1. Chemistry for Aggies or “Everythin’ I ever needed for CHEM 111, I learnt in HORT 100” A/H 100G J.G. Mexal

2. Chemistry for Aggies Stuff ya’ll need to know! Woolworth’s 5 and 10 ¢ Metric units Fertilizer calculations Irrigation requirements pH

3. Metric Scale Things to memorize: –M = 1,000,000 –k = 1,000 –c = 0.01 = 1/100 –m = 0.001 = 1/1000 –  = 0.000001

4. Metric Scale YottaY1 000 000 000 000 000 000 000 00010 24 ZettaZ1 000 000 000 000 000 000 00010 21 ExaE1 000 000 000 000 000 00010 18 PetaP1 000 000 000 000 000 00010 15 TeraT1 000 000 000 000 00010 12 GigaG1 000 000 000 (billion)10 9 MegaM1 000 000 (million)10 6 kilok1 000 (thousand)10 3 hectah1 00 (hundred)10 2 decada10 (ten)10 1 1 decid0.1 (tenth)10 -1 centic0.01 (hundredth)10 -2 millim0.001 (thousandth)10 -3 microµ0.000 001 (millionth)10 -6 nanon0.000 000 001 (billionth)10 -9 picop0.000 000 000 00110 -12 femtof0.000 000 000 000 00110 -15 attoa0.000 000 000 000 000 00110 -18 zeptoz0.000 000 000 000 000 000 00110 -21 yoctoy0.000 000 000 000 000 000 000 00110 -24

5. Chemistry for Aggies Cipherin’ the easy way Things to memorize: –M = 1,000,000 –k = 1,000 –c = 0.01 = 1/100 –m = 0.001 = 1/1000 –  = 0.000001 Things to help: –1 m ~ 1 yd –1 L ~ 1 qt –1 kg ~ 2 lb

6. Length EnglishMetric (approx.) MetricEnglish (approx.) 1 in2.5 cm1 mm0.04 in 1 ft30 cm1 cm0.4 in 1 yd0.9 m1 m3.3 ft 1 mi1.6 km1 km0.6 mi

7. Area EnglishMetric (approx.) MetricEnglish (approx.) 1 in 2 6.5 cm 2 1 cm 2 0.16 in 2 1 ft 2 0.09 m 2 1 m 2 1.2 yd 2 1 yd 2 0.8 m 2 1 m 2 11 ft 2 1 mi 2 2.6 km 2 1 km 2 0.4 mi 2 1 acre0.4 ha1 ha2.5 ac

8. Volume EnglishMetric (approx.) MetricEnglish (approx.) 1 in 3 16 mL1 mL0.06 in 3 1 ft 3 0.03 m 3 1 m 3 35 ft 3 1 yd 3 0.76 m 3 1 m 3 1.3 yd 3 1 fl oz30 mL1 mL0.03 fl oz 1 cup0.24 L1 L4.2 cups 1 pt0.47 L1 L2.1 pt 1 qt0.95 L1 L1.06 qt 1 gal3.8 L1 L0.26 gal

9. Mass EnglishMetric (approx.) MetricEnglish (approx.) 1 oz28 g1 g0.035 oz 1 lb454 g1 kg2.2 lb 1 lb0.454 kg1 ton1.1 short ton 1 ton (short)0.9 metric ton 1 ton1 long ton 1 ton (long)1 metric ton

10. Chemistry for Aggies Doin’ it the hard way How many square inches in a square yard? How many ounces in a gallon? How many square feet in an acre? The K-Mart parking lot is one acre in size and receives 1” or rain. How many gallons will run off into the sewer?

11. Chemistry for Aggies Why use metric units? Metrics involves the judicious movement of zeros to determine the answer. There are no complicated conversions to learn. RoT: If the answer looks right--it probably is! The question you must always ask yourself is: “Does the answer make sense?”

12. Chemistry for Aggies Cipherin’ the easy way How many cm 2 in a m 2 ? How many mL in a L? The WalMart parking lot is 1 ha in size and receives 1 cm of rain. How many L of water will run off into the sewer?

13. Chemistry for Aggies Rule of Thumb 1 g = 1 mL = 1 cc or 1 cm 3 This rule really holds only for pure water at 4 O C. But remember, this is Chemistry for Aggies, and this estimate is “close enuf”. This will allow you convert from weight (g) to volume (mL) to length (cm).

14. Chemistry for Aggies Cipherin’ the easy way 1 m = 100 cm = 1,000 mm –1 km = 1,000 m –1 ha = 100 m x 100 m 1 L = 1,000 mL 1 g = 1,000 mg –1 kg = 1,000 g –1,000 kg = 1 ton = 1 Mg

15. Chemistry for Aggies Cipherin’ the easy way How many cm 2 in a m 2 ? Answer: =1 m = 100 cm =1 m x 1 m = 100 cm x 100 cm =1 m 2 = 10,000 cm 2

16. Chemistry for Aggies Cipherin’ the easy way How many mL in a L? Answer: =1 L = 1,000 mL

17. Chemistry for Aggies Cipherin’ the easy way The WalMart parking lot is 1 ha in size and receives 1 cm or rain. How many L of water will run off into the sewer? Answer:

18. Chemistry for Aggies Cipherin’ the easy way The WalMart parking lot is 1 ha in size and receives 1 cm or rain. How many L of water will run off into the sewer? Answer: * 1 ha = 100 m x 100 m = (100 m x 100 cm/m) x (100 m x 100 cm/m) = 100,000,000 cm 2 * 1 cm of rain on 100,000,000 cm 2 = 100,000,000 cm 3 * 100,000,000 mL x 1 L/ 1,000 mL = 100,000 L

19. Chemistry for Aggies Woolworth’s 5 and 10 ¢ To estimate the temperature in O C, beginning at the freezing point of water (32 O F and 0 O C), the Celsius scale changes about 5 O for every 10 O change in the Fahrenheit scale.

20. Chemistry for Aggies Woolworth’s 5 and 10 ¢

21. Chemistry for Aggies More Cipherin’ the easy way The Rio Grande has 300 ppm dissolved salts. If a farmer applies 5 cm of irrigation water to one ha, how may tons of salts is the farmer adding? Answer:

22. Chemistry for Aggies More Cipherin’ the easy way The Rio Grande has 300 ppm dissolved salts. If a farmer applies 5 cm of irrigation water to one ha, how may tons of salts is the farmer adding? Answer: Calculate the amount of water * 1 ha = 100 m x 100 m = (100 m x 100 cm/m) x (100 m x 100 cm/m) = 100,000,000 cm 2 * 5 cm of water on 100,000,000 cm 2 = 500,000,000 cm 3 * 500,000,000 mL x 1 L/ 1,000 mL = 500,000 L

23. Chemistry for Aggies More Cipherin’ the easy way Now calculate the amount of salts in the water: 300 ppm is 300 mg/L (How do I know this?) 300 mg/L x 500,000 L = 150,000,000 mg 150,000,000 mg x 1 g/1,000 mg x 1 kg/1,000 g x 1 t/1,000 kg = 150,000 g 150 kg 0.15 t of salt

24. Chemistry for Aggies %, ppm, mg/L % = parts/100 ppm = parts/million ppb= parts/billion (hazardous materials) mg/L= 0.001 g/1,000 g (or 1,000 mL) = 1 g/1,000,000 g = ppm

25. Chemistry for Aggies %, ppm, mg/L The atmosphere is 21% oxygen and 310 ppm carbon dioxide. Express carbon dioxide in % Express oxygen in ppm

26. Chemistry for Aggies %, ppm, mg/L The atmosphere is 21% oxygen and 310 ppm carbon dioxide. Express carbon dioxide in % =_____310 ____31 ___3.1 _0.31 0.031or 0.031% 1,000,000 100,000 10,000 1,000 100 Express oxygen in ppm = 21 210 2,100 21,000 210,000 or 210,000 ppm 100 1,000 10,000 100,000 1,000,000

27. Chemistry for Aggies %, ppm, mg/L How much nutrient is in a fertilizer? Example: N in ammonium sulfate Answer: Calculate molecular weight –ammonium sulfate = (NH 4 ) 2 SO 4 – H = 1, N = 14, O = 16, S = 32 –What % is N of AmSul? –is / of= 14 + 14 / 28+8+32+64 = 28/132 –N% = 21%

28. Chemistry for Aggies %, ppm, mg/L Fertigate with 150 ppm N, using ammonium sulfate (NH 4 ) 2 SO 4, which is 21% N. The injector dilutes the stock solution 100:1. How much fertilizer should be added to the stock solution? Answer:How much nitrogen do you need? 150 mg N/L -I x 100 -I / 1L -S = 15,000 mg N/L -S = 15 g N/ L -S How much ammonium sulfate do you need? RoT = % = is/of 21% = 15 g/ ?.21 x ? = 15 g ? = 15 /.21? = 71 g AmSul

29. Chemistry for Aggies %, ppm, mg/L Conclusion: You will need to add 71 g of ammonium sulfate per liter of stock solution to deliver 150 ppm N from your 1:100 injection fertigater.

30. More problems A cabbage crop is 1.5% N. If 1 ha yields 2,000 boxes of 24 heads each weighing 0.1 kg (DW), how much ammonium nitrate should be added to meet the minimum nitrogen requirement? –Hints: H = 1, N = 14, O = 16

31. More problems A homeowner irrigates the yard with city water (150L/min). If his yard is 800 m 2 (40 m x 20 m), how long will it take to apply 1 cm?

32. More problems Your roommate decides to kill you for eating all the Oreos. The only thing toxic in the house is some A-rest from your HORT 100 lab. If the plant growth regulator A-rest is toxic to humans at a rate of 5000 mg A-rest to 1 kg human body weight, what % A-rest will be found in your body at the time of death? Alternatively, could he force feed you Oreos (sucrose LD 50 = 5,440 mg/kg)?

33. More problems How much ammonium nitrate should be applied to fertilize a crop at the 35 kg N/ha rate. Ammonium nitrate = NH 4 NO 3, N = 14, H = 1, O = 16, S = 32, K = 39. How much hydrogen are you applying?

34. More problems A farmer applies 40 t dry cow manure to 1 ha. If the manure contains 1% N and 7% salts (Ca, Mg, K, Na, etc): Hint: 1 t = 1,000 kg, 1 ha = 100 m 2 x 100 m 2 How much N is added to the field? How much total salt is added?

35. pH pH is the measure of the hydrogen ion concentration in solution pH is calculated as =-- log [H + ] =-- log (10 -7 g H + /L H 2 0 ) =-- (--7) =-- 7 (neutral) Neutral means there are equal amounts of H + and OH -

36. pH pH 6 = 0.000001 g H + / L= 10 -6 g H + /L pH 7 = 0.0000001 g H + / L= 10 -7 g H + /L pH 8 =0.00000001 g H + / L= 10 -8 g H + /L Change from pH 8 to pH 6? –pH 6= 0.00000100 g H + / L= 10 -6 g H + /L –pH 8 =-- 0.00000001 g H + / L= -10 -8 g H + /L –Add 0.00000099 g H + / L = 9.9 x 10 -7 g H + /L

37. pH Therefore, to change 1L of pH 8 solution to pH 6, you need to add 0.00000099 g H + or add 1 L of pH 6.004 solution.