Oh, dude, you'd need a lot of water for that. Like, a crazy amount. To dilute copper chloride solution from 100000 ppm to less than 1 ppm, you'd need to dilute it by a factor of 100000. So, you'd need to add 99999 parts water for every part of the original solution. Good luck with that math!
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.
DISSOLVE 1.2 Gm OF THYMOL CRYSTALS IN 20mL OF ETHYL ALCOHOL. THEN, BRING TO FINAL VOLUME WITH ETHYL ALCOHOL.
To make an 8-fold dilution, you would mix 1 part of the substance you want to dilute with 7 parts of the diluent (usually water or buffer solution). This results in a total of 8 parts, with 1 part being the original substance and 7 parts being the diluent. This dilution reduces the concentration of the original substance by a factor of 8.
HELP! HELP! HELP! i need to know this question
to convert ratio to percent just multiply the ratio by 100 (sometimes, see below)e.g. suppose a ratio a:b=4:5, so percent of a will be;4/5×100=80or a=80%The above works if the ratio is comparing something to the total, like if the 4:5 ratio represented "4 out of 5 dentists prefer ___", then 80% of the dentists were favorable to the product.But what if you have 4 apples and 5 oranges and the total fruit is 9, then you do not have 80% apples, you have 4/9 = 44.4 % (a big difference).So it is important to know the context of the ratio, before you can just apply one rule and plug and chug, as one professor used to tell me.One real life example from my work. Everybody wants to see real-life examples. We get 2 different chemical which is used to process printing plates. The chemicals are shipped concentrated, and we dilute with water here before use. One of the chemicals states to dilute at a 1:1 ratio, which means 1 gallon of concentrate to 1 gallon of water (it even clarifies it by stating that this bottle will make 2 gallons of mixed solution). The other chemical states to dilute 3:1 water to concentrate, and states this gallon of water will make 4 gallons.
You would add powdered copper carbonate to dilute hydrochloric acid to produce copper chloride solution and carbon dioxide gas.
To dilute the copper chlorine solution from 100,000 ppm to 1 ppm, you would need to dilute it by a factor of 100,000. This means you would need 100,000 liters of water to dilute 1 liter of the original solution.
It forms copper chloride and water.
Mixing copper with hydrochloric acid would produce copper chloride and hydrogen gas. This reaction would dissolve the copper, forming a blue-green solution of copper chloride. The release of hydrogen gas could be observed as bubbles.
The acid that reacts with copper(II) carbonate to give a blue solution is hydrochloric acid (HCl). This reaction forms a solution of copper(II) chloride, which appears blue due to the presence of copper ions.
Yes, copper chloride is soluble in dilute hydrochloric acid because it forms a complex ion, [CuCl4]^2-, which increases its solubility in the acidic solution.
When dilute hydrochloric acid is added to copper oxide in a beaker, a chemical reaction occurs where the acid reacts with the oxide to form copper chloride, water, and releases carbon dioxide gas. This reaction will be observed as bubbling or fizzing due to the release of the gas. The color of the solution may change as well, from blue-green to a light blue due to the formation of copper chloride.
Sodium chloride is a crystalline solid but can be dissolved in water to form a solution.
Sodium chloride (and other compounds) are diluted only if it is necessary, this depends on each application.
When electrolysis is performed on a dilute sodium chloride solution using copper and zinc electrodes, chlorine gas will be produced at the anode (positive electrode) and hydrogen gas will be produced at the cathode (negative electrode). At the same time, sodium hydroxide solution will be formed in the solution. The copper electrode will remain unaffected, while the zinc electrode may corrode due to hydrogen evolution reactions.
Yes its solution in water is a mixture of hydrogen chloride and water.
To extract copper using dilute acid, the copper-containing ore is typically crushed and then leached with a dilute sulfuric acid solution. This process helps dissolve the copper from the ore, forming copper sulfate solution. The copper can then be extracted from the solution using various methods such as electrolysis or cementation.