To determine the grams of sodium sulfate needed, you first need to specify the molarity (M) of the sodium sulfate solution. Once you have the molarity, you can use the formula: grams = molarity (M) x volume (L) x molar mass (g/mol). This will give you the amount of sodium sulfate in grams needed to make the solution.
Na2SO4 10.0 grams Na2SO4 (1 mole Na2SO4/142.05 grams)(2 mole Na/1 mole Na2SO4)(22.99 grams/1 mole Na) = 3.24 grams of sodium -------------------------------
Amount of sodium sulfate required = 0.683 x 350/100 = 0.239The formula mass of sodium sulfate, Na2SO4 is 2(23.0) + 32.1 + 4(16.0) = 142.1 Therefore mass of sodium sulfate required = 0.239 x 142.1 = 34.0g Approximately 34 grams of sodium sulfate would be needed.
To calculate the mass in grams of sodium sulfate, we need to know the number of moles. Once we have the number of moles, we can multiply it by the molar mass to find the mass in grams. For example, if we have 2 moles of sodium sulfate, the mass would be 2 moles * 141.98 grams/mole = 283.96 grams.
The mass of the solution would be the sum of the mass of the sodium chloride and the mass of water. So, the mass of the solution would be 17.8 grams (NaCl) + 217 grams (water) = 234.8 grams.
The concentration of a saturated solution of copper sulfate is approximately 25% by weight, which means that 25 grams of copper sulfate are dissolved in 100 grams of water at a specific temperature. At room temperature, this solution is around 47-50 grams of copper sulfate per 100 milliliters of water.
There are 2.42 grams of sodium in 10 grams of sodium sulfate (Na2SO4). This is calculated based on the molecular weight of sodium sulfate and the molar ratio of sodium to the sulfate ion in the compound.
Benedict's solution typically contains copper(II) sulfate, sodium citrate, and sodium carbonate. The exact composition can vary, but a common formulation is 17.3 g copper(II) sulfate, 100 g sodium citrate, and 173 g sodium carbonate per liter of water.
Na2SO4 10.0 grams Na2SO4 (1 mole Na2SO4/142.05 grams)(2 mole Na/1 mole Na2SO4)(22.99 grams/1 mole Na) = 3.24 grams of sodium -------------------------------
Amount of sodium sulfate required = 0.683 x 350/100 = 0.239The formula mass of sodium sulfate, Na2SO4 is 2(23.0) + 32.1 + 4(16.0) = 142.1 Therefore mass of sodium sulfate required = 0.239 x 142.1 = 34.0g Approximately 34 grams of sodium sulfate would be needed.
To calculate the mass in grams of sodium sulfate, we need to know the number of moles. Once we have the number of moles, we can multiply it by the molar mass to find the mass in grams. For example, if we have 2 moles of sodium sulfate, the mass would be 2 moles * 141.98 grams/mole = 283.96 grams.
The mass of the solution would be the sum of the mass of the sodium chloride and the mass of water. So, the mass of the solution would be 17.8 grams (NaCl) + 217 grams (water) = 234.8 grams.
The concentration of a saturated solution of copper sulfate is approximately 25% by weight, which means that 25 grams of copper sulfate are dissolved in 100 grams of water at a specific temperature. At room temperature, this solution is around 47-50 grams of copper sulfate per 100 milliliters of water.
To calculate the mass of anhydrous sodium sulfate needed, you first need to determine the total moles of Na+ required. In this case, 60 ml * 0.1 mmol/ml = 6 mmol of Na+. Anhydrous sodium sulfate has a molecular weight of 142.04 g/mol, so you will need 6 mmol * 142.04 g/mol = 852.24 mg or 0.85224 grams of anhydrous sodium sulfate to prepare the 60ml solution.
To solve this stoichiometry problem, first calculate the number of moles of sodium hydroxide (NaOH) present in 200 grams. Then, using the balanced equation, determine the moles of sodium sulfate (Na2SO4) that will be formed. Finally, convert the moles of Na2SO4 to grams using the molar mass of sodium sulfate.
To prepare a 0.01N solution of sodium metabisulfite, you would need 2.31 grams of sodium metabisulfite per liter of solution.
To calculate the total amount of sodium chloride needed for a 13 L solution at 4 grams per liter, multiply the concentration by the volume of the solution: 4 grams/L x 13 L = 52 grams of sodium chloride. Therefore, you will need 52 grams of sodium chloride to make the 13 L solution.
To find the mass of sodium sulfate, we need to know the molar mass of Na2SO4, which is about 142.04 g/mol. Multiply the number of moles by the molar mass to get the mass: 0.150 mol * 142.04 g/mol = 21.31 grams. So, the mass of 0.150 mol of sodium sulfate is 21.31 grams.