Fill in the blank where the pink is. Sulfur-38Original amount: 5.0 gAmount left: rTotal time: 8.61 hHalf-life: 2.87 hNumber of half-lives:

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Sulfur-38 Original amount: 5.0 g Amount left: c Total time: 8.61 h Half-life: 2.87 h Number of half-lives:

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Sulfur-38
Original amount: 5.0 g
Amount left: r
Total time: 8.61 h
Half-life: 2.87 h
Number of half-lives:

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Question 4 1. The absorbance of a M2*solution containing 2 mg M²*/mL was reported as 600 at 732 nm. What will be the absorbance if the solution is diluted to twice its original volume? 2. If absorbance of the sample is 1.00 at a specific wavelength, what is the percentage amount of light that is transmitted?
em1.png Part 3: Q9 Average absorbance Flask 8 Flask 9 Flask 10 1:1 Ratio so, Flask 11 Part 3: Q Flask 8 Flask 9 Flask 10 Flask 11 Part 3: Q10 Flask 8 Flask 9 Flask 10 Flask 11 Part 3 0.146 0.166 [Fe] -0.002mol/L 0.215 0.294 [FESCN³mol/L) 4.59% 10 5.18*10* 6.64*10* 9.08*10* [FXmol/L) 5.00 *10* 5.00*10* Y=3362x-0.0083 R² = 0.9993 5.00*10* 5.00*10* (Amounts of flask 8 (9 - 11's values are in the spreadsheet) 71% volume 5.00mL Molarity of Fe(NO3))3 = 0.2M Volume of Fe(NO3))3 = 5.00mL Moles of Fe³= Molarity x volume(L) Moles of Fe¹ = 0.2M X 0.005L = 1x10-³mole x [SCN][mol/ 4x10 5×10-4 7x10 1×10 1 Concentration of Fe³ = mole of Fe³ Volume mole of Fe³1x10³ mole G.O05L 8, Average absorbance for solution 8 (same for 9 - 11) Average absorbance for flask 8 = sum of all absorbances / number of readings (0.145 +0.147 +0.148) 3 Using equation of calibration curve's line(slope) of best fit, the average absorbances of the unknown concentrations to determine [FeSCN-Jat equilibrium in solution 8:…
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