and no E2 = 0,535 - 0,059/2 *log( 1/(0,1)^2) ?E2 = 0,535 + 0,059/2 *log( 1/(0,1)^2)
Search found 31 matches
- Thu Mar 28, 2024 1:56 am
- Forum: Chemistry forum
- Topic: Redox reactions
- Replies: 5
- Views: 3130
Re: Redox reactions
Ok, but why did you write:
- Thu Mar 28, 2024 12:43 am
- Forum: Chemistry forum
- Topic: Redox reactions
- Replies: 5
- Views: 3130
Re: Redox reactions
Thanks, but I have some questions: 1. Why did you use plus and not minus in the second Nernst equation (for iodine). 2. Why did you assume that the activity of elemental iodine is 1? Iodine could also dissolve a little. When I was solving this, I thought as if I wanted to calculate the voltage durin...
- Wed Mar 27, 2024 12:22 pm
- Forum: Chemistry forum
- Topic: Redox reactions
- Replies: 5
- Views: 3130
Redox reactions
Please help: The standard electrode potential for the reduction of nitrate(V) ions to nitrate(III) ions is 0.940 V. The standard electrode potential for the reduction of iodine to iodide ions is 0.535 V. If we have a solution that has the same concentration of nitrate(V) and nitrate(III) ions and a ...
- Mon Mar 25, 2024 11:35 am
- Forum: Chemistry forum
- Topic: Redox reactions and standard electrode potentials
- Replies: 1
- Views: 1706
Redox reactions and standard electrode potentials
Hello, please help me with the following matter: I have given the standard electrode potential for the reduction of silver triiodide to silver and three iodide ions. The value is -0.029 V. Then I have given other standard electrode potentials such as E0(H+/H2) = 0.000 V etc. The question is which of...
- Fri Mar 22, 2024 2:34 am
- Forum: Chemistry forum
- Topic: Liquid-liquid extraction of weak electrolytes
- Replies: 1
- Views: 1706
Liquid-liquid extraction of weak electrolytes
Im asking for help:
Calculate the distribution coefficient D if the partition coefficient K = 50.0 for the benzene-water system if the pKb of the base (solute) is 5 and pH is 8.00.
Thank you!
Calculate the distribution coefficient D if the partition coefficient K = 50.0 for the benzene-water system if the pKb of the base (solute) is 5 and pH is 8.00.
Thank you!
- Sat Feb 10, 2024 10:54 am
- Forum: Chemistry forum
- Topic: Titration of mixture of KOH and K2CO3
- Replies: 11
- Views: 2420
Re: Titration of mixture of KOH and K2CO3
But still, if we consider the aliquot, the mass is too high.
- Sat Feb 10, 2024 1:13 am
- Forum: Chemistry forum
- Topic: Titration of mixture of KOH and K2CO3
- Replies: 11
- Views: 2420
Re: Titration of mixture of KOH and K2CO3
That's how I calculated it: c(HCl) = (6*0.1250)/(0.0357*214)= 0.09817 M n(K2CO3) = 0.0261*0.09817 = 0.002562 mol V(HCl for KOH) = 33.7 - 26.1 = 7.6 mL n(KOH) = 0.0076 * 0.09817 = 0.000746 mol m(K2CO3) = 0.002562 mol *138.21 g/mol * 5 = 1.7705 g 1.7705 g is much more than 0.7341 g. So where did I go ...
- Fri Feb 09, 2024 3:26 am
- Forum: Chemistry forum
- Topic: Titration of mixture of KOH and K2CO3
- Replies: 11
- Views: 2420
Titration of mixture of KOH and K2CO3
Weigh 0.7341 g of technical KOH, which also contains K2CO3, quantitatively transfer it to a 250 mL volumetric flask, dissolve and make up to the mark. Prepare two Erlenmeyer flasks, pipette 50 mL of the prepared solution into each, add the indicator phenolphthalein to the first, and methyl orange to...
- Fri Feb 02, 2024 3:02 pm
- Forum: Chemistry forum
- Topic: Electrochemistry
- Replies: 3
- Views: 2103
Re: Electrochemistry
I understood the task as if at the beginning we only have lead and H2SO4 and when they react PbSO4 is formed, and therefore the concentration of free SO4 2- ions must be less than 2 M. But this is obviously not the case and I understand it wrong. Can you please explain to me how to imagine this.
- Fri Feb 02, 2024 8:51 am
- Forum: Chemistry forum
- Topic: Electrochemistry
- Replies: 3
- Views: 2103
Electrochemistry
Please help me with this task, the result should be -0.366 V. I know how to come up with the result, but in the calculation I took into account that the concentration of sulfate ions is 2 M, which does not seem right to me. Calculate the potential of the lead electrode in 2 M H2SO4 if E0 = -0.126 V ...
- Sun Dec 17, 2023 7:53 am
- Forum: Chemistry forum
- Topic: Neutron activation analysis
- Replies: 3
- Views: 3451
Re: Neutron activation analysis
Determining the amount of mercury in hair. The basic method would be CVAAS, but I have to mention an alternative. I was thinking of taking NAA. I have to compare according to analysis time, detection limit, analysis type and other specifics.
- Sat Dec 16, 2023 11:12 am
- Forum: Chemistry forum
- Topic: Neutron activation analysis
- Replies: 3
- Views: 3451
Neutron activation analysis
Hi, do we need a calibration curve for NAA analysis or not? I also wonder how long this analysis takes?
Thank you!
Thank you!
- Sun Dec 03, 2023 2:17 am
- Forum: Chemistry forum
- Topic: Determination of Hg in human hair
- Replies: 10
- Views: 6604
Re: Determination of Hg in human hair
Thanks for the reply. I'm curious about one additional aspect of the hair sample processing. As I've read, the decomposition procedures are rather distinct, but because I've never done this before, I'm wondering if the decomposition with nitric acid and hydrogen peroxide is so good that no filtratio...
- Sat Dec 02, 2023 2:59 am
- Forum: Chemistry forum
- Topic: Determination of Hg in human hair
- Replies: 10
- Views: 6604
Determination of Hg in human hair
Hello, I have to write a term paper on the determination of mercury in human hair. I've read quite a few professional articles, but none describe the process in great detail. After reading everything, I somehow came to the conclusion that it would be best to use cold-vapor-atomic-absorption-spectome...
- Sat Oct 14, 2023 12:32 pm
- Forum: Chemistry forum
- Topic: Gravimetric determination of magnesium
- Replies: 1
- Views: 768
Gravimetric determination of magnesium
Hello,
can you please tell me how to write the chemical reaction that takes place between magnesium ion and (NH4)2HPO4 as a precipitating reagent.
Thank you.
can you please tell me how to write the chemical reaction that takes place between magnesium ion and (NH4)2HPO4 as a precipitating reagent.
Thank you.