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post-lab 7 ch 204 - obennoskey(blo242 Post-lab 7 Lyon(52635...

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obennoskey (blo242) – Post-lab 7 – Lyon – (52635) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. These questions are randomized and are different from the questions in the post-lab of other lecture sections. If you request help via the Facebook page, please indicate your lec- ture section and do not refer to a question number because the number of your question will be different than the number of the ques- tion on a peer’s assignment. Instead, write out the question and be specific about what you do not understand in your inquiry. Note: Some numerical questions in this assignment do not consider the number of significant figures for the correct answer. It will be noted at the beginning of the question how many sig figs should be used in the answer IF the answer does not require the correct number of sig figs. 001 (part 1 of 3) 10.0 points A compound with a molecular weight of 259 . 3 g was dissolved in a 5 . 0 mL volumetric flask (flask 1) to a volume of 5 . 0 mL. A 1 . 000 mL aliquot was removed from flask 1 and trans- ferred to a 10 . 0 mL volumetric flask (flask 2) where it was diluted to the mark. A sample from flask 2 had an absorbance 0 . 45 at 340 nm in a 1 . 000 cm cuvet. The molar absorp- tivity for this compound at 340 nm is ǫ 410 = 428 M - 1 cm - 1 . Calculate the concentration of compound in the cuvet. 1. 2 . 8 × 10 - 3 M 2. 1 . 1 × 10 - 3 M correct 3. 3 . 4 × 10 - 4 M 4. 2 . 5 × 10 - 4 M 5. 1 . 5 × 10 - 5 M Explanation: A = ǫ Cl C = A ǫ l = 0 . 45 (428 M - 1 cm - 1 )(1 cm) = 1 . 1 × 10 - 3 M 002 (part 2 of 3) 10.0 points What was the concentration of compound in flask 1? 1. 1 . 5 × 10 - 4 M 2. 2 . 5 × 10 - 2 M 3. 2 . 8 × 10 - 2 M 4. 1 . 1 × 10 - 2 M correct 5. 1 . 7 × 10 - 3 M Explanation: A 1 . 000 mL aliquot of the substance in the flask was diluted to 10 . 0 mL: M flask 1 =? V 1 = 1 . 000mL M flask 2 = 1 . 1 × 10 - 3 M V 2 = 10mL M flask 1 V 1 = M flask 2 V 2 M flask 1 = (1 . 1 × 10 - 3 M)(10mL) 1 . 000mL M flask 1 = 1 . 1 × 10 - 2 M 003 (part 3 of 3) 10.0 points How many milligrams of compound were used to make the 5 . 0 mL solution in flask 1? 1. 14 mg correct 2. 33 mg 3. 28 mg 4. 18 mg 5. 24 mg Explanation: (0 . 0050 L) × 1 . 1 × 10 - 2 mol L
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obennoskey (blo242) – Post-lab 7 – Lyon – (52635) 2 × 259 . 3 g mol × 10 3 mg g = 14 mg 004 10.0 points If a 0.0200 M solution exhibits 50.0% trans- mittance, what will be the percent transmit- tance for a 0.0400 M solution of the same substance measured at the same wavelength? Correct answer: 25 . 0% transmittance. Explanation: Percent transmittance is not linearly re- lated to concentration, but ansorbance is, so first you have to convert percent transmit- tance to absorbance, then double the ab- sorbance (because you’re doubling the con- centration), and then convert the new ab- sorbance back to percent transmittance.
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