Q: 10.5 mol equals how many mmol?

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moles of Al=4.40 g/26.9815 g/mol=0.163 moles cl2=15.4g/70.906g/mol=0.217 the ratio is 2:3 cl2 is the limiting reagent

Formula for table salt = NaCl 1.00 g = 1.00 / 58.45 mol = 1.71 x 10-2 mol Mole Na required = 1.71 x 10-2 mol Mass of Na = 1.71 x 10-2 mol x 23.0 g /mol = 0.393 g = 393 mg

Because it takes 2a to make 1mol of a2b, the ratio is 1mol a2b/2mol a. If you only have 1.0 mol of a to react, then 1mol a x (1mol a2b/2 mol a)= 0.5 mol of a2b Because it takes 1 mol of b to make 1 mole of a2b, the ratio is 1mol a2b/1mol b. So, 1.0 mol b x (1mol a2b/1mol b)= 1.0 mol of a2b. Since you run out of a to react with b, a is the limiting reagent meaning that you can only produce 0.5 mol of a2b even though you have an excess amount of b. So, only 0.5 mol of a2b can be produced.

•96485 coulomb/mol of electrons (for J) •23054 C/mol (for Cal) •23.054 C/mol (for kCal) •

[H+] = Kw / [OH-] = 1.0*10-14 / 2.5*10-4 = 4.0*10-11 mol/L

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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

3.5 mol x (1000mmol/1mol) = 3500mmol

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

Molarity = moles of solute/Liters of solution ( 450 ml = 0.450 liters) 5M C6H12O6 = moles C6H12O6/0.450 liters = 2.25 moles C6H12O6 (180.156 grams/1 mole C6H12O6) = 405.351 grams of glucose ( you do significant figures )

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

8 mol x (4 mol / 2 mol) x 133.5 g / 1 mol = 2136 grams

Carbon: 105 kJ/mol = 105 (kJ/mol) / 0.012(kg/mol) = 8750 kJ/kg Carbon

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

0.063 g of oxalic acid * (1 mol H2C204*2H2O / 126.07 g) = 0.0004997 mol H2C2O4*2H2O 0.0004997 mol H2C204 / 0.250 L = 0.001999 M of H2C2O4