วีดีโอ

Thank you so much for your kind words,, stay tuned 🙂

Xylenol Orange Indicator; [3H-2,1-Benzoxathiol-3-ylidenebis-(6-hydroxy- 5-methyl-m-phenylene )methylenenitrilo ]tetra acetic acid S,S-dioxide Tetrasodium Salt: C3IH28N2NH2OI3S=760.60 General laboratory reagent grade of commerce. Reddish brown, crystalline powder; gives a violet color with mercury, lead, zinc and certain other metal ions, in alkaline solutions. When metal ions are absent, for example in the presence of an excess of disodium edetate, the solution is yellow. XylenolOrange Mixture Indicator; Xylenol Orange Triturate Indicator: Triturate 1 part of xylenol orange with 99 parts of potassium nitrate. Complies with the following test. SENSITIVITY - Add 50 mg to a mixture of 50 ml of water, 1 ml of 2 M acetic acid and 0.05 ml of lead nitrate solution. Add sufficient hexamine to change the color from yellow to violet red. Add 0.1 ml of 0.1 M disodium edetate; the color changes to yellow. Xylenol Orange Indicator Solution: Mix 0.1 g of xylenol orange with 100 ml of water and filter, if necessary. Read more at: www.pharmaguideline.com/2010/09/preparation-of-indicator-solutions.html?m=1 please check for details on this website

Sir please give me sometime because I am suffering from some health problem. Video has already been recorded, probably by this Sunday I will try, thanks for watching and showing your concern

Thank you very much

🙏

Thanks 🙏

Thank you!

Thank you very much!

Thank you so much for your kind words 🙏

It is given in th-cam.com/video/ggDTykVWYtk/w-d-xo.htmlsi=7Sqc05tSpGtsITl5 this video

Thank you very much!

Thanks for watching, I am not sure but please go through the concentration written over there 1. if it is in less concentration then you can go without dilution, if it is in high concentration then you can dilute it, 2. solvent you have to check if it is soluble in water or in any other solvent that you have to check first because in organic solvents may be you will not get the desired results, I hope it will work, please share your results so that we all will get benefited with your experience, thanks

Pehle se pdh liya hota to aaj yahan nhi aana tha tujhe ..🎉

Glad to hear that, thanks

Thank you so much

I am planning to perform this experiment but due to other assignments I couldn't, if you didn't dilute the solution then no need to multiply, here I would suggest you should dilute because it is very concentrated, thanks for watching, please like share and subscribe

Thank you so much, stay tuned

Thank you so much

Your words really makes me happy 😊, thank you

Thank you so much for your kind response 💛

Glad to see your response 🙏

You'll not get exact N/10 it should ranges between 0.97 N/10 or 1.01N/10 You've to calculate with that, it's not at all problematic

Thank you!!

Thanks for watching, you please elaborate your question here is the video for preparation of solutions th-cam.com/video/H7yR2DzVOTU/w-d-xo.htmlsi=1wOdKqfrs5zLVa28 If you are asking for other places write me back thanks 🙏

Thank you for your kind response 🙏

Thanks for watching, actually when I made this video I forgot to mention that in this video we are just comparing the energy levels that in both LS and jj coupling same number of energy levels are observed, here energy terms taken w.r.t LS coupling. Since coupling intermediate between the two extremes can occur but terms can always be identified by their J values because the total angular momentum (J) of the atom is conserved as the coupling is changed. Thanks for highlighting the point this certainly helps to other viewers.

Many many thanks

Thanks for watching, you please use the formula wt = (n/44)x molecular weight of sodium thiosulfate x volume in litre; the weight you get from this formulae dissolve that in to the volume you need to prepare e.g you wish to prepare in 100 ml then put (100 ml/1000 ml i.e., 0.1 litre) put 0.1 litre in the formula. Here important point is in case of normality we use equivalent weight (= molecular weight/ oxidation state) in case of sodium thiosulfate vs iodine solution the change in oxidation state= 1, therefore in this case equivalent weight= molecular weight; if you are doing some other titration or for some other purpose then you can write me back the equivalent weight may be changed, if this is helpful to you give me a 👍, thank

Glad to know that you like it 🙂

Thank you

Thank you so much for your nice response, stay tuned 🙂

Thanks for watching and sorry for the delayed response, in case of adsorption it is always written solute adsorbed per gram of adsorbent as I have gone through many questions so in my opinion it is not necessary to convert to mg as in the answer we are mentioning the amount per gram of adsorbent, if I will find this somewhere else I will certainly mention here., thanks and stay tuned

Both relaxation process are responsible for NMR spectra

😊

Thank you so much for watching and your concern, sorry for the delayed response. For super saturated solution there are dissolved and undissolved species present that are in equilibrium, it can be better understand by knowing the fact that in solution, NaCl is completely dissolved in water that means it is completely dissociated in the water which is written as NaCl----) Na+ + Cl- ; means there is no equilibrium (equilibrium exists for partially dissociated salt). But in case of super saturated solutions undissociated and dissociated both kind of species present for NaCl and the reaction now written as NaCl= Na+ + Cl-, I hope I am able to explain the point you asked, if not then you write me back, thanks

Thanks