exam2spr07form1expl-1

exam2spr07form1expl-1 - Second exam, FORM 1 page 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Second exam, FORM 1 page BIO201C/Hon—Janicke—Spring 2007 If you have not already done so, IMMEDIATELY fill in the bubbles on the back of the answer sheet for (1) your name (LAST NAME FIRST) (2) 8-digit person number, (3) seat number (4) FORM 1 (under “GRADE or EDUCATION”) On the front of your scantron on the top line, sign your name and the code word “second time.” You will not be given extra time to do this at the end. You receive a ZERO if you hand in a scantron with no name on it, fail to fill in your form number, use notes of any kind, or copy another student’s work. Other procedural errors in your scantron or in exam rules (such as hats, calculators of your own, cell phones heard or seen, or talking) will result in deduction of 5 points (if cheating is NOT suspected). Be sure erasures are complete. You may ask for a new scantron if you’ve ruined yours. CHOOSE ONE AND ONLY ONE ANSWER PER QUESTION. 0 6 0 1. Andrew measures out some 1M HCl with his P200 pipetman set as above, then expels the acid into 540 ul of water. What is the weight, concentration, and pH of the resulting mixture? weight [HCl] pH A 600 mg 0.1M 1 B 600 ug 1M 0.2 C 546 ug 0.01M 0.01 D 546 mg 0.2M 2 E 60 g 2M 0.1 60 ul + 540 ul is 600 ul total volume. (600 x10^-6 liters water)(1 kg water/1 liter water)(1x10^3g/kg) = 600 x10^-3g = 600mg 60 ul of 1M in 600ul total volume is a 1/10 dilution. So the HCl has been diluted from 1M to w now 0.1 M = 1x10^-1M. HCl is a stong acid that dissociates virtually completely into H+ and Cl-. So 1x10^-1M HCl is 1x 10^-1 H+ pH = - log [H+] = - log(10^-1) = - (-1)= 1 2. Before measuring a concentration of DCIP with a spectrophotometer, you should A. Place a concentrated solution of DCIP in the sample chamber and set %T to 100. B. Leave the sample chamber empty, close the lid, and set Absorbance to zero. C. Place the solvent (without any DCIP) in the sample chamber, and set %T to zero. D. Place the solvent (without any DCIP) in the sample chamber, and set Absorbance to zero. E. Leave the sample chamber empty, close the lid, and set Absorbance to zero. 3. Kevin has 10 ml of 1 mM DCIP. If one mole of ascorbic acid reduces one mole of DCIP, what is the minimum amount of 1 M ascorbic acid stock that will convert his DCIP entirely to DCIPH2? He has 10 ml (1 millimoleDCIP/1000 ml) = 1x10^-2 millimoles of DCIP. He needs 1x10^-2 millimoles of ascorbic acid to reduce that much DCIP. How many ml of 1 M DCIP do you need to get 1x10^-2 millimoles? 1x10^-2 millimole DCIP (1x10^3ml/1x10^3moles DCIP)= 1x10^-2 ml = 0.010 ml = 10 ul. 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Second exam, FORM 1 page A. 10 ml B. 1 ml C. 0.1 ml D. 10 ul E. 1 ul 4. If 3 ml of a 1 mM solution with an absorbance of 1 is mixed with 3 ml of water, the new absorbance will be A. 2 B. 0.5 C. 0.3 D. 1.3 E. 1.5 5. The Hill reaction can be assayed by conversion of DCIP (which absorbs at 600 nm) to DCIPH2, which does not absorb at 600 nm. Rick has a chloroplast sample with an absorbance at 600 nm of 0.3. He mixes 3 ml of his chloroplast sample with 1 ml of 0.4 mM DCIP in the presence or
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/08/2008 for the course BIO 201 taught by Professor Janicke during the Spring '08 term at SUNY Buffalo.

Page1 / 9

exam2spr07form1expl-1 - Second exam, FORM 1 page 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online