If you are asking what percent of H2SO4 that you have, I would need to know that volume of the 5M H2SO4
There are 5 moles of sulfur in 5 moles of H2SO4, as there is 1 mole of sulfur in each mole of H2SO4.
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.
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.
0.08 n
N x 5% = N/20.71000/20 = 3550.
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.
There are 5 moles of sulfur in 5 moles of H2SO4, as there is 1 mole of sulfur in each mole of H2SO4.
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.
75 = 0.05 * n 75/0.05 = n n = 1500
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.
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.
0.08 n
If 15 percent of a number N is 0.3 what is 45 percent of N
N stands for molality and it indicates the number of moles of a substance in a unit mass of the solution.
a lot
20%= .2, n*.2= n/5, 100*.2=100/5, 100/5=20 20% of 100 is 20