I'm not sure about the 37 thing but here is from NIOSH method in how to prepare 6N HCL -pipette 25.64 mL of 11.7 N (37% HCL fuming) to 50 mL volumetric flask and top up with distilled water GOOD LUCK Always add acid to water.
1N HCl is also 1M HCl because it is mono-protic. Therefore 36.5 g of HCl is required per liter or 3.65%. Simply take 100 g of 37% HCl and make up to the 1 liter mark on the volumetric flask. Check the value by titration against 1M NaOH. It should be perfect. If very slightly strong dilute very slightly (calculate) with water and re-standardize.
1N - conc will change the pH too quickly. You might want to consider 0.1molar for finer adjustments.
very strong - 6 molar
weigh out 365g of HCl pellets and dilute to 1 liter to prepare the 10N HCl solution. Reasons:N is short for NORMAL SOLUTIONS, The definition of a NORMAL SOLUTION is a solution that contains 1 gram equivalent weight (gEW) per liter solution. An equivalent weight is equal to the molecular weight divided by the valence (replaceable H ions). eg:1N NaCl = 58.5 g/L 1N HCl = 36.5 g/L 1N H2S04 = 49 g/L Problems involving normality are worked the same as those involving molarity but the valence must be considered: 1N HCL the MW= 36.5 the EW = 36.5 and 1N would be 36.5g/L 1N H2SO4 the MW = 98 the EW = 49 and 1N would be 49 g/L 1N H3PO4 the MW = 98 the EW = 32.7 and 1N would be 32.7 g/L so,u can weigh out 365g of HCl pellets and dilute to 1 liter to prepare the 10N HCl solution.
I'm not sure about the 37 thing but here is from NIOSH method in how to prepare 6N HCL -pipette 25.64 mL of 11.7 N (37% HCL fuming) to 50 mL volumetric flask and top up with distilled water GOOD LUCK Always add acid to water.
Preparing 1N HCl for 1L. 1N=1M in HCl. Conc. HCl= 12M M1V1=M2V2 12*V1=1*1000 V1=1000/12 V1=83.33ml 1N HCl= 83.33ml of Conc. HCl in 1L of water 2N HCl= 167ml of Conc. HCl in 1L of water.
Take specific volume of 3N solution and increase the volume three times by adding distilled water.
1N HCl is also 1M HCl because it is mono-protic. Therefore 36.5 g of HCl is required per liter or 3.65%. Simply take 100 g of 37% HCl and make up to the 1 liter mark on the volumetric flask. Check the value by titration against 1M NaOH. It should be perfect. If very slightly strong dilute very slightly (calculate) with water and re-standardize.
1N - conc will change the pH too quickly. You might want to consider 0.1molar for finer adjustments.
very strong - 6 molar
1N HCL is the same as 1 Molar HCl. You take the # of H ions and multiply by the molarity to get the Normality. Usually you buy HCl in concentrated form which is 12 Molar or 12 Normal HCL. You need to dilute the concentrated HCl to get the reduced concentration. Use the formula Molarity Initial x Volume Initial = Molarity Final x Volume Final ex. 12 M HCL x 10 ml = 1 M x 120 ml. So take 10 ml of concentrated HCl and add enough water to make 120 ml. This will give you 120 ml of 1 M (which is 1N) HCl. Venkat Reddy
weigh out 365g of HCl pellets and dilute to 1 liter to prepare the 10N HCl solution. Reasons:N is short for NORMAL SOLUTIONS, The definition of a NORMAL SOLUTION is a solution that contains 1 gram equivalent weight (gEW) per liter solution. An equivalent weight is equal to the molecular weight divided by the valence (replaceable H ions). eg:1N NaCl = 58.5 g/L 1N HCl = 36.5 g/L 1N H2S04 = 49 g/L Problems involving normality are worked the same as those involving molarity but the valence must be considered: 1N HCL the MW= 36.5 the EW = 36.5 and 1N would be 36.5g/L 1N H2SO4 the MW = 98 the EW = 49 and 1N would be 49 g/L 1N H3PO4 the MW = 98 the EW = 32.7 and 1N would be 32.7 g/L so,u can weigh out 365g of HCl pellets and dilute to 1 liter to prepare the 10N HCl solution.
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7n -1n,0n,1n,2n,3n,4n,5n,6n,7n 1 2 3 4 5 6 7 8 9
take the Mr of HCl to be 36.46. to prepare 1.0M we should take 36.46g and dilute to a litre to prepare 0.10M we should take 3.646g and dilute to a litre From density =mass/volume, we get volume=mass/density. So ideally we should take that volume and dilute to a litre. However we know that most conc HCl solution are not 100% HCl. thus we must multiply that volume by a factor 100/x for an HCl solution labelled x% Tauya Chigayo
By dilution (1000x) with water: Take 1.0 mL 1.0N HCl and add up to 1000 mL with pure water.