90. mL NaOH, remember there are two sig figs in this problem not one, hence the decimal point after the 90.
35 mL
38.5*0.15 = 5.775 mol
5.775 / 3 = 1.925 mol
1.925/ 15.0 = 0.128 M
2
We need 158,3 mL HrPO4 5M.
71.8 M
[600 mL]*[1 L / 1000 mL]*[0.1 mol/L]*[97.995 g/mol] = [0.06 mol]*[97.995 g/mol] = 5.81 g H3PO4
97.8 - 98.2 98 worked for me
0.0532
How many grams of KHP are needed to exactly neutralize 36.7 mL of a 0.328 M barium hydroxidesolution
0.0932 L
2
We need 158,3 mL HrPO4 5M.
A concentration of 110 M or 106 M doesn't exist.
98g
262 - 266
71.8 M
262 - 266
[600 mL]*[1 L / 1000 mL]*[0.1 mol/L]*[97.995 g/mol] = [0.06 mol]*[97.995 g/mol] = 5.81 g H3PO4
97.8 - 98.2 98 worked for me