J. A. CAMPBELL Harvey Mudd College Cloremont, California 9171 1
Questions 4 3 2 . Many biological cells are about 75% by volume or veight water. How many other molecul?~, raoh containing an amragc of 10,000 atoms ~ o u l bd r in an average bacterial cell 10W6m in diameter? 4 3 3 . Estimate the mp and bp of thorium which are missing in thc tablc.
4 3 4 . Many aquatic organisms have float bladders containing fixed quantitirs of gas (mainly N1) whosc volumrs adjust to rqualize inside and outside pressure. At n-hat d q t h in a lake of density 1.00 g/cm3 and 6'C \\.ill an organism of 1.35 g veight (density 1.35 g/cm3) ncit,her risr nor sink if the float bladder contains 0.001 molr of gas?
4 3 5 . Fumigation ~ v i t hhydrogen cyanide usually usrs one pound of sodium cyanide per 1000 ft3 of air spacr. What conccntration of hydrogen cyanide docs this produce? 4 3 6 . Death by sufforation in a sealed container is normally duc to CO? poisoning (n.liicl1 occurs a t about 7% COXby volume) not oxygen deficiency. For n-hat length of time ~vouldit be safe t o bc in a sealed room lox 1 0 X 2 0 f t . 4 3 7 . An aut,omobile cylinder has a volume of 1 1. What weight of gasoline, CsHls, should be inject~d into tlie cylinder to give a stoicliiometric mixture 11-it11 the air burning to CO? and H20. What can you say about the relationship of engine shaft rcvolutions t o whcd rcvolutions if this car gets20 mpg? Questions (plus possible, but certainly oot uniquely satisfactory, answers) 1,equiring no more than a concurrent first-year, college level course, a data handbook, and a willingness to apply fundamental chemical ideas to the systems which surround us (or even are inside us) make up this column. Conlributions for pmsible inclusion are solicited. Initiated in the January, 1972 issue of THIS JOURNAL.
Answers =
A32. Only one significant figure so assume density of cell I g/om3 3m3 I l r y wt in cell = 0.25 X l(g/cma) X (4/3)s(10-8/2)a X
I06cm"ma 10-l3 g/eell Ave. wt of an atom (mtio E C,HaOzN, from problem 11) 8 avoprams Ave. molecular wt, = 10,000 X 8 = 80,000 avograms = (8 X 10V6 X 1023)g/molecule = 1 X 10-'9 g/molecule No. of molecules/cell = 10-'3(g/cell)/l X 10-"(g/moleeule) = lo6 Aetuxlly there are many more smaller molecules plus a few larger molecules in any one cell. But water molecules are the most cornrnon by far; there would be about 10'"f them, ten thousand waters per each large 10,000 atom molec~de,or one water molecule per each other atom present. A33. Any guesses must be extrapolations (always doubtful) since T h is not bracketed in any property. Bp seems easiest since i t varies little from column t o column, seems t o hit. a maximnm in the H i column and be slightly less in the Ti than in the V or Cr row. So, judging from U, guess just under 3600 z t 100'K for boiling. Np's vary more hoth in the rows and columns but they are $ways least in the T i row when examined in rows and are =
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changing less rapidly in the Ti column. Thus T h m p should be less than U, less than Hf and around 1000 i 100°K. Recent data suggest 2020 and 4000°K respect,ively for mp and bp. Thus, the guess on m p is terrible, that on the bp not. so bad, but not good. This result not only warns about extrapolations, i t suggests that inanium, especially its m p which is widely variant. from all others listed, is not typical in its phase changes. Furlher study shows that uranium has an unusual crystal sfrncture, less tightly packed than most metals, thus lowering AH of melting, and, hence, the mp. A34. If t,he organism has a weigh? of 1.35 p a n d a density of 1.3.5 g/em3, its volume must be 1.33 g/1.33(g/rma) = 1 ema, presumably exclosive of float bladder. Now 0.001 mole of air weighs 0.001 mole X 28 g/mole = 0.03 g so total wt = 1.35 0.03 = 1.38 g. The volume must he (1.00 V&)cm3, where V,,, = volume of the air in the float bladder. I n order to float, the over-all density must be 1.00 so 1.38 gi(1.00 V,,,)cm3 = 1.00 g/cm3 and Vrn = 0.38 cm3 = n R T / P = 0.001 mole X 82 (cm" atm/K male) X 279K/P,t,.P,c, = 60 stm. A column of mercury ( p = 13.6 g/cm3) 76 cm deep gives P = 1 atm; so 60 s t m of water = 60 a i m X 76(cm Hg/atm) X 13.6(cm H20/cm Hg) = 6.2 X 1 0 h m H.0 or 0.62 km in depth. Note that the organism must have positive control over volume in the hladder. Otherwise any rise would let the bladder increase in volume, thus increasing the rate of rise, and vice versa.
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A35. 1 mole NaCN (fwt 49) 1 mole HCN (fwt 27), a 453149 = 9 moles HCN gas. 1 lb = 453 g, so 1lb NaCN produced (1 sig. fig.). 1f t 3 = 27 1 so 1000 fts X 27(l/ft3) = 27,000 1 = 3 X lo41 to 1 sig. fig. 9 moles in 3 X 10'1 = 3 X lo-' moles/l as t,he concentration of gaseous HCN during fumigation. This figure certainly justifies assuming ideality in the gas. The lethal concentration of HCN for human beings is less then this, about 7 X 10-5 M. A36. 1 mole COXformed requires 1 mole 0% used up, assuming little H is oxidized to H1O by inhaled oxygen. (Actually the ratio is about 1.2 02/1CO.). Thus, the air becomes lethal (due to COI) after 7% of the air, as 01,is removed (to 1 sig. fig.). 10 ft X 10 f t X 20 ft X 0.07 = 140 ft3 of On required or 140ft5 X 271/ft3 X 1/2,5 (Ijmole) = 140moles of 02. You breathe about 0.20 X 0.3 1 of 0.20 times per minute and for a rate of about 0.02 mole/min. ingest about 30% of the 0% If you were quiet, half this amount would be enough, or 0.01 mole/min. So each human requires about 0.01 mole Os/min or 0.01 X 60 X 24 = 10 mole/day. Thus one individual could last 14 days, or 14 individuals 1day in such a sealed room before the CO. became lethal, if they remained quiet.
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A37. CaHlar., 12.5 Oar., = 8 COpw 9 HzOw by volume, 1 mole of Use 1 sig. fig. Air is 80 Nz, 20 0% gaseous CsHa will require 12.5 moles Oz plus 50 moles N.. Cylinders actually vary in P and T but let us assume gasoline and air both enter as gases a t P = 1 atm, T = 300°K (prohably good to 1 sig. fig.). The molar volume will be 20 1 (1 sig. fig.) or the cylinder will accept 1/20 = 0.03 mole of gas. CaH,. constitutes 1/(1 12.5 30) = 1/60 of the total or (1/60)0.05 = 0.001 mole. Molar wt of C8HI8= (8 X 12) (18 x 1) = 114 gfmole. So 0.001 mole = 0.1 g CsH,sper full cycle of the cylinder. This means about 1000 cylinder cycles per mole of gasoline, or, in an 8 cylinder car, somewhat more than 200 revolutions of the drive shaft. A gallon is ahout 4 1, or about 3000 g, or 30 moles. So such a car should get about 6000 revolutions of the drive shaft per gallon. A tire tread is &bout 6 feet (2rr) around its perimeter so revolves shout 1000 times per mile. I n order to obtain 20 miles/ gallon a wheel must make about three revolutions for each revolution of the engine shaft.
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Volume 49, Number 8, August 1972
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