well this is what i have to do basically put on top of the compound or element what the oxidation # is. Ex: CL2 has a zero oxidation number na+ has a positive one axidation #.Some of you might be pros at this thats why im askin cause im not..... H3PO4, H2O2, Si(OH)4, KH2PO4, Fe(HCO3)2 , Cu2CrO4, KMnO4 , NO3, HNO3, BeH2, yeah so i need the oxidation number for each, if you guys can help with more email me ill appreciate it. thankss
128196
in H3P04 u need the oxidation number of what? all these elements have different oxidation numbers in different compounds. the procedure for u to find out the oxidation number is .... if u want the oxidation number of P then write there as x, for H as u know it is +1, for O it is -2.\n then, (+1*3)+(x*1)+(-2*4)=0; 3+x-8=0; x=5 thus, the oxidation number of P in H3P04 is 5.\nin the same way, in H2O2; O is -1\nSi(OH)4 ; Si is x+(4*-2)+4=0; x=4\nin KH2PO4, P is  +1+(+1*2)+x+(-2*4)=0; x=5\nin Fe(HCO3)2 if Fe is +2, H is +1, O is -2 then C is \n+2+2+2x-12=0; x=4\nCu2CrO4, Cu is +2, O is -2 and Cr is 4+x-8=0; x=4\nKMnO4, K is +2, O is -2 and Mn is +2+x-8=0 x=6\nN03, Ois -2, x+(-2*3)=0, x=6\nin HN03, H is +1, O is -2,then N is +1+x-6=0, x=5\nBeH2, H is +2\n\nany doubts u can ask again.
287607
The H's are +1, and the O's are -2, and the whole thing is 0, so...\n4(+1) + 2x + 7(-2) = 0\nx=5, so the oxidation number of phosphorus is +5.
69662
I not sure what style you learned to use to balance chemical equation but here is one way to do it. \nWe have four elements to balance (Al,H,Cl,O)\nLeave the compounds with aluminum with one of each. (Al done)\nThe simplest compound Al is AlCl3. To match the Cl on both sides put a 3 in front on HCl (Cl done)\nIf you add up the H (hydrogen) on the left side of the equation you'll see 6 of them. Add a 3 in front of the H20 to balance that. (H done)\nDo a quick check and you'll see there are 3 oxygen on each side (0 oxygen)\nAll the elements have been balanced, yielding a final equation of:\n3HCl + Al(OH)3 ---> 3H20 + AlCl3
405123
The oxidation number of sulfur in SO3 is +6. The oxidation number of oxygen is -2. So 6 + 3(-2) = 0. The oxidation number of oxygen is always -2, except in peroxides like H2O2, when it is -1. The oxidation number of hydrogen is always +1, except in hydrides like NaH, when it is -1. After that, everything adds up.\n\nDon't worry too much about oxidation numbers. They have no physical meaning. They are a system of bookkeeping to balance oxidation-reduction equations.
490423
Ba(NO3)2 and sodium phosphate\n\n1.  Ba(NO3)2 + Na3PO4 --> NaNO3 + Ba3(PO4)2\n\n2.  Ba(+2) + NO3(-1) + Na(+1) + PO4(-3) --> Na(+1) + NO3(-1) + Ba(+2) + PO4(-3)\n\n3.  No reaction (All spectator ions cancel out)\n\nBa(NO3)2 and sodium carbonate\n\n1.  Ba(NO3)2 + Na2CO3 --> BaCO3 + NaNO3\n\n2.  Ba(+2) + NO3(-1) + Na(+1) + CO3(-2) --> BaCO3 + Na(+1) + NO3(-1)\n\n3.  Forms BaCO3 (Barium Carbonate)\n\n\nBa(NO3)2 and pottasium iodide\n\n1.  Ba(NO3)2 + KI --> BaI2 + KNO3 \n\n2.  Ba(+2) + NO3(-1) + K(+1) + I(-1) --> Ba(+2) + I(-1) + K(+1) + NO3(-1)\n\n3.  No reaction (All spectator ions cancel out)\n\nBa(NO3)2 and sodium sulfate\n\n1.  Ba(NO3)2 + Na2SO4 --> NaNO3 + BaSO4\n\n2.  Ba(+2) + NO3(-1) + Na(+1) + SO4(-2) --> Na(+1) + NO3(-1) + BaSO4\n\n3.  Forms BaSO4 (Barium Sulfate)\n\nHopes this helps you!
173595
3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n                                        +11H2O\n\nFirst you must calculate the (average)oxidation number of central atom of main compounds in reactants.\nEx;Cr oxidation no=+6 in K2Cr2O7\n   C  oxidation no=-2 in C2H5OH\nThen in the products.\nEx;Cr oxidation no=+3 in Cr2(SO4)3\n   C  oxidation no=0 in CH3COOH\n\nThen you must evalute how many electrons must be transferred.\nEx;Oxidation no of Cr is changing from +6 to +3 \n   So it is reduced.\n   SO each atom of Cr must receive 3 electrons.\n   So totally 6 electrons must be received by K2Cr2O7\n   \n   Oxidation no of C is changing from -2 to 0\n   SO each atom of C must donate 2 electrons.\n   So totally 4 electrons must be donated by C2H5OH\n\nThen you must write now a explantion reaction.\nEx;you must write oppsite (simple)propotion of electrons in front of each compound\n  3C2H5OH + 2K2Cr2O7\n \nThen you can calculate how many Cr,C(,K) atoms should be there\n So, 3C2H5OH + 2K2Cr2O7=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n\nThen you can calculate the medium(SO42-) should be there\n i think 8 SO4 should be there\n\nSo,3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n\nFinally you must equilize H,O atoms(balanced by water)\nSo,3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n                                        +11H2O\n\nSecond one ................\nC2H5OH+4I2+6NaOH=CHI3+HCOONa\n                  +5NaI+5H2O
395740
F.  Oxygen's is -2\n\nAtoms are assigned numbers, called valence numbers, oxidation numbers, or oxidation states, which range in value from -4 through 0 to +7 and describe the combining behavior of the atoms in chemical reactions, particularly oxidation-reduction reactions.  Metals, which commonly donate electrons and form compounds in which they exist in the positive, or cationic, state, are assigned positive oxidation numbers. For a metal such as zinc, which donates two electrons to achieve a stable electron configuration, the oxidation number is +2. Nonmetals, which commonly accept electrons and in compounds exist in the negative, or anionic, state, are assigned negative oxidation numbers. The oxidation number is -1 for chlorine and the other halogens, which accept one electron to complete their valence shell.  The oxidation (valence) number is -2 for oxygen.\n\nYou can figure the oxidation (valence) number by looking at the column of the periodic table.
498478
okay. i just took the test on this and passed with a B. so i promise i know what im doing.\n\nK, look at your elements. then look them up on the periodic table. lets pick, oh say the compout Cl(subscript)2O. that is 2 chlorines, and 1 oxygen. well, chlorine is in group 17, which means that each have 7 valence electrons, and oxygen is in group 16, which means that it has 6 valence electrons. following me so far? okay, take those valence electrons, and add them up. that equals 20 valence electrons. now, position the lewis dot diagram so that the oxygen is between the two chlorines. now, distribute the 20 electrons amongst the three letters, so that each has eight. now each letter should be surrounded by exactly eight dots. keep in mind that it should look like this -- :Cl:O:Cl:, but dots on the tops and bottoms. that is the lewis dot. luckily, this one did NOT have double or triple bondage. but in the event of double bondage, it might look like this -- :Cl::O:Cl, without dots all the way around each letter.\n\nand the VSEPR theoryis quite simple. it is just talking about the potential shapes that chemicals can make. for example, there is linear, or trigonal planar, or tetrahedral. this is vital. i promise that, if you contact me, i can give you a simple cheat sheet (not really for cheating, but it helps your memorization) to remember VSEPR. it gives you the shapes, lewis dot examples for the shapes, and angle measurements.\n\nhope this helps!\n~ally
351316
It is neutral too.  In the NaClO case the oxidation number of Cl is different from the oxidation number of Cl in NaCl\n\nNaClO\nNa is +1, O is -2, so Cl must be +1\n\nNaCl\nNa is +1 so Cl must be -1.\n\nGo, for example, to the following sites for some good rules to find out the oxidation numbers:\n\nhttp://chemistry.boisestate.edu/rbanks/inorganic/oxidation-reduction/redox4.htm
488559
The overall charge is the number that is at the top right-hand side, in this case it is -1.\nTo find the oxidation number of Cl... Let's assume it is x. Then the sum of all the oxidation numbers of all the atoms in this group must be equal to the overall charge. Oxygen has usually oxidation number -2 , so we must have\nx+4*(-2)= -1 => x-8= -1 => x=+7
497170
C, Al(OH)3...\nWould you like an explanation? Okay, I'm supposed to be learning about this in Chemistry right now, so, I think this is how it works...\n\nIf you study the periodic table of elements, then you will find the number of valence (or outer) electrons each element has. Al has 3, O has 6, and H has 1. Before we start, it is important to know that a filled orbital has 8 valence electrons... okay? \nO and H combine, so they have a total of 7 valence electrons. You can imagine this by thinking of the letters OH surrounded by 7 dots in a circle, 2 pairs of dots on each side and 1 alone (a filled orbital with have four pairs opposite each other). The lone pair needs an electron, so it can borrow one from aluminum, which has 3 electrons. Aluminum loses its electron, so now it has 2, and the OH is now full, because it has 8 electrons. However, the aluminum has to get rid of its other two electrons, so another OH is brought in. The aluminum gives up an electron and the second OH is now full. Finally, the last OH pair is brought in and the aluminum gives away its last electron. So now, it is empty and the OH's are full. I guess they're balanced. Anyway, you are now left with 1 Al and 3 OH's, making Al(OH)3. Do you understand now? \n\nAl (3), and OH (7) (because O=6 & H=1, which equals 7)...\nAl - 1 = (2), so now OH = (8) (now it's full)\nAl - 1 = (1), so a 2nd OH (7) is brought in, making it OH (8)\nAl - 1 = (0), so a 3rd OH (7) is brought in, making it OH (8)\n\nIt took 1 Al and 3 OH's, so the new compound is Al(OH)3\nHope this helps!!!\nP.S. sorry for any confusion earlier when I tried to "draw" them. The computer changed the spacing so all you could see was a mess of stars and stuff. : )
44600
Haven't had chem in a while but maybe it would help to write out all the individual symbols.  Your second example would look like\n\nHHHPOOOO + MgOOHH  = MgMgMgPPOOOOOOOO + HHO\n\nthen you can see that the left side needs three Mg so you'd need 3 of the Mg(OH)2 to balance.  The left side also needs two P so there needs to be two H3PO4 there.  This makes a total of 12 H on the left, so to balance that you need 6 H2O on the right.  This also gives 14 O on the right to balance the 14 O you'd now have on the left.  So\n2 H3PO4 + 3 Mg(OH)2 = Mg3(PO4)2 + 6 H2O
546211
You should probably need your periodic table.\nGenerally chemical formulas are for ionic compounds, hence you need to know the charges of the ions formed.\nAll Group I metals form ions of +1 charge, Group II metals forms +2 charged ions, Group III (like Al) forms +3 charge, Group IV prefers to share, seldom will they form ions, hence we shall keep them out of the picture.\nGroup V non-metals form -3 charge, Group VI -2 and Group VII -1. Group O elements have full outermost shell, hence they seldom participate in bonding.\nSo if you are looking at magnesium oxide, the ionic compound is formed between magnesium ion (Group II, hence Mg2+) and the oxide ion (from oxygen, Group VI, hence O2-).\nYou will realise that to balance the charges, you need exactly 1 Mg2+ and 1 O2-. Hence the chemical formula of magnesium oxide is MgO.\nAnother example, for sodium chloride. Sodium ion (Na+) and chloride ion (Cl-). Hence 1 of each is needed to balance the charges and the formula is NaCl.\nAnother example, lead (IV) oxide. Lead is a transition metal and transition metals form ions of variable valencies. Lead (IV) means the charge of the lead ion here is Pb4+. Thus you need 1 Pb4+ and 2 O2- (so that the charges are cancelled) for lead (IV) oxide, PbO2.\nYou have also polyatomic ions, like hydroxide (OH-), sulphate [(SO4)2-], nitrate [(NO3)-] and carbonate [(CO3)2-]. Similarly, ionic compounds formed will require the balance of charges.\nExample, with sodium sulphate.\nYou have Na+ and sulphate [(SO4)2-].\nYou will need 2 sodium ions to balance the -2 charge of the sulphate ion. Hence the chemical formula is Na2SO4.\nHope this helps. :)
509645
phosphate\nsulfate\nnitrate\nperchlorate\n\nThe oxidation state of the anion will not change.  These anions simply have a proton (or a couple protons) as the cation instead of a salt.  The name of the acid directly reflects the anion's name.  All the acids you listed had an -ic suffix which implies a fully oxidized state.  The per- prefix on perchloric sugggest it is one step beyond fully oxidized.  \n\nFor example, another oxoacid is sulfurous acid.  The -ous suffix informs us that the oxidation state of the non-oxygen element is lower than those with an -ic ending
362091
start with the O... they are always -2\nsince there are 2, and the compound must balance...\nPb must be +4\n\nso you get +4 -2
28552
Wow, I hope these are right.  I wish i had a solubility chart or something (Which you probably have)  \n\nCuSO4(aq) + Na2S(aq) ---> CuS (s) + NaSO4 (aq)\nNet ionic - Cu2+ (aq) + S2- (aq)  ---> CuS (s)\n\nNiCl2 + 2KNO3 --> Ni(NO3)2 (aq) + 2KCl (aq)\nnet ionic - none (if NiCl2 isn't aq, then this is wrong)\n\nCa(OH)2 + 2HC2H3O2 ---> Ca(C2H3O2)2 (aq) + 2H2O (l) \nnet ionic - H+ (aq) + OH-(aq) ---> H2O (l)\nactually, since the acetic acid is a weak acid, this might be a little different.  If you learned it this way, then your net ionic is:\nHC2H3O2 (aq) + OH- (aq) ---> H2O (l) + C2H3O2- (aq)\n\nHope this helped
524127
a) [O 2-]\nb) [ I - ]\nc) [Na+]\nd) [Al 3+]\ne) [Ni 2+]\nf) [Mg 2+]