The formula for sodium acetate is CH3COONa and the equivalent weight is the same as the molecular weight (molar mass) which is 82.0343 g/mol.
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β 7y agoThe equivalent weight of anhydrous sodium acetate is 82 g/mol. This value is calculated based on the atomic weights of sodium (Na = 23) and acetate (CHβCOO = 59).
The molecular weight of sodium acetate is 82.03 g/mol.
The equivalent weight of zinc acetate is 136 g/mol.
To calculate the assay of Ceftriaxone sodium BP on an anhydrous basis, you first need to determine the molecular weight of Ceftriaxone sodium. Next, you would divide the weight of the anhydrous Ceftriaxone sodium by the total weight of the sample and multiply by 100 to get the percentage on an anhydrous basis. This calculation helps ensure accurate dosing and purity of the compound.
To convert 50 millimoles of sodium acetate to grams, you would multiply the number of moles by the molar mass of sodium acetate. The molar mass of sodium acetate is 82.03 g/mol. Therefore, 50 millimoles * 82.03 g/mol = 4101.5 grams of sodium acetate.
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.
The molecular weight of sodium acetate is 82.03 g/mol.
The equivalent weight of zinc acetate is 136 g/mol.
To calculate the assay of Ceftriaxone sodium BP on an anhydrous basis, you first need to determine the molecular weight of Ceftriaxone sodium. Next, you would divide the weight of the anhydrous Ceftriaxone sodium by the total weight of the sample and multiply by 100 to get the percentage on an anhydrous basis. This calculation helps ensure accurate dosing and purity of the compound.
To convert 50 millimoles of sodium acetate to grams, you would multiply the number of moles by the molar mass of sodium acetate. The molar mass of sodium acetate is 82.03 g/mol. Therefore, 50 millimoles * 82.03 g/mol = 4101.5 grams of sodium acetate.
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.
The equivalent weight of anhydrous citric acid is the molar mass divided by the number of acidic hydrogens it can donate in a reaction. Citric acid has three acidic hydrogens that can be donated, so you would divide the molar mass by 3 to calculate the equivalent weight.
The gram-equivalent weight of sodium bicarbonate is calculated by dividing the molar mass by the number of equivalents of the compound. For sodium bicarbonate (NaHCO3), the molar mass is approximately 84 grams per mole. Since it has one equivalent of bicarbonate ion (HCO3^-), the gram-equivalent weight of sodium bicarbonate is 84 grams.
If you mean what percentage of sodium sulfate is Na, the answer is about 32.4. That's for the anhydrous material. It can also be hydrated. In Glauber's salt, for example, a molecule of Na2SO4 is associated with 10 molecules of water. %Na = 14.3.
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To prepare 100mM ammonium acetate solution, weigh out the appropriate amount of ammonium acetate powder based on its molecular weight and add it to a known volume of water, such as 1 liter. The final volume should be adjusted by adding more water and then mix well to dissolve the powder completely.
The milligram equivalent weight of sodium sulfate is 142.04 mg, calculated by dividing the molecular weight of the compound (142.04 g/mol) by 1000 to convert it to milligrams.
The equivalent weight of Sodium Hypochlorite (NaClO) is 74.44 g/mol. This is because one mole of NaClO contains one mole of available chlorine, which has an atomic weight of 35.45 g/mol and one mole of sodium with an atomic weight of 22.99 g/mol. Therefore, the equivalent weight is the sum of the atomic weights of chlorine and sodium in NaClO.