Since HCl is a monoprotic acid, its normality is the same as its molarity. A 4 N solution of HCl is a 4 M solution of HCl as well. If you want to make a liter of a 4 N solution of HCl, you need to dilute a certain volume of a standard solution you already have to one liter (1,000 mL) of the 4 M solution. You could use a standard concentrated solution of HCl that is usually 12 M using the following equation:
(V1)(M1) = (V2)(M2)
V1 = ?
M1 = 12 M
V2 = 1,000 mL
M2 = 4 M
V1 = [(V2)(M2)] / ((M1)
V1 = [(1,000 mL)(4 M)] / (12 M)
V1 = 83 mL
According to the above calculations, to make 1,000 mL (1 L) of a 4 M solution of HCl, you would need to dilute with water 83 mL of 12 M solution of HCl to 1,000 mL (1 L) volume. As stated before, a 4 M solution is also a 4 N solution of HCl.
1N = 40.0g in 1000mL (1M or 1mol/L)
3N = 120.0g in 1000mL (3M or 3mol/L).
Therefore 3N (3M) in 500mL = 60g NaOH.
Take 28.21 ml Hcl make up to 1000 ml Dis.Water its 3N HCl
25.5ml hcl in 100ml of water
The concentration of this solution (in NaOH) is 40 g/L.
I think you may have missed a decimal point somewhere. 125M of NaOH would be a solution of sodium hydroxide containing 125 moles per litre. One mole of a compound is the same number of grams as the molecular weight of the molecule. Sodium hydroxide has a molecular weight of 40 ( sodium 23, oxygen 16, and hydrogen 1), so a one molar solution would have forty grams of NaOH per litre. 500ml of a 1M solution would contain 20g. 500ml of a 125M solution would need 2 500g. 1L of a 125M solution would need 5 000g of sodium hydroxide in the litre. The maximum solubility for NaOH in water at 20 degrees is 1110g per litre, so if you tried to dissolve 5 000g in a litre you would be left with 3 890g undissolved. A 1.25M solution would have 1.25 times 40g per litre, which is 50g per litre. 500ml of this solution would have half this amount of NaOH, or 25g.
molarity equals moles of solute /volume of solution in litres . moles of NaOH equals 5g/40g = 0.125 and volume of solution will be volume of water + volume of NaOH = 0.5 litre+0.002 l which is nearly 0.5 litre . (volume of NaOH is calculated by its density) so molarity = 0.125mol/0.5litre = 0.25 M
1M contains 43.1ml in 500ml, 2M contains 86.2ml in 500ml
It is possible only if you evaporate the water.
(100ml)(0.125M NaOH) = (500ml)(X Molarity) Molarity = 0.025 M
The concentration of this solution (in NaOH) is 40 g/L.
I think you may have missed a decimal point somewhere. 125M of NaOH would be a solution of sodium hydroxide containing 125 moles per litre. One mole of a compound is the same number of grams as the molecular weight of the molecule. Sodium hydroxide has a molecular weight of 40 ( sodium 23, oxygen 16, and hydrogen 1), so a one molar solution would have forty grams of NaOH per litre. 500ml of a 1M solution would contain 20g. 500ml of a 125M solution would need 2 500g. 1L of a 125M solution would need 5 000g of sodium hydroxide in the litre. The maximum solubility for NaOH in water at 20 degrees is 1110g per litre, so if you tried to dissolve 5 000g in a litre you would be left with 3 890g undissolved. A 1.25M solution would have 1.25 times 40g per litre, which is 50g per litre. 500ml of this solution would have half this amount of NaOH, or 25g.
(3n)(3n-1) = 3n * 3n - 3n * 1 Now, perform the multiplication: (3n * 3n) = 9n^2 (3n * 1) = 3n So, (3n)(3n-1) simplifies to: 9n^2 - 3n
molarity equals moles of solute /volume of solution in litres . moles of NaOH equals 5g/40g = 0.125 and volume of solution will be volume of water + volume of NaOH = 0.5 litre+0.002 l which is nearly 0.5 litre . (volume of NaOH is calculated by its density) so molarity = 0.125mol/0.5litre = 0.25 M
3n + 2 + (3n + 3 - 3n + 1) = 3n + 2 + (3n + 3 - 3n + 1) = 3n + 2 + (4) = 3n + 6
Is it a. 3n square and 3n ans. 3n b. 3n *2 and 3n ans.3n ans. 3n for both!!
1M contains 43.1ml in 500ml, 2M contains 86.2ml in 500ml
what is 3n times 3n-1
The nth term of this sequence is 3n + 4
No. 3n is not a factor of 3n + 7.
No. 3n is not a factor of 3n + 7.