To do this, you need to know the molecular weight of the element you're dealing with, by adding up the atomic weights of the elements involved (found on any Periodic Table). The molecular weight is the mass in grams of the compound in one mole - this will provide you with a conversion factor. So take the measurement in grams and divide it by the molecular weight to convert to moles. Really what you're doing is multiplying the number by 1 mole, and dividing it by the equivalent of one mole, the molecular weight. That's the thought process behind unit analysis and how you get your "units to cancel".
In this case, the answer is about 0.137 moles HCl.
Find moles HCl. 5 g HCl (1 mole HCl/36.450 grams) = 0.1372 moles HCl Now, Molarity = moles of solute/Liters of solution Molarity = 0.1372 moles HCl/1 liter = 0.1372 M HCl Then. -log(0.1372 M HCl) = 0.9 pH ( you might call it 1, but pH can be off the scale ) -----------
31 g of HCl divided by the molar mass of HCl to find how many mols. HCl --> H(1) + Cl(1) = molar mass of HCl 1.008(1) + 35.45(1)= 36.428 molar mass 31/36.428= 2.97 which is the mols of solute The mols of solute divided by kg of solute = molality 2.97/.500= 1.7 molality
One step at a time.1/103 = 0.001 M HCl, so.....Molarity = moles of solute/Liters of solution ( 25 ml = 0.025 Liters )0.001 M HCl = X moles HCl/0.025 Liters= 2.5 X 10 - 5 moles HCl========================now, balanced eqiationNaOH + HCl --> NaCl + H2O ( all one to one )( now drive reaction towards mass NaOH )2.5 X 10 - 5 moles HCl (1 moles NaOH/1 mole HCl)(39.998 grams/1 mole NaOH)= 10 -4 grams caustic soda needed==========================
5 moles
85 grams of AgNO3 represents 0,.5 moles.
Find moles HCl. 5 g HCl (1 mole HCl/36.450 grams) = 0.1372 moles HCl Now, Molarity = moles of solute/Liters of solution Molarity = 0.1372 moles HCl/1 liter = 0.1372 M HCl Then. -log(0.1372 M HCl) = 0.9 pH ( you might call it 1, but pH can be off the scale ) -----------
31 g of HCl divided by the molar mass of HCl to find how many mols. HCl --> H(1) + Cl(1) = molar mass of HCl 1.008(1) + 35.45(1)= 36.428 molar mass 31/36.428= 2.97 which is the mols of solute The mols of solute divided by kg of solute = molality 2.97/.500= 1.7 molality
One step at a time.1/103 = 0.001 M HCl, so.....Molarity = moles of solute/Liters of solution ( 25 ml = 0.025 Liters )0.001 M HCl = X moles HCl/0.025 Liters= 2.5 X 10 - 5 moles HCl========================now, balanced eqiationNaOH + HCl --> NaCl + H2O ( all one to one )( now drive reaction towards mass NaOH )2.5 X 10 - 5 moles HCl (1 moles NaOH/1 mole HCl)(39.998 grams/1 mole NaOH)= 10 -4 grams caustic soda needed==========================
The limiting reagent in a reaction is the reactant that runs out first. For example, if you are reacting 10 moles of HCl and 5 moles of NaOH, you will get 5 moles of H20, 5 moles of NaCl, and 5 moles of HCl, because the remaining HCl had nothing to react with. Therefore, the NaOH is the limiting reagent.
5
5 moles
85 grams of AgNO3 represents 0,.5 moles.
5
5 moles of lead is equal to 1 036 g.
You have to use the equationnumber of moles = mass/Molar massn = 5/(39.1+32.1+64)n = 5/135.2number of moles = 0.04:)
number of moles=mass/molar mass =5/(28+32) =0.83 mol
The mass of sulfuric acid is 490,395 grams.