0.084%
About 13M. You can assume it is 13M if you don't need an exact concentration (like if you need a ~1M HCl solution for an extraction or whatever) but if you need an exact concentration (for a titration, for example) then you will need to standardize your HCl first.
To determine the number of moles in 8.63 mL of HCl, you need to know the concentration of the HCl solution. Once you have the concentration, you can use the formula: moles = concentration (mol/L) x volume (L) to calculate the moles of HCl present in the given volume.
Depends on the concentration of HCL but 255 to 264 gallons
Assuming that the 20% refers to the HCl concentration, the solute is water.
To prepare 0.5N HCl from 37% HCl solution, you can use the formula C1V1 = C2V2 where C1 is the initial concentration, V1 is the initial volume, C2 is the desired concentration (0.5N), and V2 is the final volume. Calculate the volume of 37% HCl needed and dilute it to the desired volume with water.
A 0.1N (Normal) HCl solution is equivalent to a 0.1M (Molar) concentration of HCl. This means that there is 0.1 moles of HCl per liter of solution. So, the percentage concentration of a 0.1N HCl solution would be 0.1%.
A 0.1 M concentration of HCl corresponds to a pH of 1.0.
1 N HCl (hydrochloric acid) is equivalent to a concentration of 0.1 M (molarity), which means it contains 1 mole of HCl per liter of solution. The percentage of HCl in 1 N solution is about 3.65%, calculated based on the molecular weight of HCl.
The concentration of HCl is 11,66 g/L.
Given that the amount of NaOH is known, and the volume of HCl required for titration can be measured, you can calculate the concentration of HCl by using the balanced chemical equation and stoichiometry. The concentration of HCl in the original solution can be determined by dividing the moles of NaOH used in the reaction by the volume of HCl used. This calculation will yield the molarity of HCl in the original solution.
Density of HCl = 1.186 g/mL Molar Mass HCl = 36.46g/mol We want a concentration in mol/L so we will first convert density into g/L (1.186 g/mL)(1000mL/1L) = 1186 g/L We must now know what density 37% of that is (0.37)(1186 g/L) = 438.82 g/L Now divide this density by the molar mass to cance out the g and give you mol/L (concentration) C = (438.82g/L)/(36.46g/mol) C = 12.04 M Since HCl is monoprotic, 1M = 1N. Therefore, 37% HCl is ~12N
Yes, when HCl is neutralized with NaOH, the concentration of HCl decreases as it reacts with NaOH to form water and NaCl. The concentration of the resulting NaCl solution will increase as the reaction progresses.