1 molar (aka 1M) is a concentration of 1 mole/liter. This means there are 36.4609g (1 mole) of HCl ions dissolved in every liter of solution. In particular, 1.0079g H+ cations (which will form an ionic bond with a water molecule to form H3O+, hydronium) and 35.453g Cl- anions.
This comes out to roughly 3.65% HCl, 96.35% H2O.
Now, I don't believe there is 100% pure HCl around, but one could start with ~36% aqueous HCl with a density of ~1.18 g/mL, and calculate the volume needed. Be sure to add the acid to the water or there could be a violent reaction.
Amount of HCl is 500cm^3 of .8M HCl = .8 x 500/1000 = 0.4 mol. Volume of 10M acid to give 0.4 mol = .4/10 = 0.04dm^3 = 40cm^3 Add 500 - 40 = 460 cm^3 of water. Caution : 10M HCl is highly corrosive and should be handled with care.
To prepare a 10M solution of acetic acid, dissolve 60.05g of glacial acetic acid (CH3COOH) in enough water to make a final volume of 1 liter. The molar mass of acetic acid is 60.05 g/mol. Make sure to wear appropriate safety gear, as acetic acid is corrosive.
To prepare 10M ammonia solution, slowly add 35.5 mL of concentrated (28-30%) ammonia solution to approximately 50 mL of water in a fume hood. Then, dilute the solution with water to a final volume of 100 mL. Remember to handle ammonia solution with caution due to its hazardous properties.
37% HCl = { [ 37 (gHCl) / 100 (g sol'n) ] / 36.46 (gHCl/molHCl) } * 1,18 (g sol'n/mL sol'n) * 1000 (mL/L) = about 12 M HClTo prepare V Litres of 10M sol'n you should dilute [10*V]/12 Litre of 37% HCl solution to the total volume of V. Be very carefull with this addition of water.
To prepare 10mL of a 25M HCl solution, you would need to dilute the concentrated HCl solution with the appropriate amount of water. For example, to make a 25M solution, you could start with a 10M HCl solution and dilute it appropriately. To determine the specific volume of each solution needed for dilution, you can use the formula C1V1 = C2V2, where C1 is the initial concentration, V1 is the volume of the initial solution, C2 is the final concentration, and V2 is the final volume.
Amount of HCl is 500cm^3 of .8M HCl = .8 x 500/1000 = 0.4 mol. Volume of 10M acid to give 0.4 mol = .4/10 = 0.04dm^3 = 40cm^3 Add 500 - 40 = 460 cm^3 of water. Caution : 10M HCl is highly corrosive and should be handled with care.
To prepare a 10M solution of acetic acid, dissolve 60.05g of glacial acetic acid (CH3COOH) in enough water to make a final volume of 1 liter. The molar mass of acetic acid is 60.05 g/mol. Make sure to wear appropriate safety gear, as acetic acid is corrosive.
you need 10m to make a guild
Acids are vigourous in nature. Therefore, it is completely ridiculous to pour water in acids( as this reaction is exothermic). Therefore, we take a beaker of water and pour the acid drop by drop to the water. Care must be taken and solution must be continuously stirred.
10M 10M 10M
1 dam3 = 10m*10m*10m = 1000 m3.
1.5-10m = -8.5
16m-10m = 6
I would place this triangle in the category of isosceles triangles, because the 10m side and the 10m side have equal lengths.
It could be something like 10m x 10m
10m X 10m
100cm = 1M 1,000cm = 10M