At standard temperature and pressure, nitrogen exists as diatomic molecules. Therefore the number of atoms in 3.4 moles is 2 X 3.4 X Avogadro's Number, or 4.1 X 1024 atoms, to the justified number of significant digits.
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 mass of nitrogen needed to make 34 g of ammonia, we first need to calculate the molar mass of ammonia (NH3), which is 17 g/mol. From this, we can see that 1 mole of ammonia contains 1 mole of nitrogen. Therefore, the mass of nitrogen needed would also be 34 g.
17.32 moles H2O (2 moles H/1 mole H2O) = 34.64 moles hydrogen
5.0 grams gold (1 mole Au/197.0 grams)(6.022 X 1023/1 mole Au) = 1.5 X 1022 atoms of gold ===================
(34 g NH3) x (14 g N ) / (17 g NH3) = 28 g N(28 g N) x (1 g N) / (2 g N2) = 14 g N2
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 mass of nitrogen needed to make 34 g of ammonia, we first need to calculate the molar mass of ammonia (NH3), which is 17 g/mol. From this, we can see that 1 mole of ammonia contains 1 mole of nitrogen. Therefore, the mass of nitrogen needed would also be 34 g.
17.32 moles H2O (2 moles H/1 mole H2O) = 34.64 moles hydrogen
5.0 grams gold (1 mole Au/197.0 grams)(6.022 X 1023/1 mole Au) = 1.5 X 1022 atoms of gold ===================
The molar mass of nitrogen triiodide is 394,719 g.So 88 moles is equivalent to 34 735,272 g.This strange compound has the curious property to explode when it is irradiated with alpha particles.
To determine the number of moles in 34 grams of KOH, we first need to calculate its molar mass. The molar mass of KOH is approximately 56.11 g/mol (potassium: 39.10 g/mol, oxygen: 16.00 g/mol, hydrogen: 1.01 g/mol). Divide the given mass by the molar mass to find the number of moles: 34 g / 56.11 g/mol ≈ 0.61 moles of KOH.
(34 g NH3) x (14 g N ) / (17 g NH3) = 28 g N(28 g N) x (1 g N) / (2 g N2) = 14 g N2
1 mole H2SO4 x 4 moles O/mole H2SO4 x 6.02x10^23 atoms of O/mole O = 2.4x10^34 oxygen atoms
There are 34 atoms in C16H12O6: 16 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.
A non cyclic alkane always has a number of hydrogen atoms equal to 2c + 2, where c is the number of carbon atoms. Therefore, hexadecane, an alkane with 16 carbon atoms, will have 34 hydrogen atoms.
To find the mass of nitrogen needed to make ammonia, first determine the molar mass of ammonia (NH3) which is 17 g/mol. Since there is one nitrogen atom in ammonia, the nitrogen mass is 14 g/mol. To make 34 grams of ammonia, you would need 14 grams of nitrogen.
For this you need the atomic mass of Al. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel.54.0 grams Al / (27.0 grams) = 2.00 moles Al