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.
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.
you need 10m to make a guild
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