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Specific Process Knowledge/Etch/KOH Etch/ProcessInfo: Difference between revisions

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DCH-Employees-701, NLAB-Employees-701, NLAB-LabmanagerAllUsers
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Kabi (talk | contribs)
DCH-Employees-701, NLAB-Employees-701, NLAB-LabmanagerAllUsers
1,146 edits
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<math>(0.85X)/(X+1000g)=Y</math>
(0.85∙X)/(X+1000g)=Y




<math>X=(Y∙1000g)/(0.85-Y)</math>
X=(Y∙1000g)/(0.85-Y)
 


So for a 10wt% solution you will need 133 g pills for each liter of water.
So for a 10wt% solution you will need 133 g pills for each liter of water.
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If you use the 50wt% premade solution you can calculate the amount of solution X you need to add 1 liter of water to achieve a concentration of Y wt% by:
If you use the 50wt% premade solution you can calculate the amount of solution X you need to add 1 liter of water to achieve a concentration of Y wt% by:


<math>(0.5∙1.509g/ml∙X)/(X∙1.509g/ml+1000g)=Y</math>


<math>X=(Y∙1000g)/(1.509g/ml∙(0.5-Y))</math>
(0.5∙1.509g/ml∙X)/(X∙1.509g/ml+1000g)=Y
 
 
X=(Y∙1000g)/(1.509g/ml∙(0.5-Y))
 


where 1.509 g/ml is the density of a 50wt% solution KOH at 20 °C.  
where 1.509 g/ml is the density of a 50wt% solution KOH at 20 °C.  
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