:40:17 UTC

200 x 1000 / 2500 = 80 mL/g

group 1

2014-11-07 03:40:20 UTC

svi= 200/2500 x1000
=80mg/l

g2

2014-11-07 03:41:08 UTC

((8.5-5.1) x 300)/9 =113mg/L

question 4

g2

2014-11-07 03:41:40 UTC

question 8

245-22 / 245 x 100 = 91%

g2

2014-11-07 03:43:05 UTC

question 11

200

g2

2014-11-07 03:43:06 UTC

question 11

200 x 1000 / 2500 = 80mg