3.5 mol x (1000mmol/1mol) = 3500mmol
25 mmol equates to 0.025 mol
1 mEq=1 mmol/valence e.g.For sodium, 1 mEq=1mmol/1 (valence of sodium=1) means, 1 mmol sodium=1 mEq of sodium take for calcium,valence=2 1 1 mEq of calcium=1mmol/2=0.5 mmol of calcium
12.5 mL * 5.0 (m)mol/(m)L HCl = 62.5 mmol spilled HClneeds62.5 mmol NaHCO3 = 62.5 mmol * 84.01 (m)g/(m)mol NaHCO3 = 5250 mg NaHCO3 = 5.25 g pure NaHCO3
Using V * M = constant at dilution = amount of H2SO4 [mol] in both of the solutionsV= volume [L] of the solutionM= molarity [mol/L] of the solutionSo: V *18.0 = 24.9 * 0.195 gives V = ( 24.9 * 0.195 ) / 18.0 = 0.270 L
The answer is 1,31 moles carbon.
1 mol = 103 mmol Conversely, 1 mmol = 10-3 mol For example: 25 mol x 103 mmol/1 mol = 25000 mmol and, 3.2 mmol x 10-3 mol/1 mmol = 0.0032 mol
25 mmol equates to 0.025 mol
10500
1000 mmol = 1 mol. So, what you do is 2.55mmol*(1mol/1000mmol). The mmol's cancel and you are left with mol. The "m" is a metric prefix. So, 1000mN = 1N just like 1000mmol = 1mol.
1 mEq=1 mmol/valence e.g.For sodium, 1 mEq=1mmol/1 (valence of sodium=1) means, 1 mmol sodium=1 mEq of sodium take for calcium,valence=2 1 1 mEq of calcium=1mmol/2=0.5 mmol of calcium
12.5 mL * 5.0 (m)mol/(m)L HCl = 62.5 mmol spilled HClneeds62.5 mmol NaHCO3 = 62.5 mmol * 84.01 (m)g/(m)mol NaHCO3 = 5250 mg NaHCO3 = 5.25 g pure NaHCO3
Using V * M = constant at dilution = amount of H2SO4 [mol] in both of the solutionsV= volume [L] of the solutionM= molarity [mol/L] of the solutionSo: V *18.0 = 24.9 * 0.195 gives V = ( 24.9 * 0.195 ) / 18.0 = 0.270 L
The answer is 0,02552 mmol.
2 MMOL
4.12 mmol is 2.48112242e+24mg
We must first figure out the amount of NaCl in moles: M = mol/L = mmol/mL 6 = mmol/25 mL 150 mmol NaCl Now divide by the total volume to get the final concentration: 150 mmol/100 mL = 1.5 M NaCl
The answer is 1,31 moles carbon.