1. The symbol for strontium is Sr. If the mixture was 14 percent strontium and 688 grams of the mixture was not strontium, what was the weight of the strontium?\n\n2. Two kilograms of iron was melted with 7 kilograms of other metals to make the alloy. If 1440 grams of the alloy was required, how many kilograms of iron should be used?\n\n3. The formula for iron sulfide is FeS. If the iron (Fe) in a batch of iron sulfide weighed 448 grams, how much did the iron sulfide weight? (Fe, 56; S, 32)
199667
1) 14% + 688 = 100%\n86% = 688\n688/ 86 = 8\n8 X 14 = 112\n\n2) 1440 / (7+2)\n = 160\nX2 = 320\n\n3) 448 / 56 = 8\n8 X 32 = 256\n256 + 448= 704\n\n\nthere ya go..  i did ur homework :)
362083
(NH4)2Cr2O7 -> Cr2O3 + 4*(H2O) + N2\n\n(NH4)2SO4 + 2*(NaOH) -> 2*(NH3) + 2*(H2O) + Na2SO4\n\nFe2(SO4)3 + 6*(NH3) + 6*(H2O) -> 2*(Fe(OH)3) + 3*((NH4)2SO4)
458499
The balanced equation is:\n4Fe + 3O2 --> 2Fe2O3\n\nNow use the stoichiometric relationship between Fe and Fe2O3 to find how many moles of Fe is required. For every 2 moles of Fe2O3 produced, you need 4 moles of Fe.\n\n.0648 mol Fe2O3 * (4 mol Fe / 2 mol Fe2O3) = .1296 mol Fe
499913
Ok, normally, metals retain their spelling when they form compounds. Non-metals will have their names spelt a little differently, usually now ending with -ide...\nLike oxygen, form compounds --> oxide\nchlorine, form compounds --> chloride\nfluorine, form compounds --> fluoride\nI'm sure you get the idea.\n\nSo for your question,\na) copper (II) chloride or copper (I) chloride \nCopper comes in 2 valencies - 1 and 2. Since you did not specify the valency of the copper ion involved here, I gave you both possible answers.\nb) iron (II) oxide or iron (III) oxide\nSimilarly, iron comes in iron (II) (valency 2) or iron (III) (valency 3).\nc) sodium fluoride\nd) er... codine? do you mean iodine?\nlithium iodide\n\nHope this helps :)
442105
First let's find how many moles of each one we have\n\nmole= M*V, so for \nFeCl3 =0.234*25*10^-3= 5.85*10^-3\nNaOH =0.453*42.5*10^-3 =19.25*10^-3\n\nFeCl3 + 3NaOH -> Fe(OH)3 + 3NaCl\n\nso\n\n1mole FeCl3 reacts with 3mole NaOH\n5.85*10^-3 mole react with x\n\nThus x =3*5.85*10^-3=17.55*10^-3 mole NaOH but we have 19.25*10^-3 mole NaOH, therefore NaOH is in excess\nTherefore the amount of Fe(OH)3 will be calculated according to the limiting reageant FeCl3.\n\nso from the reaction\n\n1 mole FeCl3 gives 1 mole Fe(OH)3\nor better yet, since the molecular weight of Fe(OH)3 is 106.85 \n1 mole FeCl3 gives 106.85 g Fe(OH)3\n5.85*10^-3 mole give x g\nx= 106.85*5.85*10^-3= 0.625 g\n\nWe calculated that 17.55*10^-3 mole NaOH reacted.\nSo we have remaining (19.25-17.55) *10^-3= 1.7*10^-3 mole NaOH\n\nThese are in a total volume of 25.0+42.5=67.5 ml= 67.5*10^-3L.\n\nSo the concentration of the excess NaOH is\nC=mole/V= 1.7*10^-3/67.5*10^-3= 0.025 M
292320
a. 2k+ = .20moles , SO42-=.10moles\n\nb.Fe3+=.10moles,3NO3-= .30moles\n\nc.2Al3+=.20moles,3SO42-=.30moles\n\nd. Ni2+=.10moles,SO42-=.10moles
534322
1.\nMn(s) + SnCl2(aq) --> MnCl2(aq) + Sn(s)\nCr(s) + 2HBr(aq) --> CrBr2(aq) + H2(g)\nCo(s) + CaBr2(aq) --> N.R\n2Al(s) + 3Ni(NO3)2(aq) --> 2Al(NO3)3(aq) + 3Ni(s)\n\n2.\nMg + 2AgNO3 --> Mg(NO3)2 + 2Ag\nmole of Ag = 3.580 / 108 = 0.033 mole\nmole of AgNO3 = 0.033 mole = 33 mmole\nmolarity of AgNO3 = 33 / 50 = 0.66 M\n\n3.\na.soluble\nb.soluble\nc.soluble\nd.insoluble\ne.soluble\nf.soluble\ng.insoluble in water, soluble in acid\nh.soluble\n\n4.\n3NaOH(aq) + Fe(ClO4)3(aq) --> Fe(OH)3(s) + 3NaClO4(aq)\n3Na+ + 3OH- + Fe3+ + 3ClO4- --> Fe(OH)3 + 3Na+ + 3ClO4-\nFe3+ + 3OH- --> Fe(OH)3\n(NH4)2SO4(aq) + Pb(NO3)2(aq) --> PbSO4(s) + 2NH4NO3(aq)\n2NH4+ + SO42- + Pb2+ + 2NO3- --> PbSO4 + 2NH4+ + 2NO3-\nPb2+ + SO42- --> PbSO4\n\nHope it helps\npeace\nvixklen
345221
The atomic weight is the mass in grams per mole (atomic mass of iron = 55.845).\n\nA mole is 6.02 x 10^23 atoms, therefore the mass of 500 atoms would be:\n\n500/(6.02x10^23) x 55.845 = 4.64 x 10^-20 grams.\n\nNotice that's 10 to the -20 power.
398757
I think these are the correct answers.  I won't show you how I got my answers, but I will tell you one integral piece of information:  remember that the atomic weight is the # of grams in one mole of the element of interest.  (I would like to see what you worked out.)\n\n1)  1.663 g of Pb(NO3)2 in 0.005 mol\n     0.211 g of K2CrO in 0.005 mol\n\n2)  417.210g of CO2 can be removed.\n\nPlease keep in mind that I could be wrong...it's been a bit since I've regularly done chemistry.  Again, I want to know what you did, if you don't mind.
506550
I'm not directly giving you the answers, you have to know how to do this stuff, but I'll give you the process to get to each.\n\n9: Americium and Curium (wikipedia!)\n\n10: Potassium ion wants to be like the noble gas closest to it (the one right before it), so it loses 1e- to get there and be K+.  If that's not a choice, then look for another ion, positive or negative, that adds up to the electronic configuration of the aforementioned noble gas. (1s2 2s2 2p6 3s2 3p6), 18 total e-'s...\n\n11: Mg is in column 2, things in the same period tend to be chemically similar.  Find the other one in the choices that's in that column\n\n12: U235 and U238 are isotopes, same atomic number, different weight.  If you change the number of protons, you change the atomic number, so that wouldn't be U anymore.  Changing electrons can make ions, but not change weight.  So different number of neutrons for the same element makes an isotope.\n\n13: Same explanation as above, it'll also be Hydrogen's atomic number because it's an isotope.\n\n14: shell 2 contains the 2s and 2p subshells.  2s can hold 2 because it's an s orbital.  2p can hold 2 per p orbital, and a p subshell has 3 p orbitals, so 6.  Total would then be...\n\n15: If an atom loses electrons, it's LOSING -1 per electron lost, so it's becoming more positive.\n\n16: Which element in that list only has 1s shell? Which one has a  stable ion as just a proton or with a configuration like Helium (1s2)?\n\n17: Inert elements are on the far right of the periodic table.  They are inert because they have 8 valence (outer shell) electrons (or 2 in Helium's case) so they can't take anymore and are stable.  Look at your periodic table or count up electrons for each of those atomic numbers.  Only one of them will have 8 in the outer shell...
329417
fe 2+ , SO4 2-, Ag+, NO3 -\n\nthe products:\n\nFe(NO3)2 + Ag2SO4
503709
a is the mass of alloy with 25% copper\nb is the mass of alloy with 50%copper\na+b=1000\na*25+b*50=1000*45\nb=1000-a\n25a+50*1000-50a=1000*45\n\nFor part 2, If the triangle is a right triangle (a special case), then the sides are:\nb\nb-4\nsqrt(2b^2-8b+16)\n\nIf it is not a right triangle, then you have to also know one angle.  If you know the angle subtended between the base and the side that is length b-4, call it a, then you can compute the height, h of the triangle as\nh=(b-4)*sin(a)\nand the base of the right triangle formed by the height and b-4 as the hypotenuse, call it x\nx=(b-4)*cos(a)\nSince the base of the triangle is known to be b,\nthen the base of the right triangle with the unknown side is b-x\nor\nb-(b-4)*cos(a)\n\nIt shares the height of the triangle as a side with the unknown side as the hypotenuse\n\nSo, using Pythagorean theorem,\nS, the unknown side, can be computed as\nS^2=\n(b-(b-4)*cos(a))^2+\n((b-4)*sin(a))^2\n\n\nj
509283
This isn't the final answer, but it is a clue:\n\nhttp://www.thesheetmetalshop.com/pdf_manual/Problem1.pdf
531535
It seems that you have problems with moles and Avagadro's number. You will need to refer to your periodic table to get the molecular weights that your teacher wants you to use.\n\nSo get the molecular weights of the elements you are working with from your periodic table and substitute in the formula.\n\nI am using round numbers to show the substitution you should know how many decimal points you need to work to.\n\nHG atomic weight 200\nO atmic weight 16\n\nSubstitute in the formula to get the masses involved\n\n2 HgO(s) - 2 Hg(l) + O2(g) Hrxn = 181.6 kJ\n 2 (200g + 16g)   =  2(200g) + 2(16g) = 181.6 kj  \n\n432g oxide decomposes releasing 181.6Kj\n\nQuestion a)\nStraightforward ratio. X = the no of KJ required\n\n572g/432g = xkj/181.6kj\n\nYou'll need to calculate the molecular mass from you periodic table substitute it for the 432 and then refine the equation.\n\nQuestion b) Straightforward ratio. X = the no of KJ required\n\nYou'll need to calculate the molecular mass from you periodic table substitute it for the 432 and then refine the equation.\n\n289/181.6 = xg/432\n\nA mole is the amount of pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12 (i.e., 6.023 X 1023). This involves the acceptance of two dictates -- the scale of atomic masses and the magnitude of the gram. Both have been established by international agreement. Formerly, the connotation of "mole" was "gram molecular weight." Current usage tends to apply the term "mole" to an amount containing Avogadro's number of whatever units are being considered. Thus, it is possible to have a mole of atoms, ions, radicals, electrons, or quanta. This usage makes unnecessary such terms as "gram-atom," "gram-formula weight," etc.
279993
(NH4)2 SO4  + (FeSO4)2 + 2H2O____(\nNH4)2SO4Fe2(SO4)3.24H2O\n\nAmmonium ferric sulphate
198870
You must first determine the maximum amount of silver that can be formed from 12.7g of copper.  I don't have a periodic table in front of me, but you would determine how many moles of copper you have and use that to determine how many moles of silver could be formed in a perfect reaction.  example:  you find that 12.7g copper is 0.25mol and from that you know that it would form 0.50mol of silver because every mol of copper yields two moles of silver.  With the 0.50mol silver you multiply by the atomic weight of silver to get the maximum yield.  In order to get the percent yield you simply divide the actual yield by the max yield.\n\nNow I'm going to get the numbers for you :)\n\nAtomic weight of Cu is 63.54g/mol\nAtomic weight of Ag is 107.9g/mol\n\nSo... 12.7g Cu is 0.200mol which means the max yeild is 0.400mol of Ag (from the balanced equation 1mol Cu forms 2mol Ag.)  \n\nNext find the max yeild which is 0.400mol Ag times the atomic weight, 107.9g/mol to get 43.16g Ag.\n\nFinally, percent yeild is actual over max yield or 38.1/43.16 x 100% = 88.3% yield
192128
Never trust materials science majors ...\n\n1. NaCl (s) --> Na+ (aq) + Cl- (aq)\n2. Ca (s) + 2H2O (l) --> Ca(OH)2 (s) + H2 (g)\n3. Ca(OH)2 (s) --> H2O (g) + CaO (s)