add 5 ml of 37% HCl to 495 ml Water. This is 0.12 N ;)
CuO + 2 HCl -> CuCl2 + H2O Please see the link.
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.
Molar mass of HCl = 1 + 35.5 = 36.5 Number of moles of HCl = mass / molar mass = 85.6 / 36.5 = 2.345 moles Molarity = number of moles / Volume = 2.345 / 0.385 = 6.091 M
To dilute it.
g HCl solution = 2500 mL of HCl * 1 liter/1000 mL * 1190 g/L = 2975 g 37% solution (37 g HCl/100 grams of solution) gives you the grams of HCl: g HCl = 0.37 * 2975 g = 1100.8 g HCl Moles HCl = 1100.8/(36.46 g/mole) = 30.2 moles Therefore the molarity, which equals the normality in this case = 30.2 moles/2.5 L = 12.07 M = 12.07 N If you want to make 100 mL of a 0.1 N solution, Volume of HCl solution needed = (0.1 N * 100 mL) /12.07 N = 0.83 mL Take 0.83 mL of the 37% HCl, and dilute it with water to 100 mL.
1M contains 43.1ml in 500ml, 2M contains 86.2ml in 500ml
3.6ml of HCl is diluted and make it into 500ml with distilled water. --------------------------------------------------------------------------------- Add 50 mL of HCl 1 N in a 1 L volumetric flask, class A or B; add ca. 900 mL distilled water to the flask. Place the flask in a thermostat at 20 0C. After 30 min add slowly distilled water to the mark (1 L) and stir well the closed flask. Pour the solution in a bottle. Place a label with the date, concentration, name of the solution on the bottle.
6g Tris HCl + 100ml dH2O, pH 6.8
520 ml of HCl in 480 ml of water=1000ml = 5 N
Prepare HCl 1 M by HCl concentration 37 % HCl concentration 37 % have density =1.19 g/ml HCl 1 M use HCl 37 % 82.81 ml make volume with water to 1 liter
25 mL
1.21 g Tris-HCl, QS water to 1L. Scale appropriately.
Mix approx. 12 mL of HCl 30 % in 1 L water.
dilute your HCl solution to 0.2 M HCl solution and then follow above mentioned link : http://delloyd.50megs.com/moreinfo/buffers2.html
Image result for You prepare a less concentrated H C l solution from a stock solution with 12m concentration. If you too 100g of the stock solution to prepare 4 MHCl solution how much water is needed to prepare o find solution 9density HCL(12) = 1,89/ml? The concentration would be 0.76 mol/L.
44.5 ml HCl TAKE AND DILUTE UP TO 1000 ML WATER MAKE A 0.5 M HCl SOLUTION
Yes, it is possible to prepare oxalic acid by adding hcl to a solution og sodum oxalate. The balance equation would be C2O4Na2 + 2HCl -----> 2NaCl + C2O4H2.