2006mcb200ftest1answerswb

2006mcb200ftest1answerswb - 13. e (2 marks) pH =pKa + log...

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MCB200F Answers Test 1 Ch1 1. e 2.b 3.d 4. b +/ - interaction is modified by salt (Na + Cl -) 5.e 6. b Ch 2 1. b 2. d liquid water has on average 2.4 depending on temp but the maximum is 4 hydrogen bond per H 2 O 3. d 4. a 5. e 6. d grape f [H + ]=10 -3.2 =6x10 -4 orange [H + ]=10 -4.3 =5x10 -5 ratio = 12 or [H + ] g /[H + ] o = 10-3.2-(-4.3) =10 1.1 =12 7. a 8. b 2.1 7.2 12.4 H 3 PO 4  H 2 PO 4 -  H 1 PO 4 2-  PO 4 3- 1mole 1mole pH= pK a3 + log PO 4 3- / H 1 PO 4 2 = 12.4 + log 1/1 = 12.4 or when HA= cB pH= pKa =12.4 9. b the carbonate/bicarbonate system would buffer near 10. (Carbonic acid /bicarbonated would buffer at pH 4) 10. d at pH 7 get about 60 % H 2 PO 4 - and 40 % H 1 PO 4 2- (very little of others) 11. d 12. a +/- 1 or 1 pH unit on either side of pKa
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Unformatted text preview: 13. e (2 marks) pH =pKa + log cb/a : .log cb/a= 4.5- 4.8= -0.3 or cb/a=10 4.5-4.8 =0.5=1/2 loga/cb =0.3 :. Acid/base= 2 14.b A varies with wavelength (absorption spectrum) 15 a A=ecd :. C =A/el = 0.4/(8000 M-1 cm-1 x0.5cm ) = 1x10-4 M (moles/L) 16.b amount of DNA in one bacterium= 600x 10-12 /10 6 = 6x10-16 gram Moles of base pairs 6.10-16 g /(600g /mole bp) = 1. 10-18 moles bp Number of base pairs 1. 10-18 moles bp x 6. 10 23 molecules/mole= 6. 10-5 molecules bp 3 bp/amino acid 1000 amino acids/protein :. No of proteins = 6. 10-5 molecules bp /(3 bp /amino acid x1000 amino acids /protein = 200 protein molecules...
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This note was uploaded on 08/27/2008 for the course MCB 200f taught by Professor Staff during the Spring '08 term at Gwinnett Technical College.

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