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2.2 mol water = 2.2 (mol) * 18 (g/mol) water = 39.6 (mol*g/mol) = 40 g18 g/mol = mol mass of H2O = 2*H + 1*O = (2*1 + 16) g/mol
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You have :N2 + O2 -----> 2NOmoles NO = ( 2 mols N2 reacted ) ( 2 mol NO / mol N2 reacted )moles NO = 4 moles NOmass NO = ( 4 mol NO ) ( 30 g NO / mol NO ) = 120 g NO
In 1 molecule of H2O, there are 2 atoms of H. Therefore, in 1 mol of H2O, there are 2 mol of H. So, in 4.51 mol of H2O, there would be 4.51 * 2 = 9.02 mol of H.
1 mol of octane (C8H18) reacts with 25 mol of oxygen (O2) to produce 8 mol of CO2 and 9 mol of H2O. Therefore, 0.74 mol of oxygen can react with 0.0744 mol of octane.
There are 3500 mmol in 3.5 mol. To convert mol to mmol, you multiply by 1000 since there are 1000 mmol in 1 mol.
To find the empirical formula, we need to determine the ratio of each element in the compound. First, find the moles of each element: K = 0.104 mol C = 0.052 mol O = 0.156 mol Next, divide each mole value by the smallest mole value to get the ratio: K = 0.104 mol / 0.052 mol = 2 C = 0.052 mol / 0.052 mol = 1 O = 0.156 mol / 0.052 mol = 3 Therefore, the empirical formula is K2CO3.
First, calculate the moles of each gas using their molar masses: He (4.0 g / 4.0 g/mol = 1.0 mol), Ar (6.5 g / 39.9 g/mol = 0.163 mol), Ne (10.0 g / 20.2 g/mol = 0.495 mol). Then, calculate the total moles of gas in the solution: 1.0 mol + 0.163 mol + 0.495 mol = 1.658 mol. Finally, find the mole fraction of He by dividing the moles of He by the total moles of gas: 1.0 mol / 1.658 mol ≈ 0.602.
Molar Mass of Al: 2(27.0g/mol) = 54.0g/mol Molar Mass of O: 3(16.0g/mol) = 48.0g/mol Molar Mass of compound: 102.0g.mol (54.0g/mol / 102.0g/mol) x 100% = 52.9%
1 mol of Ag and 1 mol of Au can be said to contain the same amount of atoms of each element.