How will you know if you have to remove interfering anions from an unknown?
967315
say u are going to perform a test for anion A in some sample. but you know that this test may also have a positive result with other anions B and C. \n\nyou then have to take into consideration that B and C might give you false positives as u test for A. if the background of the sample suggests it may contain B and C then find a way to remove these without affecting A.\n\nthen test for A using your originally proposed method.\n\n\nnow for the example:\nyou are to test for chloride ions. the reagent that comes to mind is aq silver nitrate which will result in white ppt of AgCl. then you realise that a ppt will form with any halide ion such as Br- and I-. you also know that AgBr and AgI solids will dissolve in ammonia while AgCl will not. so you first add ammonia to the sample, then test with silver nitrate. voila, now if a ppt forms, you will know it is AgCl only, and of course the conclusion that Cl- was present.
1174256
I don't see how it could, but that doesn't mean much.  Doesn't DI mean that the solution is neutral in charge?  It probably would corrode with an oxidizer present.
1024740
The question makes no sense.  A single, isolated anion has no pH value.  pH refers to the H+ concentration in aqueous solutions.
904474
Do not crush the rock salt as this will make the impurities harder to remove.\n\nDissolve the rock salt in very hot water and filter.  If the filtrate (the water that comes through the filter) is colored, then you might need to treat it with charcoal to remove these soluble impurities.\n\nOnce you have a clear solution, allow some of the water to evaporate until the solution becomes saturated.  A single seed crystal will start the formation of sodium chloride crystals from the solution.  Stirring will keep the crystals smaller, otherwise you can end up with one very large crystal of salt.\n\nThis crystallization process is what does the final purification.  The crystals formed will be almost pure sodium chloride.
294143
add:\n\nH2S2O8(ag)+ NH3(aq) + ? -->(NH4)2S208(aq)  + ?\n\nAmmonia can be prepared in the laboratory by treating ammonium chloride with sodium hydroxide:\n\nthe net ionic equation to this reaction is:\n\n NH4+(aq) +  OH-(aq) ---> NH3(g) + H2O(l) in system 1\n \nand\n\n NH3(g)   = NH3(aq) in solution in system 2\n\nyou finish it ,if you need more help just e-mail me
409824
When what is mixed is not fully disolved into the liquid medium.  It usually settles.  A cenrifuge will not separate out the components that are not part of the solution.
773115
you can surely rule out some things based on the type of plate you are using.  assuming that you have covered that base, you can separately pick a colony that you think may be a contaminant and look for differences in growth patterns, color, morphology, gram stain, growth on other media, and if all else fails, you can sequence it to find out exactly what it is.
98382
You will not find ions along by themselves, they are paired with other ions, forming ionic compounds.\nSome ionic compounds are soluble, some are.  The ones that are can range in solubility based on the solvent, temperature, and other ions in solution.\n\nFor instance, NaCl is very soluble in water at room temperature, where as under the same conditions, Silver Chloride is considered insoluble.\n\nTherre are actually a set of solubility rules to determine whether an ionic compound is or is not soluble in water based on the ions that make it up.\nhttp://www.csudh.edu/oliver/chemdata/solrules.htm
378767
You would complete dissolve it in strong acids and then separate the metals by qualitative analysis techniques.  Not simple and very expensive.
842976
20 grams\ndissolving compounds and then driving of the solvent is a common way to purify a compound. You will not lose any of the compound unless it is reactive with the solvent you dissolve it in. In the case of water and salt they are not reactive so when you drive off the water you will be left with the same amount of salt that you started with.
740901
You need to split the ionic compounds apart to find the individual ions you would find in these structures.\n\nMgO would produce Mg^2+ & O^2- ions.\n\nK2S would produce K^+ & S^2- ions.\n\nYou can find the charge values directly from the periodic table trends for the charges atoms assume when they are ionized.
887064
The main thing you need to know to figure out the ionic equation is whether the product is soluble or not.  In this case, PbSO4 and Mg(NO3)2 are your products, and lead salts are insoluble.  Thus, PbSO4 and any ion that is used to make this molecule is removed from the ionic equation:\n\nPb(+2) + 2(NO3)(-1) + Mg(+2) + (SO4)(-2) -> PbSO4 + Mg(NO3)2 (aq)\n\nbecomes:\n\n2(NO3)(-1) + Mg(+2) -> Mg(NO3)2 (aq)
947749
Without knowing exactly what you did I can't answer this question!  Dissolution in dilute sodium hydroxide solution, filtration to remove any fillers etc, re-precipitation with hydrochloric acid and isolation by filtration should have given you an answer of some sort. Not exactly accurate though.
723747
The results may be affected by not cleaning the sides with distilled water.  Assume that your titrant has pH 10.  that will be 1000 times more concentrated that distilled water.  Losing some of this titrant to the wall may affect your result.  Putting some more water will not.
272575
let me explain this with an example.take Cl atom which has 17 protons in the nucleus and equal no. of electrons(17)revolving around the nucleus.this means tht the 17 positive charges are attracting the 17 negative charges.\nnow let us consider the chloride ion Cl(-1). the protons are again 17 here but one negative charge is extra ie there are 18 electrons.so,in this case the 17 positive charges are pulling the 18 negative charges.\nnow u can imagine in which case is the pull less efficient atom or anion?\nobviously it is less efficient in case of anion because of the inequality of charges thus making it bigger in size.\nhope ,i ve made it clear with an example...
696442
It would depend on what the nature is of the unwanted impurities. One could use chemical separation or distillation.  Distillation would be the easiest.
151863
Either:\n\n(a) you're not getting a very clean separation, as the peaks should be vivid\n\nor\n\n(b) your detector is insuitable for the acids you're studying.\n\nAtleast, that's my analysis. Any A-Chem experts here?