Na2O + H2O = 2NaOH
Sodium Oxide (Na2O) is a white crystalline solid.
Note the molar ratios are 1:1::2 or 1/2: 1/2 ::1
Hence
Its Mr is
2 x Na = 2 x 23 = 46
1 x O = 1 x 16 = 16
46 + 16 = 62
Using the equation
Moles = mass(g) / Mr
Then mass (g) = moles X
The molar mass of sodium (Na) is 22.99 g/mol, and the molar mass of hydroxide (OH) is 17.01 g/mol. To find the molar mass of sodium hydroxide (NaOH), you can add the molar masses of sodium and hydroxide together, which equals 39.00 g/mol.
The density of a 1 M solution of sodium hydroxide is approximately 1.04 g/mL at room temperature.
To neutralize the sulfuric acid completely, you need a 1:2 molar ratio of sodium hydroxide to sulfuric acid. Therefore, you would need to add twice the amount of sodium hydroxide compared to the amount of sulfuric acid, which is 40.0 mL of the sodium hydroxide solution.
You cannot answer this without also knowing the VOLUME of the 2 M solution. For example, if you have 1 liter of 2 M NaOH, the the mass contained in that bottle will be 1L x 2 mol/L x 40 g/mol = 80 g.If, however, you have 2 liter, then the mass of NaOH contained in that bottle will be 160 g. And so on.
To prepare 0.2M solution of anhydrous sodium thiosulfate (Na2S2O3), you dissolve 24.6g of anhydrous Na2S2O3 in distilled water and dilute it to 1 liter. This is the molar mass method, where molar mass of Na2S2O3 is 158.10 g/mol.
the molar mass of sodium hydroxide is 40g/mol mike
16.5g 97% pure NaOH pellets dissoved in 1 litre of distilled
The molar mass of sodium (Na) is 22.99 g/mol, and the molar mass of hydroxide (OH) is 17.01 g/mol. To find the molar mass of sodium hydroxide (NaOH), you can add the molar masses of sodium and hydroxide together, which equals 39.00 g/mol.
The molar mass of sodium hydroxide (NaOH) is approximately 40 g/mol. To prepare a 0.10 M solution in 100 mL, you would need 1.0 g of NaOH. This can be calculated using the formula: mass (g) = molarity (M) x volume (L) x molar mass (g/mol).
The density of a 1 M solution of sodium hydroxide is approximately 1.04 g/mL at room temperature.
20 ml
To neutralize the sulfuric acid completely, you need a 1:2 molar ratio of sodium hydroxide to sulfuric acid. Therefore, you would need to add twice the amount of sodium hydroxide compared to the amount of sulfuric acid, which is 40.0 mL of the sodium hydroxide solution.
Sodium hydroxide does not have a pH number. The pH of a solution of sodium hydroxide depends entirely on the concentration of it in that solution. To learn how to determine the pH of a sodium hydroxide solution, see the Related Questions links.
You cannot answer this without also knowing the VOLUME of the 2 M solution. For example, if you have 1 liter of 2 M NaOH, the the mass contained in that bottle will be 1L x 2 mol/L x 40 g/mol = 80 g.If, however, you have 2 liter, then the mass of NaOH contained in that bottle will be 160 g. And so on.
To prepare 0.2M solution of anhydrous sodium thiosulfate (Na2S2O3), you dissolve 24.6g of anhydrous Na2S2O3 in distilled water and dilute it to 1 liter. This is the molar mass method, where molar mass of Na2S2O3 is 158.10 g/mol.
So you want 0.04M but you have 400ml, not a litre. 0.04/1000*400 is 0.016 moles wanted. 0.016*40 (molecular weight) is 0.64g
To calculate the grams of sodium hydroxide present in the solution, first calculate the number of moles using the formula: moles = Molarity (M) x Volume (L). Then, use the molar mass of sodium hydroxide (NaOH) to convert moles to grams. The molar mass of NaOH is 40 g/mol. Thus, in this case, you have 0.3375 moles of NaOH and if you convert this to grams, it would be 13.5 grams.