What is the importance of writing correct chemical formulas?
922214
If you write the formula wrong, then you will calculate the formula weight wrong (too many or too few of one atom in a formula will translate to an incorrect formula weight). Then, when you try to figure out how many moles of the compound you have, your moles will be wrong too. And THEN, if you try to work out the stoichometry of a reaction using the incorrect formula, your mole RATIOS will be wrong, and thus you will get a wrong number when you try to determine the expected yield of a product in a reaction. (Not to mention it probably won't balance with an incorrect formula). If you're making up a solution using a wrong number of moles (see above), your molarity (moles per liter solution) will be wrong too. Everything will be wrong if you don't get your chemical formula right!
142662
the best way to remember chemical equations is to take notes or just wright it on your hand.
498666
Aaaaaaaaalright... Formulas depend on what you're solving for but since they all involve using the ratio of moles it shouldn't be too different from equation to equation.\n\nLet's go over the mole first to help you better understand what's going on in stoichiometry.  The mole is the amount of a substance that contains as many particles as there are atoms in exactly 12g of carbon-12.\n\nWHOA there... hold on for a second.  The mole is simply a counting unit, just like a dozen is.  We don't usually buy 12 or 24 doughnuts; we order half dozen, one dozen, or two dozen.  Similarly, a chemist may want 1 mol of carbon, or 2 mol of iron, or 2.567 mol of calcium.  \n\nThe number of particles in a mole is known as Avogadro's number, named after an Italian chemist Avogadro. (Go figure)\n\nAvogadro's number is 6.022 x 10^23... which is the number of particles in exactly one mole of a pure substance.  One mol of a substance is going to equal its AMU on the periodic table, or, the mass. \n\nFor example, one mol of Helium is 4g.  If you wanted to find out how many grams are in 3 mols of Helium, you would do a conversion like this:\n\n3 mols He x 4g He/1 mol He\n\nIn words that would be... 3 mols of Helium multiplied by however many grams per mol Helium is, which is 4 grams per mole.  The mols would cancel out leaving us with the units in grams.  Do 3x4 and you get 12 grams of Helium in 3 mols of it.\n\nIf you want to find out how many mols of something you have you would take the amount of grams you have of it and set up another conversion like this:\n\nAmount of g x 1 mol/gram in that mol\n\nFor example, if you have 20 grams of Helium and you want to find out how many mols of it you have, the equation would look like this:\n\n20g Helium x 1 mol/4g He per mol\n\nThe grams would cancel out, leaving the answer in mols.  You would then take 20 divided by 4 and get 5 mols of Helium.\n\n(Hoping this is making sense)\n\nNow stoichiometry is just pretty much comparisons of stuff. (Mols)\n\nAlright, when you're asked to calculate the amount in moles of one substance that will react with or be produced from the given amont in moles of another substance, you must use this simple mole conversion problem.\n\namount of given substance (mols) ----> amount of unknown substance (mols)\n\nIn order to solve this type of equation, use this conversion factor.\n\ngiven quantity x mole ratio = unknown quantity\n\n\nLet's try an example now.\n\nCOsub2 + 2LiOH ---> Lisub2COsub3\nsub=subscript\n\nLet's calculate how many moles of Lithium Hydroxide are required to react with 20 mol of COsub2\n\nGiven information = amount of COsub2 = 20 mols\nUnknown = amount of LiOH in moles\n\nSince the equation is balanced, we need to use the ratio that would give us LiOH in the final answer, therefore, meaning LiOH must be on the top of the ratio.\n\nThe correct ratio would be (mols LiOH/mols COsub2)\n\nIn order to calculate this, set up the equation now. (given q x ratio = unknown)\n\nmols COsub2 x (mols LiOH/mols COsub2) = mols LiOH \n\nsince the mols of COsub2 are on the bottom, it cancels out with the mols of COsub2 on the other side of the multiplication side because that's the rule in math, therefore leaving you with the units of LiOH.\n\nsubstitute the values into the equation and calculate the answer.\n\n20 mols COsub2 x (2mols LiOH/1 mol COsub2) = 40 mol LiOH\n\nsince the mols of COsub2 cancel out, you're left with 20 x 2 mols LIOH which equals 40 mols of LiOH.\n\nThe ratio of LiOH/COsub2, 2:1, was made from the balanced equation because for every mol of COsub2 you have you need 2 mols of LiOH to react with it to produce the right amount of products.\n\nHope this helped, good luck.
1134846
stoichiometry is easy, there are many websites that can tell you exactly what do to for different types of problems\nhttp://www.chemtutor.com/\nhere is the specific website for help on mols and stoichiometry\nhope that helps!\nhttp://www.chemtutor.com/mols.htm
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And if you need more than just a formula, here's some background information on the writing process that may help:\n\nhttp://www.angelfire.com/wi/writingprocess/\nhttp://www.csuohio.edu/writingcenter/writproc.html\nhttp://freedom.up.net/~msbones/ww01000.htm\n\nGood luck with your essay!\n~Tia~
811223
Generally, the jobs that use scientific notation are jobs that require a knowledge of science.  :)  In science, the goal of experimentation is to quantify  the object of study.  For instance, in biology we count up numbers of molecules of genes, proteins, carbohydrates, lipids, etc...In chemistry, we quantify the number of atomic particles, ie; protons, neutrons, electrons, etc... The same is also true of physics, both astro-as well as quantum.  In all of these disciplines, we are generally working with quantities greater than one million, or 1  x 10 ^6 .  Conversely, we are also generally working with quantities less than 1 one millionth, or 1 x 10^-6.  When working with numbers greater than or less than these numbers, scientific notation makes routine calculations much easier to do in manner which is calculator free and less prone to error. \n\nThe biggest problem we have is being off by a factor of 10 or a 1x10^1, as it very easy to be off by one zero :)   \n\nHope this helps. \n\nSincerely,\n\nCurt\n\nP.S.  1x10^0 = 1
183585
at least write them correctly
707973
You have to start with a balanced equation.  The subscripts will give you the correct charge and a prefix will give you the right amount of atoms on each side of the equation.  Some elements, like O, F, and I are diatomic and come in pairs.
109334
Think of the chemicals u consume every day....a disproportionate amount of some chemical will be harmful to you...arithmetic & algebra is used to determine what qntity of what to use...don't forget expiration dates are calculated based on the "life expectancy" of used products....ofcourse statistics comes in with the financial mgt (be it the food industry or what they have to account for their expenditure)!
225876
Balancing Chemical Equations is absolutely essential if you want to determine quantities of reactants or products. An unbalanced Chemical Equation gives only the identify of the beginning reactants and the final products using the appropriate formulas as well as the conditions of temperature, physical state, and pressure conditions under which the reaction is to operate under. However an unbalanced equation can say nothing about the quantities involved until the equation has been balanced. A balanced equation assures that the Conservation Law of matter is obeyed. The total mass of reactants must equal the total mass of products. A balanced equation is like a recipe. It tells you the proportional quantities of each substance involved. The coefficients that appear to the left of each formula can be interpreted as mole units or molecules, but for practical purposes, the coefficients will be interpreted as mole units. No coefficient is interpreted as 1 mole. The following principles should be employed when balancing a Chemical Equation by inspection:\n\n   1. Never touch subscripts when balancing equations since that will change the composition and therefore the substance itself.\n   2. Check to be sure that you have included all sources of a particular element that you are balancing on a particular side since there may be two or more compounds that contain the same element on a given side of an equation.\n   3. I would suggest that you adjust the coefficient of mono atomic elements near the end of the balancing act since any change in their coefficient will not affect the balance of other elements\n   4. When there are a group of atoms that are acting as a unit such as a polyatomic ion and they appear intact on both sides of the equation, it is best to balance them as a self contaned group. For example, if there are Phosphate groups, PO4-3, that appear on both sides, balance the phosphates as a group instead of separating the Phosphorus and Oxygens. It can be done either way, but there is less likely of a mistake if they are balanced as Phosphate groups.\n\nConsider the example of:\n\n(NH4)2CO3 ---> NH3 + CO2 + H2O\n\nThe first thing to do is choose a starting point. If we choose the carbon as the starting point, we go nowhere, since there is already one carbon on each side. Choosing oxygen is not good, because the oxygen is divided between two species on the right side. The interesting part is in the ammonium (NH4) ion. If we start with the nitrogen on the left (in the ammonium ion), we see that there are two nitrogens and therefore we must put at "2" in front of the ammonia on the right side.\n\n(NH4)2CO3 ---> 2 NH3 + CO2 + H2O\n\nSince we are already working with the ammonia, now consider the hydrogens that make up the rest of the molecule: the six hydrogens in the ammonia plus the two hydrogens in the water make eight hydrogens. Bouncing back across to the left side, we count eight hydrogens; since no other atomic species have been changed, we can count up everything and see that the reaction is now balanced:\n\n2 N, 8 H, 1 C, and 3 O on each side.\n\nNotice how the reaction itself led us through the steps of the balancing process. A more complex reaction would have involved more steps, but the process would have been the same.\n\nThis "inspection method" is the first one tried by chemists in the real world. Only when it fails to work, as in the case of many oxidation-reduction reactions, for example, are more complex procedures used.\n\nHere is another example:\n\nFe2(SO4)3 + K(SCN) ----> K3Fe(SCN)6 + K2SO4\n\nK3Fe(SCN)6 is pretty complicated. Now look at it and see what ions are in it that are not shared on the same side of the reaction. These are SCN and Fe.\n\nWe start to balance these\n\nFe2(SO4)3 + 6 K(SCN) ---> K3Fe(SCN)6 + K2SO4 Now we go left to right.\n\nThe irons are messed up, two on the Left, and one on the right. We adjust\n\nFe2(SO4)3 + 6 K(SCN) -- 2 K3Fe(SCN)6 + K2SO4 .\n\nNow t
574078
This might help :)\nhttp://www.chemicalforums.com/index.php?board=8;action=display;threadid=3263
1187055
As mentioned above, Mg(HCO3)2\n\nIt was so easy.. in which grade u study?\n Do u know how to make formulae? Just learn the valencies and symbols of most commonly used elements... Here are some for your help...\n\nMagnesium (Mg): 2\nAluminium (Al): 3\nCarbonate (CO3): 2\nBicarbonate (HCO3): 1\nSulphate (SO4): 2\nBisulphate (HSO4): 1\nSodium (Na): 1\nBoron (B): 3\nBeryllium (Be): 2\nCalcium (Ca): 2\nPotassium (K): 1\nChlorine (Cl): 1\nBromine (Br): 2\nFluorine (F): 1\nNitrogen (N): 1\nCarbon (C): 4\nHydrogen (H): 1\nOxygen (O): 2\nHydroxyl (OH):1\nNitrate (NO3): 1\n\nI think they are enough for you...\n\nNow see how to make a formula:\nStep No. 1: Write the symbol of each element in a compound. \nE.g. You want to make the formula of magnesium bicarbonate. Write their symbols...\n\nMg HCO3\n\nStep No.2: Write their valencies.. \nThe valency of Mg is 2\nThe valency of HCO3 is 1\n\nMg (2) HCO3 (1)\n\nStep No.3: Now cross the valencies of each element.. It means write the valency of Mg with HCO3 and write the valency of HCO3 with Mg\n\nand here you go with the formula...\nMg (1) HCO3 (2)\n\nNo need to write 1 with any element..\nso it becomes,\nMg(HCO3)2\n\nHere are some more examples..\nPotassium hydroxide\nK OH\nK (1) OH (1)\nKOH\n\nCalcium hydroxide\nCa OH\nCa (2) OH (1)\nCa(OH)2\n\nMagnesium chloride\nMg Cl\nMg (2) Cl (1)\nMgCl2\n\nSodium sulphate\nNa SO4\nNa(1) SO4(2)\nNa2SO4\n\nAluminium nitrate\nAl NO3\nAl (3) NO3 (1)\nAl(NO3)3\n\nHere is one last rule for you:\nIf the valencies are equal, do not cross..\n\nExample:\nCalcium sulphate:\nCa SO4\nCa (2) SO4 (2)\nNow both the valencies are equal, so just cancel the valencies...\nand the formula becomes,\nCaSO4\n\nAhhhh... got tired... Hope this helps.. good luck for you submission.. Now u can also explain your teacher that how u got the formula...\ncya\nAllah Hafiz
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Today, chemistry plays an important role in nearly every area of commercial and residential construction. Virtually every component of construction involves the use of chemistry -- from vinyl drainage pipes below the foundation to roofing materials. Chemistry is used to make pipes, siding, windowsills and frames, roofing, carpeting, paints, sealants, wall coverings, insulation and adhesives. It plays an important role not only in construction, but also in preserving a building from damage caused by time and nature.\n\nThe average new house in the United States contains tens of thousands of dollars of chemistry-based products. Chemistry is used to make building materials that are fire and rust retardant. Chemistry is used to supply energy-saving materials like home insulation and thermal windows. Stronger materials better protect us from high winds and even earthquakes, and more flexible materials contribute to design, function and comfort.
1097015
chemical reactions are the basis for life on this planet, and the fact that everything in the world that we know of realtes to chemistry it carries a very high importance in understanding the world around us. The greater understanding of the world the better the chance that we can manipulate the world to suit our needs and desires.
285998
Depends on the active ingredient and its physiological action.  Why not ask your local pharmacist to pick a product and recommend a dosage?  When you explain you're writing a book I'm sure he'll know how to help.
297119
Chemisty, A Formula For Answers
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something about chemistry