There are 3 moles of nitrogen in 3 moles of ammonium nitrate. Ammonium nitrate contains 2 nitrogen atoms in its chemical formula NH4NO3. Each mole of ammonium nitrate contains 2 moles of nitrogen atoms.
6,00 g of water contains how 0,33 moles.
To find the volume of solution needed, you can use the formula: moles = Molarity × Volume. Rearranging the formula to solve for volume: Volume = Moles / Molarity. Plugging in the values, you get Volume = 0.50 moles / 0.25 M = 2 liters of solution needed.
The number of moles is 18.56.
Boron trifluoride is BF3. So each mole of BF3 contains 1 moles of boron (B) and 3 moles of fluorine (F). Thus, 3 moles of BF3 contains NINE moles of fluorine.
Some conversion required. (mmolar into mol, or moles into mmol ) Molarity = moles of solute/Liters of solution 100 millimolar = 0.1 M glycine Molarity = moles of solute/Liters of solution manipulate algebraically Liters of solution = moles of solute/Molarity 0.005 mole glycine/0.1 M glycine = 0.05 Liters ( 1000 ml/1 L) = 50 milliliters of solution --------------------------------
1 gram NaCl (1 mole NaCl/58.44 grams) = 0.0171 moles NaCl Molarity = moles of solute/volume ( liters) of solution 0.171 M = 0.0171 moles NaCl/volume 0.1 Liters, or more to our point; 100 milliliters
The answer is 2,09 moles.
Since 14 (4+10) moles of P4O10 contains 4 moles of Phosphorus, 8 moles of P4O10 will contain :: (8 x 4)/14 = 2.286 moles of Phosphorus
There are 9.12 moles of hydrogen atoms in 4.56 moles of NH2NH2. Each NH2NH2 molecule contains 2 hydrogen atoms.
To determine the moles of LiF required, use the formula: moles = molarity (M) * volume (L). Given a 7 M solution with a volume of vol L, the moles of LiF needed would be 7 * vol.
The answer is 13,89 moles.