For example a solution with the concentration of sodium chloride less than 10 g/L.
When chlorine gas (Cl2) reacts with dilute sodium hydroxide (NaOH), it forms sodium hypochlorite (NaOCl) and sodium chloride (NaCl), along with water (H2O). The reaction can be represented as Cl2 + 2NaOH -> NaOCl + NaCl + H2O.
You would need to dilute the 2.50M NaCl solution with water. To prepare 260 mL of 0.340M NaCl solution, you would need to use the formula C1V1 = C2V2, where C1 is the initial concentration (2.50M) and V1 is the volume of the initial solution needed. Using this formula, you would calculate the volume of the 2.50M solution needed, then add water to dilute it to a total volume of 260 mL.
The concentration of NaCl in the solution is 9 parts per million (ppm), which is equivalent to 9 mg/L (milligrams per liter). This concentration indicates a very dilute solution of NaCl.
You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.
When dilute hydrochloric acid mixes with sodium hydroxide, a neutralization reaction occurs, producing sodium chloride (table salt) and water. The equation for this reaction is: HCl + NaOH -> NaCl + H2O.
No, the dilute solutions of highly water soluble compounds are unsaturatd as solution of NaCl but dilute solution of AgCl or BaSO4 are saturated because they are very little soluble in water.
Oh, dude, it's like making a fancy cocktail but with salt. So, for 0.1mM, you just take 1 part of the 10mM NaCl and mix it with 9 parts of water. For 0.3mM, it's 3 parts NaCl and 7 parts water. And for 1mM, it's just 1 part NaCl and 9 parts water. Easy peasy, lemon squeezy!
When chlorine gas reacts with a cold dilute solution of sodium hydroxide, it forms a mixture of sodium hypochlorite (NaOCl) and sodium chloride (NaCl). This reaction is used in the production of household bleach.
To prepare a 0.9% solution take 0.9grams NaCl and dilute with 100mls of water.
Dilute Nitric acid when reacted with Sodium hydroxide will produce Sodium nitrate and Water. NaOH + HNO3 = NaNO3 + H2O.
It means you dilute your sample in a volume that is as great as the one you current sample has. Ex: you dilute 50 ml NaCl-solution in 50 ml MQ-water. The result of this is that the concentration will always be halved, seeing as the volume increases twofold.
Sodium hydroxide (NaOH) react with dilute Hydrochloric Acid (HCl) to form Sodium chloride (NaCl) and water (H2O).