Since the molar mass of is 210.14 g·mol−1, there are
0.358 (mol) * 210.14 (g/mol) = 75.2 g trimellitic acid in 0.358 mole of it.
(.05)X(grams of total solution) = grams of acetic acid (grams of acetic acid)/ (mol. wt. of acetic acid(=60g/mol)) = mol. acetic acid (mol. acetic acid)/ (Liters of total solution) = molarity(M)
The formula of acetic acid is NOT CHCOOH but CH3COOH which also is found back in the corrected molar mass:The total molecular mass. 12.01 gC/mol + (3 *1.008) gH/mol + 12.01 gC/mol + 2*16.00 gO/mol + 1.008 gH/mol = 60.05 grams per mole
A mol = 6,022x10^23 atoms 1,5 mol = 9,033x10^23 atoms Always.
First you need the molar mass of citric acid. It is 192 g/mol. Now you can calculate the number of moles: 0.1 g x (1 mol/192 g) = 0.00052 mol
The number of carbon atoms which are present in 0,062 mol acetic acid HC2H3O2 is 0,747.10e23.
(.05)X(grams of total solution) = grams of acetic acid (grams of acetic acid)/ (mol. wt. of acetic acid(=60g/mol)) = mol. acetic acid (mol. acetic acid)/ (Liters of total solution) = molarity(M)
The formula of sulfuric acid is H2SO4, showing that each mole of sulfuric acid contains four mol of oxygen atoms. Therefore 750/4 or 187.5 mol of sulfuric acid will be sufficient.
The formula of acetic acid is NOT CHCOOH but CH3COOH which also is found back in the corrected molar mass:The total molecular mass. 12.01 gC/mol + (3 *1.008) gH/mol + 12.01 gC/mol + 2*16.00 gO/mol + 1.008 gH/mol = 60.05 grams per mole
A mol = 6,022x10^23 atoms 1,5 mol = 9,033x10^23 atoms Always.
First you need the molar mass of citric acid. It is 192 g/mol. Now you can calculate the number of moles: 0.1 g x (1 mol/192 g) = 0.00052 mol
The number of carbon atoms which are present in 0,062 mol acetic acid HC2H3O2 is 0,747.10e23.
M = mol L-1 so 2.5M of H2SO4 means in one Litre of a 2.5M solution of Sulfuric acid, there is 2.5 mol of dissolved NaCl
AnswerpH=1 --> [HCl] = 0.10 mol/L , and this is why: By definition [H+] = 10-pH or pH = -log [H+]So at pH = 1.0 level the [H+] concentration is [H+] = 10-pH = 10-1.0 = 0.10 mol/LIf the acid is a strong, monoprotic acid like HCl and HNO3 then the acid concentration equals the [H+] = 0.10 mol/L HClIf the acid is a strong but diprotic acid like H2SO4 then the acid concentration equals half of [H+] , so acid conc. = 0.050 mol/L H2SO4
62.03 g/mol
98.08 g/mol
Assuming that you mean that your pure acid is twice as concentrated as your 50% acid. Pretending your pure acid is at 1 mol/gallon and the 50% acid is 0.5 mol/gallon1 molgallon-1 * x gallons = x mol0.5 molgallon-1 * 4 gallons = 2 mol(x + 2) mol for (x+4) gallons = 0.8(x+2) mol / (x+4) gallons= 0.8 molgallon-1x + 2 = 0.8x + 3.20.2x = 1.2x = 6add 6 gallons of pure acid to the 4 gallons to make 10 gallons of 80% acid solution
2.25 mol X 98 g mol-1 = 220.5 g