85 moles Li x 6.02x10^23 atoms/mole Li = 5.17x10^25 atoms of Li
1 mole Li = 6.94g Li = 6.022 x 1023 atoms Li 27.0g Li x 6.022 x 1023 atoms Li/6.94g Li = 2.34 x 1024 atoms Li
20,32 g of lithium nitride can be obtained.
3.977 mol
To find the number of moles in 55g of lithium chloride, we first need to calculate the molar mass of lithium chloride, which is approximately 42.39 g/mol. Then, we divide the mass given (55g) by the molar mass to get moles. Therefore, 55g of lithium chloride is approximately 1.30 moles.
1 mole of Li₂O contains 2 moles of lithium (Li) atoms and 1 mole of oxygen (O) atoms. Therefore, in 1 mole of Li₂O, there are a total of 3 moles of atoms.
85 moles Li x 6.02x10^23 atoms/mole Li = 5.17x10^25 atoms of Li
From the periodic table, lithium has an atomic weight of 6.941. The molar mass of an element is the atomic weight in grams. Therefore, 1 mole Li = 6.941g Li Therefore, moles Li = 15g Li X 1 mole Li/6.941g Li = 2.2 moles Li
There are approximately ( 2.84 \times 10^{24} ) atoms in 4.7 moles of lithium (Li). This is calculated by multiplying Avogadro's number (6.022 x 10^23) by the number of moles.
The balanced chemical equation for the reaction is 3Li + H3PO4 -> Li3PO4 + 3H2. The mole ratio is 1:1 between Li and H3PO4. Therefore, 4 moles of H3PO4 will react with 4 moles of Li.
There are 1.81 x 10^24 atoms in 3.00 moles of lithium (Li). This is calculated by multiplying Avogadro's number (6.022 x 10^23 atoms/mol) by the number of moles.
When 7.6 grams of LiBr (Lithium Bromide) is dissolved, it will produce 0.069 moles of Li+ cation. This is because in one mole of LiBr, there is one mole of Li+ cation.
For every mole of lithium reacting, one mole of lithium nitride is produced following the balanced chemical equation. Therefore, 0.450 mol of lithium reacting produces 0.450 mol of lithium nitride.
15g Li * (1mol Li / 6.941g Li) = 2.16 mol Li
For this problem, the atomic mass is not required. Take the mass in moles and multiply it by Avogadro's constant, 6.02 × 1023. Divide by one mole for the units to cancel.2.5 moles H2 × (6.02 × 1023 atoms) = 1.51 × 1024 atoms
To determine the number of Li atoms in 97.9 g of Li, you first need to calculate the number of moles of Li using the molar mass of Li (6.94 g/mol). Then, you can use Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles of Li to atoms.
No moles of BaCO3 are found in any amount of Li.