molar mass citric acid = 72 + 8 + 112 = 192g/mol 192 g x 1 mol/192g x 6.02x10^23 molecules/mole = 6.02x10^23 molecules
2.3 × 1024 atoms of Ar
334 g x 1 mol/331.6 g x 6.02x10^23 molecules/mole = answer
325 g of CBr4 have 5,9.10e23 molecules.
The answer is 8,5379.10e23 molecules.
112g NO x 1 mol NO/30 g x 6.02x10^23 molecules/mole = 2.2z10^24 molecules
molar mass citric acid = 72 + 8 + 112 = 192g/mol 192 g x 1 mol/192g x 6.02x10^23 molecules/mole = 6.02x10^23 molecules
2.3 × 1024 atoms of Ar
26.3 g of calcium hydroxide contain 2,054 molecules.
22.0 g of silver chloride contain 0,918.10e23 molecules.
334 g x 1 mol/331.6 g x 6.02x10^23 molecules/mole = answer
325 g of CBr4 have 5,9.10e23 molecules.
32 g SO2 x 1 mole SO2/96 g x 6.02x10^23 molecules/mole = 2.0x10^23 molecules
2.05×1022 molecules
The answer is 1,57.10e27 molecules.
1,4.10e23 molecules
The answer is 8,5379.10e23 molecules.