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To prepare a 1 M solution of perchloric acid (HClO4), you would need to dissolve 1 mole of HClO4 in enough water to make a final volume of 1 liter. This can be calculated by using the formula: moles = Molarity (M) x Volume (L). Please handle perchloric acid with caution and use appropriate safety measures while preparing the solution.
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A 25 ml sample of 723 m hclo4 is titrated with a 273 m koh solution the h3o concentration after 66.2 ml of koh is?
The balanced chemical equation for the reaction between HClO4 and KOH is HClO4 + KOH → KClO4 + H2O. By using the stoichiometry of the reaction, you can calculate the moles of HClO4 reacted with KOH. Then, use the remaining volume of KOH solution added to calculate the final H3O+ concentration in the solution.
How do you prepare 1 M solution from 6 M solution?
It's simple to dilute a solution, just add more water. If you have one liter of 6 molar solution, and add another 5 liters of water, it becomes a 1 molar solution. You can also use M1V1 = M2V2 formula for dilution. For example, if you want to prepare 1 M of solution in 1 litre of water, then how much volume you need from the initial 6 M solution? Simply use the equation, (6 M) x (V1) = (1 M) x (1 litre) ---> V1 = (1 M) x (1 litre)/(6 M) = 0.167 L or 167 ml
How do you prepare 1 L of 15 M sodium hydroxide solution by diluting the stock solution of 6 M sodium hydroxide?
To prepare 1 L of 15 M NaOH solution, you need to dilute the 6 M stock solution. You would need to take 500 mL of the 6 M NaOH stock solution and add water to bring the total volume up to 1 L. This will give you a final concentration of 15 M.
A 30.0 ml sample of 0.200 m koh is titrated with 0.150 m hclo4 solution calculate the pH after 39.9 ml of hclo4 is added?
To solve this problem, you first need to determine the moles of KOH present in the 30.0 mL sample. Then calculate the moles of HClO4 added after 39.9 mL. Based on these concentrations, determine the excess and limiting reagents to find the resulting pH. Consider the reaction that occurs between KOH and HClO4, and use the stoichiometry to calculate the amount of products formed. Finally, calculate the pH using the concentration of the resulting solution.
How many grams of KCl is needed to make 2 molar solution in 1 liter of water?
To prepare a 2 M solution of KCl in 1 liter of water, you would need to dissolve 149.5 grams of KCl. This is because the molar mass of KCl is approximately 74.5 g/mol, and 2 moles of KCl are needed to prepare a 2 M solution in 1 liter of water.
A 25 ml sample of 723 m hclo4 is titrated with a 273 m koh solution the h3o concentration after 66.2 ml of koh is?
The balanced chemical equation for the reaction between HClO4 and KOH is HClO4 + KOH → KClO4 + H2O. By using the stoichiometry of the reaction, you can calculate the moles of HClO4 reacted with KOH. Then, use the remaining volume of KOH solution added to calculate the final H3O+ concentration in the solution.
How do you prepare M Cal solution?
To prepare 1 M CaI aqueous solution, dissolve 29.4 g in a total volume of 100 mls, or 294 g in a total volume of 1 liter.
How do you prepare 1 M solution from 6 M solution?
It's simple to dilute a solution, just add more water. If you have one liter of 6 molar solution, and add another 5 liters of water, it becomes a 1 molar solution. You can also use M1V1 = M2V2 formula for dilution. For example, if you want to prepare 1 M of solution in 1 litre of water, then how much volume you need from the initial 6 M solution? Simply use the equation, (6 M) x (V1) = (1 M) x (1 litre) ---> V1 = (1 M) x (1 litre)/(6 M) = 0.167 L or 167 ml
How do you prepare 1 L of 15 M sodium hydroxide solution by diluting the stock solution of 6 M sodium hydroxide?
To prepare 1 L of 15 M NaOH solution, you need to dilute the 6 M stock solution. You would need to take 500 mL of the 6 M NaOH stock solution and add water to bring the total volume up to 1 L. This will give you a final concentration of 15 M.
How would one prepare 10 ml of a 25 M HCI solution if 1 M HCI was available?
To prepare a 10 ml of a 25 M HCl solution, you would need to dilute the 1 M HCl solution by a factor of 25. You will need 0.25 ml of the 1 M HCl solution and 9.75 ml of solvent (usually water) to make a total volume of 10 ml for the 25 M HCl solution.
A 30.0 ml sample of 0.200 m koh is titrated with 0.150 m hclo4 solution calculate the pH after 39.9 ml of hclo4 is added?
To solve this problem, you first need to determine the moles of KOH present in the 30.0 mL sample. Then calculate the moles of HClO4 added after 39.9 mL. Based on these concentrations, determine the excess and limiting reagents to find the resulting pH. Consider the reaction that occurs between KOH and HClO4, and use the stoichiometry to calculate the amount of products formed. Finally, calculate the pH using the concentration of the resulting solution.
How much gm of disodium edta require to prepare 0.05 m disodium edta solution?
To prepare a 0.05 M disodium EDTA solution, you would need to dissolve 3.72 grams of disodium EDTA dihydrate (Na2C10H14N2Na2·2H2O) in enough water to make 1 liter of solution.
How many grams of KCl is needed to make 2 molar solution in 1 liter of water?
To prepare a 2 M solution of KCl in 1 liter of water, you would need to dissolve 149.5 grams of KCl. This is because the molar mass of KCl is approximately 74.5 g/mol, and 2 moles of KCl are needed to prepare a 2 M solution in 1 liter of water.
How do you prepare a 0.0001 M soulution HCL using 0.1 M of HCl?
you need to make 1000 times dilutions, this could be done in multi-steps: transfer 1 ml 0.1 M HCl into 100 ml volumetric flask and complete volume with water --------(1) from solution (1) transfer 2 ml into 20 ml volumetric falsk and complete volume with water, this is 0.0001 M HCL.
What volume of a 2.5 M stock solution of acetic acid (HCHÀO) is required to prepare 100.0 milliliters of a 0.50 M acetic acid solution?
To prepare a 0.50 M acetic acid solution, you would need to dilute the 2.5 M stock solution. By using the formula M1V1 = M2V2, you can calculate the volume of the stock solution needed as: (0.5 M)(100.0 mL) = (2.5 M)(V2), where V2 is the volume of the stock solution needed. So, V2 = (0.5 M x 100.0 mL) / 2.5 M = 20.0 mL. Therefore, you would need 20.0 mL of the 2.5 M stock solution to prepare the desired 100.0 mL of 0.50 M acetic acid solution.
How do you prepare 1millimolar silver nitrate solution from 0.5 molar silver nitrate?
Dilute 1 mL of 0.5 M silver nitrate solution to a total volume of 1 L with water to make a 1 mM silver nitrate solution.
Why you use 85 ml Conc HCl for 1 M HCl?
To prepare 1 M hydrochloric acid (HCl) solution, you usually need to dilute concentrated HCl solution (~35-37% HCl) with water in a specific ratio. The specific volume of concentrated HCl needed depends on its concentration and the final volume of the 1 M solution you want to make. In this case, 85 ml of concentrated HCl is likely to be the volume needed to make 1 L of 1 M HCl solution.
