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In order to make 0.02 N NaOH from 0.2 N NaOH, one needs to dilute it by 10 x (10 fold). Depending on the volume of 0.02 N NaOH needed, that will determine the volume of 0.2 N used. For example, to make 100 ml of 0.02 N NaOH, you would dilute 10 mls of 0.2 N to 100 ml. This is seen in the following calculation: (x ml)(0.2 N NaOH) = (100 ml) (0.02 N NaOH) and x = 10 ml
To prepare a 0.02 N NaOH solution from a 0.2 N NaOH solution, you can dilute the 0.2 N solution by adding 10 times the amount of water. For every 1 part of the 0.2 N solution, add 9 parts of water. Ensure thorough mixing to obtain a 0.02 N NaOH solution.
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How to prepare 2N solution of NaOH in 10 ml of water?
To prepare a 2N solution of NaOH in 10 ml of water, you would need to calculate the amount of NaOH needed based on its molecular weight. Once you have determined the amount needed, dissolve it in 10 ml of water to make the solution. Remember to handle NaOH with caution as it is corrosive.
How make 0.5ml of 2N NaOH?
To prepare 0.5 mL of 2N NaOH, you would need to dilute a higher concentration of NaOH solution. Assuming you have a 4N NaOH solution, you would mix 0.25 mL of the 4N solution with 0.25 mL of water to obtain 0.5 mL of 2N NaOH. Be cautious when handling concentrated NaOH solutions, as they are caustic and can cause skin irritation.
How do you make 2N NaoH SOLUTION?
To make a 2N NaOH solution, you need to dissolve 80g of NaOH pellets (sodium hydroxide) in enough water to make 1 liter of solution. This will yield a solution with a concentration of 2N (normality), as the equivalent weight of NaOH is 40 g/mol. Be sure to wear appropriate protective gear and handle NaOH with care, as it is a caustic and potentially hazardous chemical.
How do you prepare 0.25N sulphuric acid from 2N sulphuric acid?
To prepare 0.25N sulphuric acid from 2N sulphuric acid, you can dilute the 2N solution by adding 7 parts of water to 1 part of the 2N solution. This will result in a final 0.25N sulphuric acid solution.
How do you prepare 2N NaOH in 20 ml?
To prepare 2N NaOH in 20 ml, first calculate the amount of NaOH needed using the formula: (Molarity x Volume) / Normality. In this case, it would be (2 x 20) / 1 = 40 grams of NaOH. Then, carefully weigh out 40 grams of NaOH and dissolve it in enough water to make a total volume of 20 ml. Make sure to wear appropriate safety gear like gloves and goggles when handling NaOH.
How to prepare 2N solution of NaOH in 10 ml of water?
To prepare a 2N solution of NaOH in 10 ml of water, you would need to calculate the amount of NaOH needed based on its molecular weight. Once you have determined the amount needed, dissolve it in 10 ml of water to make the solution. Remember to handle NaOH with caution as it is corrosive.
How make 0.5ml of 2N NaOH?
To prepare 0.5 mL of 2N NaOH, you would need to dilute a higher concentration of NaOH solution. Assuming you have a 4N NaOH solution, you would mix 0.25 mL of the 4N solution with 0.25 mL of water to obtain 0.5 mL of 2N NaOH. Be cautious when handling concentrated NaOH solutions, as they are caustic and can cause skin irritation.
How do you make 2N NaoH SOLUTION?
To make a 2N NaOH solution, you need to dissolve 80g of NaOH pellets (sodium hydroxide) in enough water to make 1 liter of solution. This will yield a solution with a concentration of 2N (normality), as the equivalent weight of NaOH is 40 g/mol. Be sure to wear appropriate protective gear and handle NaOH with care, as it is a caustic and potentially hazardous chemical.
How do you prepare 0.25N sulphuric acid from 2N sulphuric acid?
To prepare 0.25N sulphuric acid from 2N sulphuric acid, you can dilute the 2N solution by adding 7 parts of water to 1 part of the 2N solution. This will result in a final 0.25N sulphuric acid solution.
How do you prepare 2N NaOH in 20 ml?
To prepare 2N NaOH in 20 ml, first calculate the amount of NaOH needed using the formula: (Molarity x Volume) / Normality. In this case, it would be (2 x 20) / 1 = 40 grams of NaOH. Then, carefully weigh out 40 grams of NaOH and dissolve it in enough water to make a total volume of 20 ml. Make sure to wear appropriate safety gear like gloves and goggles when handling NaOH.
What is the name for 2NaOH?
One Mole of Sodium Hydroxide NaOH= 40.00g/l =1N (8.0g NaOH in 100.0ml of water)= 2N NaOH or (80g of NaOH in 1L of water)= 2N NaOH
How do you prepare HCL 2N solution?
To prepare a 2N HCl solution, you can mix concentrated hydrochloric acid (approximately 36.5% HCl by weight) with water in a specific ratio. For example, to make 1 liter of 2N HCl solution, you would need to mix approximately 167 mL of concentrated HCl with about 833 mL of water. Always add acid to water slowly while stirring, and wear appropriate safety equipment.
How do you prepare 2N ammonium hydroxide solution from 25 percent ammonium solution?
To prepare a 2N ammonium hydroxide solution from a 25% ammonium solution, you would first need to calculate the volume of the 25% solution needed based on the concentration and volume of the final solution required. Then, dilute this calculated volume with the appropriate amount of water to get a 2N solution. Make sure to follow proper safety protocols when handling and diluting chemicals.
What is the solution to (2n plus 12)?
Right now, there is no solution. The sum of 2n and 12 will vary according to the value of n. If 2n + 12 = 0, then n = -6
What mass is required to make 2N solution of Na2co3 calculate mass of sodim carbonate which will be dissolved in 500 ccc?
To prepare a 2N solution of Na2CO3 in 500 mL, you would need to dissolve 42.97 grams of Na2CO3. This is calculated by first converting N to molarity (2N = 2 moles/L), then using the molar mass of Na2CO3 (105.99 g/mol) to calculate the required mass.
What are 6 even consecutive numbers whose summation is 600?
Answerno solutionProcedureAssume the numbers are:2n, 2n+2, 2n+4, 2n+6, 2n+8, 2n+10Summation = 12n + 30 = 60012n = 597n = not integer numberno solution
What is 2N dilute sulphuric acid?
2N dilute sulfuric acid refers to a solution where the concentration of sulfuric acid (H2SO4) is equivalent to 2 moles per liter (2N). This solution is often used in chemical reactions or laboratory experiments that require a specific concentration of sulfuric acid.
