Hi,
Diamagnetic materials have negative susceptibility and all the electrons in those materials are usually paired.
As you said, the opposite alignment is because of Lenz law of electromagnetic induction. The motion of electrons in orbit forms a current loop, and this loop will be oppositely...
Hi,
Just want to know post 8 correct or not? I have done for w/w and also for w/v.
PS: in post 8, w/w calculation is correct, just verified.
But in w/v calculation I have doubts.
Hi, I will prepare NaOH solution using NaOH flakes. PhD in physical chemistry. But my research work was related to synchrotron x-ray characterization of molecules (no chemistry lab work or synthesis). In one website "Add 72.31% of water in the mixing tank and start mixing. Add 5.04% of sodium...
Hi,
I am prepare liquid detergent for my self, hobbyist! For the past 3 yrs I was preparing powder, which do not require NaOH. Now I am preparing liquid detergent, which is relatively easy than preparing powder. I am aware of the risks and danger in handling caustic soda and labsa. Already...
Hi Borek,
Yes. 20 g NaOH in 80 g water gives 20% (w/w) NaOH solution, AND
dissolving 20 g NaOH in ~60 ml water and then again adding water so that the final solution will be 100 ml. This is 20% (w/v) NaOH solution.
Correct?
Thanks
Hi, to prepare 20% NaOH solution, dissolving 20 g of NaOH in little water (~60 ml) and after cooling add further water so that final volume becomes 100 ml. Now this final solution will 20% NaOH solution (20g/100ml=20%w/v). Is this correct? I need to know this detail for liquid detergent...
It is weight percentage. Denoting the concentration by weight. Here the weight of NaOH dissolved in water, expressed in percentage. If it is confusing, please ignore it
Thanks
Dear All,
To prepare 20% (w/w) NaOH solution is it okay to dissolve 20 g of NaOH in 80 g of water.
or
20 g of NaOH in 100 ml water?
Which is correct?
Thanks for your reply,
rajini.
PS: This is not a homework question, I am already 43 yrs!
Hi blue leaf:
I think in the usual exponentially modified Gaussian (EMG) function
##y=\frac{ac\sqrt{2\pi}}{2d}\rm{exp}[\frac{c^2}{2d^2}-\frac{x-b}{d}][\frac{d}{|d|}+\rm{erf}(\frac{x-b}{c\sqrt2}-\frac{c}{d\sqrt2})]##
##ac\sqrt{2\pi}## represents area under the EMG function.
For reversed EMG one...
Hi blue leaf:
Thanks for your reply.
I just understand that to get a reversed EMG, I need to change the sign of 'd'.
So i took d as -3 and got the correct reversed EMG.
So now both (one with d as 3, and other as -3) EMGs has same height of 0.9.
But I am confused why 0.9 instead of 2?
Regards, Rajini
Hello all,
I have a data which look like reversed exponentially modified Gaussian (EMG) function and interested to fit the data with with reversed EMG function. After searching on internet I found the EMG function, which is given below...
Hi Khashishi,
An exponential decay is still possible (considered as 2 level system). The decay from 136 keV level to 14.4 keV level emit 122 keV gamma. The detection of 122 keV acts as a start signal for 14.4 keV level. Now the 14.4 keV level decays to ground state by emitting 14.4 keV gamma...
Hello Khashishi,
That data is from coincidence count experiment obtained form Moessbauer spectrometer. The source/radioactive 57Co is used in such experiments. Kindly see the decay scheme of 57Co source more details. In short the 57Co decays to the 136 keV level and then to 14.4 keV level and...