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4.0 moles Cu (6.022 X 1023/1 mole Cu) = 2.4 X 1024 atoms of copper ====================
Divide that number of atoms by 6.02x10^23 atoms/mole
Copper nitrate molecule contain nine atoms.
7 atoms, like math, if in parenthesis multiple. 2 x 3 = 6 plus one atom of Ca for a total of 7 atoms.
Copper(II) nitrate, represented as Cu(NO₃)₂, contains a total of 9 atoms. It has 1 copper (Cu) atom, 2 nitrogen (N) atoms, and 6 oxygen (O) atoms (2 nitrate ions, each containing 3 oxygen atoms). Thus, the total is 1 + 2 + 6 = 9 atoms.
There are 5 atoms in one molecule of Cu(OH)2, comprising one copper (Cu) atom, two oxygen (O) atoms, and two hydrogen (H) atoms.
Cu(NO3)2 has 9 atoms 1 Cu atom 2 N atoms 6 O atoms
There are a total of 64 oxygen atoms in 2Ca(NO3)2. This is calculated by multiplying the subscript (in this case, 16 oxygen atoms per formula unit) by the number of formula units given (2).
4.0 moles Cu (6.022 X 1023/1 mole Cu) = 2.4 X 1024 atoms of copper ====================
Divide that number of atoms by 6.02x10^23 atoms/mole
1 mol Cu Atoms (6.02x10^23 atoms)
There are 11 atoms in one formula unit of Cu(NO3)2: one copper atom (Cu), two nitrogen atoms (N), six oxygen atoms (O), and two nitrogen atoms (N) from the nitrate (NO3) group.
Copper nitrate molecule contain nine atoms.
7 atoms, like math, if in parenthesis multiple. 2 x 3 = 6 plus one atom of Ca for a total of 7 atoms.
Copper(II) nitrate, represented as Cu(NO₃)₂, contains a total of 9 atoms. It has 1 copper (Cu) atom, 2 nitrogen (N) atoms, and 6 oxygen (O) atoms (2 nitrate ions, each containing 3 oxygen atoms). Thus, the total is 1 + 2 + 6 = 9 atoms.
There is no CuSO. If you mean CuSO4, it would be 6 atoms per formula unit; one Cu atom, one S atom, and 4 O atoms.
To calculate the number of Cu atoms in 85 mol, you need to use Avogadro's number, which is approximately 6.022 x 10^23 atoms/mol. So, for 85 mol of Cu, the number of Cu atoms would be 85 mol x 6.022 x 10^23 atoms/mol = 5.12 x 10^25 Cu atoms.