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 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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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
HCL 2N solution is prepared by diluting it in a one to five ratio with water. It is not necessary to provide an outside force such as heat or pressure. Simple mixing with complete the dilution.
Reference: the density of 25% NH4OH solution = 0.907g/cm³ Therefore 1 litre of solution has mass 907g 25% = 25/100*907 = 226.75g NH3 in 1 litre solution Molar mass NH3 = 17.031 g/mol 226.75g = 226.75/17.031 = 13.31N solution To make 1 litre of 2N solution: ( Work in ml= 1000ml) N1V1 = N2V2 13.31*V1 = 2*1000 V1 = 2000/13.31 V1 = 150.2ml Measure out 150.2ml of the 25% solution and dilute to 1 litre. This is a 2N solution.
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
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.
Answerno solutionProcedureAssume the numbers are:2n, 2n+2, 2n+4, 2n+6, 2n+8, 2n+10Summation = 12n + 30 = 60012n = 597n = not integer numberno solution
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.