Asked by Anonymous
Describe how you would make 250 mL of 0.500 M HCI from 12.1 M HCI. Show calculations.
Answers
Answered by
John Wheeler
Using the dilution formula M1V1 = M2V2 you would use the following calculations.
((250 mL)x(0.500 M HCl))/12.1 M HCl
= 10.3 mL
This means you want to add 10.3 mL of the 12.1 M HCl into a 250-mL volumetric flask and then fill the rest with distilled water up to the etch line in the flask.
((250 mL)x(0.500 M HCl))/12.1 M HCl
= 10.3 mL
This means you want to add 10.3 mL of the 12.1 M HCl into a 250-mL volumetric flask and then fill the rest with distilled water up to the etch line in the flask.
Answered by
Garrett
How did you get 12.1 M HCI?
Answered by
Kimberly Avulelhon
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Answered by
Christopher chester
The materials used are as follows:tripple beam balance,evaporating basin,stop watch,HCL pellets,matches,candle, .weigh the empty basin on tripple beam balance and records the mass,then put hcl on evaporating basin,burn it until it change from blue to white,then put it again on tripple beam balance and records the mass,then calculate the mass of empty basin and that with hcl,state the sources of erros after experiments.this is how u can prepare 250 ML in 0.500 M HCL
Answered by
Anonymous
given 150ml of kcl solution of20g/l.prepare75mlof kcl solution whose concentration has to be 12g/l.
Answered by
Migos
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Answered by
Mike
The individual who put "Versace" is an answer is the pinnacle of comedic ability and should be lauded as such!
Answered by
Gabi
My answer is 80 mL HCl.
I used the dilution formula:
250 mL x .500 M HCl = "X" mL x 12.1 M HCl
= 80 mL
I used the dilution formula:
250 mL x .500 M HCl = "X" mL x 12.1 M HCl
= 80 mL
Answered by
Jason
Describe how you would make 1.50 Liters of 0.555 M magnesium chloride solution from a 3.77 M magnesium chloride solution already prepared?