First, balance the chemical equation between Na2SiO3 and HF to determine the stoichiometry. Then, use the molar masses of Na2SiO3 and HF to convert the mass of HF to moles. Finally, use the stoichiometry to calculate the mass of Na2SiO3 that can react with 0.740 g of HF.
The gram molecular mass of HF is 20.01. Therefore, 0.589 mole has a mass of 11.79 grams and is sufficient to form 0.589 mole of sodium fluoride* when sodium is present in excess. The gram molecular mass of sodium fluoride is 41.99, so that 0.589 grams of it has a mass of 24.7 grams, to the justified number of significant digits. __________________________ The formula unit of both hydrogen and sodium fluorides contains a single fluorine atom.
The balanced equation for the reaction between ammonia (NH3) and oxygen (O2) is 4NH3 + 5O2 → 4NO + 6H2O. To find the grams of oxygen needed to react with 23.9 grams of ammonia, you need to calculate the molar ratio between ammonia and oxygen using the balanced equation. Once you find the molar ratio, you can calculate the grams of oxygen required.
For every 1 gram of zinc, 3.88 grams of iodine are required to react. So in this case, with 4.2 grams of zinc, the amount of iodine needed would be (4.2 grams zinc) * (3.88 grams iodine / 1 gram zinc) = 16.296 grams of iodine.
The balanced chemical equation for the reaction is: 4Fe + 3O2 -> 2Fe2O3 From the equation, it can be seen that 3 moles of O2 are required to react with 4 moles of Fe. Therefore, to determine the grams of O2 required to react with 100 g Fe, you would need to use stoichiometry to find the answer.
To determine the amount of iron needed to react with 40 grams of iron(III) oxide, you should use the stoichiometry of the reaction. Calculate the molar mass of iron(III) oxide (Fe2O3) and determine the molar ratio between iron and iron(III) oxide in the balanced chemical equation. From there, you can calculate the amount of iron needed to fully react with 40 grams of iron(III) oxide.
The gram molecular mass of HF is 20.01. Therefore, 0.589 mole has a mass of 11.79 grams and is sufficient to form 0.589 mole of sodium fluoride* when sodium is present in excess. The gram molecular mass of sodium fluoride is 41.99, so that 0.589 grams of it has a mass of 24.7 grams, to the justified number of significant digits. __________________________ The formula unit of both hydrogen and sodium fluorides contains a single fluorine atom.
100, of course.
cannot answer without more info.
160...cant quite grasp HOW though
You need 145,337 g silver nitrate.
266,86 g aluminium chloride are obtained.
mass H2O =49.2g
The iron(III) hydroxide is not soluble in water and doesn't react with sodium chloride.
Potassium oxide violently react with water !
63 g of water are needed.
The mass of diborane is 442,7 g.
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