2.7×1024 atoms
4.5
3,00 moles of Li have 18,066422571.10e23 atoms.
The atomic weight of Li is 6.941 g/mol The mass of the sample is 55.2 g This means that: 55.2/6.941=7.95 moles (rounded number) Multiplying this with Avogadro's number (6.022*1023) 7.95*6.022*1023=4.79*1024 atoms. There you have your answer.
3.977 mol
0.106 moles Lithium (6.022 X 1023/1 mole Li) = 6.38 X 1022 atoms of Lithium ======================
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
3,00 moles of Li have 18,066422571.10e23 atoms.
85 moles Li x 6.02x10^23 atoms/mole Li = 5.17x10^25 atoms of Li
How many lithium atoms are in 10.56 g of lithium
No moles of BaCO3 are found in any amount of 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
The atomic weight of Li is 6.941 g/mol The mass of the sample is 55.2 g This means that: 55.2/6.941=7.95 moles (rounded number) Multiplying this with Avogadro's number (6.022*1023) 7.95*6.022*1023=4.79*1024 atoms. There you have your answer.
3.977 mol
3.977 mol
Li atomic mass= 6.941g/mol= 4.9 moles of Li1.00 mol = 6.02 x 1023 atoms4.9 mol Li = 2.95 x 1024 atoms= 3.0 x 1024 atoms
0.106 moles Lithium (6.022 X 1023/1 mole Li) = 6.38 X 1022 atoms of Lithium ======================
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
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