chemistry for aggies or “everythin’ i ever needed for chem 111, i learnt in hort 100” a/h 100g...
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Chemistry for Aggiesor
“Everythin’ I ever needed for CHEM 111, I learnt in HORT 100”
A/H 100G
J.G. Mexal
Chemistry for Aggies
Stuff ya’ll need to know!
• Woolworth’s 5 and 10 ¢
• Metric units
• Fertilizer calculations
• Irrigation requirements
• pH
Metric Scale
• Things to memorize:– M = 1,000,000
– k = 1,000
– c = 0.01 = 1/100
– m = 0.001 = 1/1000 = 0.000001
Metric Scale
Yotta Y 1 000 000 000 000 000 000 000 000 1024
Zetta Z 1 000 000 000 000 000 000 000 1021
Exa E 1 000 000 000 000 000 000 1018
Peta P 1 000 000 000 000 000 000 1015
Tera T 1 000 000 000 000 000 1012
Giga G 1 000 000 000 (billion) 109
Mega M 1 000 000 (million) 106
kilo k 1 000 (thousand) 103
hecta h 1 00 (hundred) 102
deca da 10 (ten) 101
1
deci d 0.1 (tenth) 10-1
centi c 0.01 (hundredth) 10-2
milli m 0.001 (thousandth) 10-3
micro µ 0.000 001 (millionth) 10-6
nano n 0.000 000 001 (billionth) 10-9
pico p 0.000 000 000 001 10-12
femto f 0.000 000 000 000 001 10-15
atto a 0.000 000 000 000 000 001 10-18
zepto z 0.000 000 000 000 000 000 001 10-21
yocto y 0.000 000 000 000 000 000 000 001 10-24
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
LengthEnglish Metric
(approx.)
Metric English
(approx.)
1 in 2.5 cm 1 mm 0.04 in
1 ft 30 cm 1 cm 0.4 in
1 yd 0.9 m 1 m 3.3 ft
1 mi 1.6 km 1 km 0.6 mi
AreaEnglish Metric
(approx.)
Metric English
(approx.)
1 in2 6.5 cm2 1 cm2 0.16 in2
1 ft2 0.09 m2 1 m2 1.2 yd2
1 yd2 0.8 m2 1 m2 11 ft2
1 mi2 2.6 km2 1 km2 0.4 mi2
1 acre 0.4 ha 1 ha 2.5 ac
VolumeEnglish Metric
(approx.)
Metric English
(approx.)
1 in3 16 mL 1 mL 0.06 in3
1 ft3 0.03 m3 1 m3 35 ft3
1 yd3 0.76 m3 1 m3 1.3 yd3
1 fl oz 30 mL 1 mL 0.03 fl oz
1 cup 0.24 L 1 L 4.2 cups
1 pt 0.47 L 1 L 2.1 pt
1 qt 0.95 L 1 L 1.06 qt
1 gal 3.8 L 1 L 0.26 gal
MassEnglish Metric
(approx.)
Metric English
(approx.)
1 oz 28 g 1 g 0.035 oz
1 lb 454 g 1 kg 2.2 lb
1 lb 0.454 kg 1 ton 1.1 short ton
1 ton (short) 0.9 metric ton 1 ton 1 long ton
1 ton (long) 1 metric ton
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?
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?”
Chemistry for Aggies Cipherin’ the easy way• How many cm2 in a m2?
• 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?
Chemistry for Aggies Rule of Thumb• 1 g = 1 mL = 1 cc or 1 cm3
• This rule really holds only for pure water at 4OC. 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).
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
Chemistry for Aggies Cipherin’ the easy way• How many cm2 in a m2?
• Answer:= 1 m = 100 cm= 1 m x 1 m = 100 cm x 100 cm= 1 m2 = 10,000 cm2
Chemistry for Aggies Cipherin’ the easy way• How many mL in a L?
• Answer:= 1 L = 1,000 mL
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:
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 cm2
* 1 cm of rain on 100,000,000 cm2 = 100,000,000 cm3
* 100,000,000 mL x 1 L/ 1,000 mL = 100,000 L
Chemistry for Aggies Woolworth’s 5 and 10 ¢• To estimate the temperature in OC,
beginning at the freezing point of water (32OF and 0OC), the Celsius scale changes about 5O for every 10O change in the Fahrenheit scale.
Chemistry for Aggies Woolworth’s 5 and 10 ¢
OF
OC (approximate)
OC (exact) 10 -10 -12 20 - 5 - 7
32 (30) 0 0
40 5 4
50 10 10
60 15 16
70 20 21
80 25 27
90 30 32
100 35 38
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:
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 cm2
* 5 cm of water on 100,000,000 cm2 = 500,000,000 cm3
* 500,000,000 mL x 1 L/ 1,000 mL = 500,000 L
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
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
Chemistry for Aggies
%, ppm, mg/L
• The atmosphere is 21% oxygen and 310 ppm carbon dioxide.
• Express carbon dioxide in %
• Express oxygen in ppm
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 0.031 or
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
Chemistry for Aggies
%, ppm, mg/L
• How much nutrient is in a fertilizer?• Example: N in ammonium sulfate• Answer:Calculate molecular weight
– ammonium sulfate = (NH4)2SO4
– 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%
Chemistry for Aggies
%, ppm, mg/L• Fertigate with 150 ppm N , using ammonium sulfate
(NH4)2SO4, 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
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.
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
More problems
• A homeowner irrigates the yard with city water (150L/min). If his yard is 800 m2 (40 m x 20 m), how long will it take to apply 1 cm?
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 LD50 = 5,440 mg/kg)?
More problems
• How much ammonium nitrate should be applied to fertilize a crop at the 35 kg N/ha rate. Ammonium nitrate = NH4NO3, N = 14, H = 1, O = 16, S = 32, K = 39.
• How much hydrogen are you applying?
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 m2 x 100 m2
• How much N is added to the field?• How much total salt is added?
pH
• pH is the measure of the hydrogen ion concentration in solution
• pH is calculated as = -- log [H+]
= -- log (10-7 g H+ /LH20)
= -- (--7)
= -- 7 (neutral)Neutral means there are equal amounts of H+ and OH-
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
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.