To prepare 0.02N H2SO4 from 0.1N H2SO4, you can dilute the 0.1N H2SO4 by adding a calculated amount of water. To calculate the dilution factor, you can use the formula: C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration (0.02N), and you can solve for V2 to find the volume of the 0.1N H2SO4 to be diluted with water to get 0.02N H2SO4.
H2SO4 releases two hydrogen ions into solution. Therefore its Normality is twice its Molarity. Or to answer the question, the molarity is half the normality.
To calculate the percent of a 5 N H2SO4 solution, you need to know the molarity (moles of solute per liter of solution) and the molecular weight of the solute. Once you have that information, you can use the formula: % = (molarity x equivalent weight) / 10. If you provide the molecular weight of H2SO4, I can help you calculate the percent.
0.08 n
Well, honey, the normality of a solution is the molarity multiplied by the number of equivalents per mole. Since sulfuric acid (H2SO4) has 2 acidic hydrogens, each molecule can donate 2 equivalents. So, a 1.25 M solution of H2SO4 would have a normality of 2.5 N. Hope that clears things up for ya!
0.1 N means you have a hydronium ion concentration of 0.1 mole/L. Since sulfuric acid is a diprotic acid, you get 2 moles of ionizable protons per mole a sulfuric acid. Meaning, you only need a concentration of 0.05 M sulfuric acid to obtain 0.1 N sulfuric acid. [10 L] * [0.1 mole H+/L] * [1 mole H2SO4/2 mole H+] * [98.08 g/mole H2SO4] * [1 mL/1.84 g H2SO4] = 26.65 mL H2SO4 required to make 10 L 0.1 N H2SO4.
If you want all the H2SO4 to react, you first need a balenced chemical equasion. Mg + H2SO4 --> MgSO4 +H2 Then you calculate using mole ratios moles is expressed as n. n Mg/1 =n H2SO4/1 n Mg= 0.2mol It's the same because there are no coefficients in front of the reactants.
H2SO4 releases two hydrogen ions into solution. Therefore its Normality is twice its Molarity. Or to answer the question, the molarity is half the normality.
To calculate the percent of a 5 N H2SO4 solution, you need to know the molarity (moles of solute per liter of solution) and the molecular weight of the solute. Once you have that information, you can use the formula: % = (molarity x equivalent weight) / 10. If you provide the molecular weight of H2SO4, I can help you calculate the percent.
0.08 n
N stands for molality and it indicates the number of moles of a substance in a unit mass of the solution.
a lot
N. Ravi was born on 1948-01-01.
Well, honey, the normality of a solution is the molarity multiplied by the number of equivalents per mole. Since sulfuric acid (H2SO4) has 2 acidic hydrogens, each molecule can donate 2 equivalents. So, a 1.25 M solution of H2SO4 would have a normality of 2.5 N. Hope that clears things up for ya!
Shaji N. Karun was born on 1952-01-01.
Charles N. Felton was born on 1832-01-01.
0.1 N means you have a hydronium ion concentration of 0.1 mole/L. Since sulfuric acid is a diprotic acid, you get 2 moles of ionizable protons per mole a sulfuric acid. Meaning, you only need a concentration of 0.05 M sulfuric acid to obtain 0.1 N sulfuric acid. [10 L] * [0.1 mole H+/L] * [1 mole H2SO4/2 mole H+] * [98.08 g/mole H2SO4] * [1 mL/1.84 g H2SO4] = 26.65 mL H2SO4 required to make 10 L 0.1 N H2SO4.
con.H2SO4 is 98%(v/v)ie 980ml/litre.or 980X1.84(specific gravity of H2So4)ie wt/litre is 1803.2Normality= wt per litre/ Eq.wtie 1803.2/49=36.8 NHence con H2So4 is 36.8 NTo prepare 5 N , It has be diluted 7.36 times with water68