You can not have 0.4 of an atom (of anything). Thus you question makes no sense.
There are 3.22 x 10^23 atoms in 0.534 moles of nickel (Ni). This is calculated by using Avogadro's number (6.022 x 10^23 atoms/mol) multiplied by the number of moles.
To find the number of moles of nickel atoms in 125 g of nickel, divide the given mass by the molar mass of nickel. The molar mass of nickel is approximately 58.69 g/mol. Therefore, 125 g Ni / 58.69 g/mol = ~2.13 moles of Ni atoms.
Ni(OH)2 1 atom of nickel 2 atoms of oxygen 2 atoms of hydrogen -------------------------------+ total atoms = 5 atoms
According to the U.S. mint the nickels currently in circulation weigh 5.000 grams and contain 25% Ni (the rest is copper). 25% of 5.000g = 1.250 grams of Ni in a nickel coin / molar mass of Ni 58.71 g/mol = 0.02129 moles of Ni times the number of particles for one mole 6.02x10^23 = 1.28x10^22 atoms of Ni in one coin. All together: (0.25 x 5.000g) / (58.71 g per mol) x (6.02x10^23 atoms per mol) = 1.28x10^22 atoms Ni per coin.
The oxidation number for Ni in NiO2 is +4. Nickel is known to have variable oxidation states, and in this compound, it is bonded to two oxygen atoms which have an oxidation number of -2 each. The overall charge of the compound is neutral.
To find the number of atoms in 26.4 g of Ni, you would first determine the molar mass of Ni (58.69 g/mol) and then use Avogadro's number (6.022 x 10^23 atoms/mol) to convert grams to atoms. Thus, there are approximately 2.24 x 10^23 Ni atoms in 26.4 g of Ni.
There are 3.22 x 10^23 atoms in 0.534 moles of nickel (Ni). This is calculated by using Avogadro's number (6.022 x 10^23 atoms/mol) multiplied by the number of moles.
The chemical formula Ni(C₂H₃O₂)₄ indicates that there is one nickel (Ni) atom, four acetate groups (C₂H₃O₂), which contain a total of 8 carbon (C) atoms, 12 hydrogen (H) atoms, and 8 oxygen (O) atoms. Therefore, the total number of atoms in Ni(C₂H₃O₂)₄ is 1 (Ni) + 8 (C) + 12 (H) + 8 (O) = 29 atoms.
To find the number of moles of nickel atoms in (8.00 \times 10^9) Ni atoms, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms/mole. The calculation is as follows: [ \text{Moles of Ni} = \frac{8.00 \times 10^9 \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mole}} \approx 1.33 \times 10^{-14} \text{ moles} ] Thus, there are approximately (1.33 \times 10^{-14}) moles of nickel atoms in (8.00 \times 10^9) Ni atoms.
To find the number of moles of nickel atoms in 125 g of nickel, divide the given mass by the molar mass of nickel. The molar mass of nickel is approximately 58.69 g/mol. Therefore, 125 g Ni / 58.69 g/mol = ~2.13 moles of Ni atoms.
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
To find the number of atoms in 0.534 moles of nickel (Ni), you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms per mole. Multiply the number of moles by Avogadro's number: [ 0.534 , \text{mol} \times 6.022 \times 10^{23} , \text{atoms/mol} \approx 3.22 \times 10^{23} , \text{atoms}. ] Therefore, there are approximately (3.22 \times 10^{23}) atoms in 0.534 moles of nickel.
2Ni3(PO4)2 Nickel phosphate. 6 nickel atoms 4 phosphorous atoms 16 oxygen atoms Now, this PO4 3- is the polyatomic ion phosphate and is internally covalently bonded, so it could be called a molecule, though it is charged. there are 2 PO4 3-
To find the mass of 5.97 x 10^27 nitrogen atoms, calculate the molar mass of nitrogen (14.01 g/mol) and then multiply by the number of atoms. Using Avogadro's number (6.022 x 10^23 atoms/mol), you would find that the mass of 5.97 x 10^27 nitrogen atoms is approximately 1.68 grams.
Ni(OH)2 1 atom of nickel 2 atoms of oxygen 2 atoms of hydrogen -------------------------------+ total atoms = 5 atoms
There are 34 atoms in C16H12O6: 16 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.
According to the U.S. mint the nickels currently in circulation weigh 5.000 grams and contain 25% Ni (the rest is copper). 25% of 5.000g = 1.250 grams of Ni in a nickel coin / molar mass of Ni 58.71 g/mol = 0.02129 moles of Ni times the number of particles for one mole 6.02x10^23 = 1.28x10^22 atoms of Ni in one coin. All together: (0.25 x 5.000g) / (58.71 g per mol) x (6.02x10^23 atoms per mol) = 1.28x10^22 atoms Ni per coin.