how do you separate a mixture of Magnesium, Strontium, and Barium?   please help me peoples of the World?
887726
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387822
Barium is a group II metal, therfore the most stable ion is Ba2+
141058
sodium sulfte is soluble, but barium sulfate would form as a precipitate. Thus\n\nBaNO3 + Na2SO4 -> BaSO4 + 2 NaNO3, or simply\n\nBa(2+) + SO4(2-) -> BaSO4
781614
Na2SO4 (aq) + Ba(NO3)2 (aq) --> 2 NaNO3 (aq) + BaSO4 (s)\n\nThe moles of Barium = 5.00 L * 0.15 mol/L = 0.75 moles\n\nThis is also the moles of sulfate needed:\n\n0.75 mol / 0.36 mol/L = 2.08 L of Na2SO4
761616
One that will ideally:\n1) dissolve the component that you wish to collect but not the others. Then you can do a filtration, collect the filtrate and evaporate off the solvent to get back the component that you want.\nOR\n2) dissolve all those that you don't wish to collect but not the component that you want. Then you can do a filtration and collect the component that you want as the residue.\n\nHope this helps :)
782238
Here's what you do:\n\n1) Write the balanced equation for the reaction.  How many moles of BaSO4 does each mole of BaX2 create? \n2) Determine the F.W. for each BaX2, and for BaSO4 (by the way, the unknown won't be BaF2. It's insoluble: http://www.ktf-split.hr/periodni/en/abc/kpt.html ) Determine the number of moles there are in 0.0944 g of BaSO4, and in 0.158 g of each BaX2.\n3) For which BaX2 does 0.158 g represent the the correct number of moles?  Note: I worked this out, and got a number that was very close, but not an exact match.  That's realistic, since the BaX2 would not be absolutely pure, SOME minute amount of BaSO4 will dissolve, etc.
745615
It's called the step rule.\n\nDraw a line that separates the metalloids from the non-metals.  The line should look like stair steps going down and right.  All non-metals will be to the right of this line.  All metalloids will be between this line and the transition metals.
1116509
analize an existing polymer with as many properties of the ones you are looking for, determine the formula ( possibly use a mass spectrometer for the recipe,)  experiment with the atomic structure of the elements in the formula. wouldn't it be great if you could manipulate the structure at subatomic level, i.e. at levels below protons and neutrons like quarks.\ntry changing the formula ( recipe?) at its basic level, with different proportions of at least carbon and hydrogen, possibly adding 'alloying' elements into the atomic structure.
892301
You do not need 2 BaCO3's.\n\n2HCl + Ag2NO3 --> H2NO3 + 2AgCl
757423
First find the mass weight of Strontium: 87.62 g/gmol\nNow find the mass weight of SrCO3: 87.62 + 12 + 3*16 = 147.62\n\nNow find the fraction of Sr in SrCO3: 87.62/147.62 = 0.594\n\nSo in order to get 100 ton of Sr which is only 59.4% of what you need SrCO3.  The the tonnes of SrCO3 you need is\n\n100 ton Sr --- 0.594\nx ton SrCO3 --- 1\nx = 100 / 0.594 = 168.4 ton SrCO3.\n\nNow, again, this is 2% of what is in the ore, so the amount of ore you need is\n\n168.4/2% = 168.4 * 50 = 8417 ton ore\n\nCheck\n8417 * 0.02 * 0.594 = 100  OK.
375562
I assume you mean alkaline earth metals and they are all metals in the second column of the periodic table and include beryllium, magnesium, calcium, strontium, barium, and radium.  They oxidize easily in air so most often pure samples are stored in oil.
867056
barium: 1s22s22p63s23p64s23d104p65s24d105p66s2\n\nradium: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s2\n\nsry if it looks a bit confusing without using exponetnts but im sure u could figure it out
793681
Those are the alkaline earth metals, technically referred to as Group 2.  They are beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba) and radium (Ra).  Radium has two valence electrons and is in Group 2, but is sometimes left off lists of the alkaline earth metals because it is radioactive.
736794
I'll teach you how to do these.  The answer may seem long at first, but just try it, and you'll see it's not as bad as it first appears.  For mass/mass problems, you only have to use the first section of this answer:\n\nToo solve these types of problems, use the following general steps:\n\n[I will use the general word equation: calcium hydroxide reacts with lithium bromide to produce     calcium bromide and lithium hydroxide; as an example, throughout each step, to aid in your learning]\n\n(1)  Write the correct chemical formula for each substance in the reaction.  Be certain each chemical equation is correct!   (ex:  calcium hydroxide:    correct formula  = Ca(OH)2;   NOT CaOH.)\n\n(2)  Write out the chemical equation;   ex:  Ca(OH)2  +     LiBr   gives      CaBr2    +       LiOH    \n\n(3)  Balance this chemical equation;    ex:   Ca(OH)2  +  2LiBr   gives   CaBr2   +   2LiOH\n\n       Note:  this balanced equation is TELLING you that one mole of calcium hydroxide is reacting with 2 moles of lithium bromide to produce (gives) one mole of calcium bromide and 2 moles of lithium hydroxide.  It is TELLING you the relative number of moles of each substance.  \n\n(4)  Determine the MOLE RATIO of the substances involved;  ex:  if the question says you have 1.0 moles of calcium hydroxide and it reacts with an excess of lithium bromide, and asks how many moles of lithium hydroxide are produced,  the answer would be 2.0 moles of lithium hydroxide.  In other words, the MOLE RATIO of calcium hydroxide to lithium hydroxide is 1 to 2, also written as a “1:2 ratio“.    ex:  same question but suppose you have 3 moles calcium hydroxide.  Because the mole ratio  is 1:2, you would produce 6 moles of lithium hydroxide.   A ratio of 3:6  =  a ratio of 1:2;  (you just reduced it to simpler terms).  \n\nStop here if you  are simply given the number of moles of product you have, and asking how many moles of some different substance are produced in a chemical reaction.  \n\n\nSame type of problems, but working with grams instead of moles:\n\nyou simply have to do a couple of extra steps:\n\nI will use the same reaction as above, but will change the question to this:  \n\n“If you have  255 grams of calcium hydroxide, how many grams of lithium hydroxide will be produced, (assuming you have an excess of lithium bromide)”.  \n\n [Note:  unless the question states otherwise, assume you have an excess of the other reactant, thus giving the maximum yield of product; see “Limiting Reagents” below to solve those types of problems]. \n\nTo solve, complete the first 4 steps above (we’ve just done that):\n\n(5)  Convert grams of substance into moles of substance!!!   This is essential in solving these types of problems.  THE KEY STEP FOR ALL PROBLEMS OF THIS TYPE IS TO ALWAYS CONVERT EVERYTHING INTO MOLES FIRST!!!  The general formula’s are as follows:  \n\n                        Moles  =  grams / molecule weight\n\nOr you can use basic algebra and say the following:  \n\n                    Grams  =  Moles   X   molecular weight\n\n[Note:  “molecular weight” is used to describe molecules; “formula weight” is used to describe ionic compounds; and “atomic weight” is used to describe elements;  ALL of them are the sum of all the atomic weights of the elements in the substance. In the case of a single element, alone by itself, this refers to the atomic weight of that element.  Note that the 7 elements:  hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine and iodine only exist as “diatomic” elements, when they are ALONE by themselves (not combined with any other element(s)).  Thus H2, is the correct way to write the chemical formula for hydrogen when it is all alone, by itself.  And it is a molecule, and it’s molecular weight would be equal to 2.0.  This is the same for the all diatomic molecules:  the molecular weight will equal double the atomic weight for that particular diatomic element (because there are TWO a
371226
Calcium(Ca+²) and Magnesium(Mg+²)
715465
1) After k half-lives, the remaining amount of Barium-131 from the m miligrams to start with is: m / 2^k (mg)\n\nIn your case, m = 10 mg and k = 3.\n\n2) Time for m miligrams of any radioactive species to decay to n miligrams is t1.\n\nHalf-life t = t1 / log(2)(m/n)\n\nIn your case, t1 = 45minutes, m = 50mg and n = 6.25mg\n\n3) Density d = p(MW)/RT\n\np = 2\nMW = 14\nR = 0.0821\nT = 273 + 127 = 400
442899
Manganese and manganese dioxide. Also potassium iodide does a good job.