All right, my friends at Cramster, I have a huge favor to askof you: I am going

Question

All right, my friends at Cramster, I have a huge favor to askof you: I am going to state a few questions and I need some helpBADLY.   I am THIS close to dropping this Chem II classI'm in.   Doing that will postpone my graduation for anentire year. So please, if you'd like to answer me, please answer sensiblyand give a complete answer.   I have read my lab manualfrom COVER to COVER and I'm about to go insane!! Please helpme, and I'd be somewhat sane again... Okay, I got as far as this question:   If the %T ofa solution is 82.7, what is the absorbance of that solution? I used A = 2.00 - log(T%) = 2.00 - log(82.7) = .0825 I figured that much was right, but if it's not, tell me. The next question is what's driving me nuts: A student mixed 4.00 mL of 1.02 x 10^-1 M Fe(NO3)3 with 100.0mL of 1.98 x 10^-4 M KSCN, using 5.0 x 10^-1 M HNO3 as the solventfor both solutions, and found the absorbance of the resultingequilibrium mixture to be 0.253.   Given these data,calulate the following: 1.   [Fe*] Okay so far, I used [Fe3+] + [FeSCN2+], but I don't know whatto do with it... 2.   Calculate: [SCN*] I used [SCN-] + [FeSCN2+], but I don't know what to do withit. 3.   Calculate: [Fe*] + [SCN*] I used ([Fe3+] + [FeSCN2+]) + ([SCN-] + [FeSCN2+]) = [Fe3+] +[2FeSCN2+] + [SCN-] but didn't know what to do with it. 4. Calculate [Fe*][SCN*] Totally lost after this... 5.   A([Fe*] + [SCN*]) 6. A/ [Fe*][SCN*] 7. A([Fe*] + [SCN*]) / [Fe*][SCN*] I know it's a lot, but I promise a rate of 10 if you helpme!!   :-) All right, my friends at Cramster, I have a huge favor to askof you: I am going to state a few questions and I need some helpBADLY.   I am THIS close to dropping this Chem II classI'm in.   Doing that will postpone my graduation for anentire year. So please, if you'd like to answer me, please answer sensiblyand give a complete answer.   I have read my lab manualfrom COVER to COVER and I'm about to go insane!! Please helpme, and I'd be somewhat sane again... Okay, I got as far as this question:   If the %T ofa solution is 82.7, what is the absorbance of that solution? I used A = 2.00 - log(T%) = 2.00 - log(82.7) = .0825 I figured that much was right, but if it's not, tell me. The next question is what's driving me nuts: A student mixed 4.00 mL of 1.02 x 10^-1 M Fe(NO3)3 with 100.0mL of 1.98 x 10^-4 M KSCN, using 5.0 x 10^-1 M HNO3 as the solventfor both solutions, and found the absorbance of the resultingequilibrium mixture to be 0.253.   Given these data,calulate the following: 1.   [Fe*] Okay so far, I used [Fe3+] + [FeSCN2+], but I don't know whatto do with it... 2.   Calculate: [SCN*] I used [SCN-] + [FeSCN2+], but I don't know what to do withit. 3.   Calculate: [Fe*] + [SCN*] I used ([Fe3+] + [FeSCN2+]) + ([SCN-] + [FeSCN2+]) = [Fe3+] +[2FeSCN2+] + [SCN-] but didn't know what to do with it. 4. Calculate [Fe*][SCN*] Totally lost after this... 5.   A([Fe*] + [SCN*]) 6. A/ [Fe*][SCN*] 7. A([Fe*] + [SCN*]) / [Fe*][SCN*] I know it's a lot, but I promise a rate of 10 if you helpme!!   :-)

Explanation / Answer

Its not as hard as it looks. All you have to do is solve for (fe) and (SCN) the rest should fall into place. You know that A=.253 because it is given to you. The next step is to solve to total volume which would be 4mL + 100 mL = 104mL Then you solve for Fe by taking 4ml x 1.02*10^-1 mmol/mL take that answer and divide it by total volume. Use the same idea for SCN. 100mL x 1.98*10^-4M = Then take that answer and divide by total volume Once you get those two then rest is plug in and answer Just be careful with your decimal placements.

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