HOW MANY ATOMS ARE IN 0.12 MOL OF CADMIUM?
Since 14 (4+10) moles of P4O10 contains 4 moles of Phosphorus, 8 moles of P4O10 will contain :: (8 x 4)/14 = 2.286 moles of Phosphorus
To find the mass of the nitrogen atoms in one mole of cadmium nitrate, calculate the molar mass of cadmium nitrate (Cd(NO3)2) and then multiply it by the number of nitrogen atoms in one mole of cadmium nitrate (2 nitrogen atoms). The molar mass of cadmium nitrate is 236.41 g/mol. Therefore, the mass of the nitrogen atoms in one mole of cadmium nitrate is 28.02 g.
There are 6 moles of oxygen atoms in 2 moles of potassium dichromate (K2Cr2O7). Each mole of K2Cr2O7 contains 7 oxygen atoms, so 2 moles would contain 14 oxygen atoms. The molar mass of oxygen is 16 g/mol, so there would be 224 grams of oxygen in 2 moles of K2Cr2O7.
In potassium dichromate (K2Cr2O7), there are 7 oxygen atoms per molecule. Therefore, two moles of K2Cr2O7 would contain 14 moles of oxygen atoms. Each mole of oxygen atoms has a molar mass of approximately 16 grams, so there would be 224 grams of oxygen in two moles of potassium dichromate.
To convert grams to atoms, you need to first convert grams of nitrogen to moles using its molar mass (14.01 g/mol). Then, use Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles to atoms. So, for 2.2 grams of nitrogen: Convert grams to moles: 2.2 g / 14.01 g/mol = 0.157 moles. Convert moles to atoms: 0.157 moles x 6.022 x 10^23 atoms/mol = 9.46 x 10^22 atoms of nitrogen.
1,0.10e9 atoms is equivalent to 0,166.10e-14 moles.
Since 14 (4+10) moles of P4O10 contains 4 moles of Phosphorus, 8 moles of P4O10 will contain :: (8 x 4)/14 = 2.286 moles of Phosphorus
Your compound's chemical formula is (NH4)2Cr2O7 - so you can see that it contains 2 moles of NH4+ per mole of the salt. So in 7 moles you have 14 moles of ammonium, containing 14 moles of N atoms which means that you have (1 mole = 6,022*10^23 atoms) 14 mols * 6,022*10^23 atoms/mole = 8,43*10^23 atoms
To find the mass of the nitrogen atoms in one mole of cadmium nitrate, calculate the molar mass of cadmium nitrate (Cd(NO3)2) and then multiply it by the number of nitrogen atoms in one mole of cadmium nitrate (2 nitrogen atoms). The molar mass of cadmium nitrate is 236.41 g/mol. Therefore, the mass of the nitrogen atoms in one mole of cadmium nitrate is 28.02 g.
The formula means, among other things, that there are 7 atoms of oxygen in each mole of the compound. Therefore, in 4.00 moles of the compound, there are 28.00 moles of oxygen atoms. Elemental oxygen usually is diatomic, so that there would be the equivalent of 14 moles of diatomic elemental oxygen.
There are 6 moles of oxygen atoms in 2 moles of potassium dichromate (K2Cr2O7). Each mole of K2Cr2O7 contains 7 oxygen atoms, so 2 moles would contain 14 oxygen atoms. The molar mass of oxygen is 16 g/mol, so there would be 224 grams of oxygen in 2 moles of K2Cr2O7.
In potassium dichromate (K2Cr2O7), there are 7 oxygen atoms per molecule. Therefore, two moles of K2Cr2O7 would contain 14 moles of oxygen atoms. Each mole of oxygen atoms has a molar mass of approximately 16 grams, so there would be 224 grams of oxygen in two moles of potassium dichromate.
To convert grams to atoms, you need to first convert grams of nitrogen to moles using its molar mass (14.01 g/mol). Then, use Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles to atoms. So, for 2.2 grams of nitrogen: Convert grams to moles: 2.2 g / 14.01 g/mol = 0.157 moles. Convert moles to atoms: 0.157 moles x 6.022 x 10^23 atoms/mol = 9.46 x 10^22 atoms of nitrogen.
In 2 moles of potassium dichromate, there are 16 moles of oxygen atoms (from the two oxygen atoms in each formula unit). The molar mass of oxygen is 16 g/mol, so in 2 moles of potassium dichromate, there are 32 grams of oxygen.
In one mole of potassium dichromate, there seven moles of oxygen. This means in two moles of K2Cr2O7, there are 14 moles of O, or 7 Moles of O2, which equals 224 grams.
Using the balanced equation 2 AlCl₃ + 3 Pb(NO₃)₂ → 3 PbCl₂ + 2 Al(NO₃)₃, the mole ratio between AlCl₃ and PbCl₂ is 2:3. Therefore, if 14 moles of AlCl₃ are consumed, 9.33 moles (14 moles / 2 * 3) of PbCl₂ will be produced.
2KClO3 + heat -> 2KCl + 3O2 14 moles KClO3 (3 mole O2/2 mole KClO3) = 21 moles oxygen made This is a common industrial method of producing oxygen.