what is the formula for aluminum and oxide?
453211
The formula for the compound of aluminium and oxygen is aluminium oxide.\n4 Al(s) + 3 O2(g) ===> 2 Al2O3(s)
927706
Al2(SO4)3 ==> Al2O3 + 3SO3
244959
ALUMINUM ACETATE BASIC \n\nFormula: Al2O(C2H3O2)4.H2O
852236
I would say Aluminum Chloride. (AlCl3)
908581
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. : )
375945
Nickel (II) oxide, sulphur dichloride, and chlorine\n\n2NiCl2 + S02 ---> SCl2 + 2NiO + Cl2
457637
Aluminum Nitrate, Al(NO3)3, since, in a neutralization you take off any acidic hydrogen's, and any hydroxyls, and you combine the remaining ions in a way that the product molecule has no net charge. So\n Al(OH)3=Al(+3) + 3OH(-) \n HNO3 = H(+) + NO3(-)\nThe H(+) ions and OH(-) ions combine to form water, and each Al(+3) ion combines with 3 NO3(-) ions to produce Al(NO3)3 \n\nFormic acid (CHOOH) and sodium hydroxide(NaOH) produce Sodium formate, CHOONa, for the same reason as above.\n\nNote: The other H in Formic Acid is non-acidic, and so is not replaced by Na
974325
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. :)
429906
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)
175516
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
783725
All right, you know what M means, right?  It's short for moles per liter.  You want to have 0.250 moles of Al2(SO4)3 in each liter of solution.  You're preparing 0.4 L of solution, so you need (0.4 * 0.250) = 0.1 moles of Al(SO4)3.\n\nNow the molar mass of each element is listed on most periodic tables.  That's how much one mole of that atom weighs.  A molecule weighs as much as all the atoms in it put together, so a mole of Al2(SO4)3 weighs as much as two moles of Al, 3 moles of S and 12 moles of O.  That's 2(26.982) + 3(32.06) + 12(15.999), according to my periodic table.  So if you add it all up, a mole of Al2(SO4)3 weighs 342.132 grams.  You only need a tenth of a mole, so you use 34.2132 grams.
1094317
aluminium because its an alkali earh metal and\nfluorine is a halogen
836848
first you'll start with Q=m*c*delta t to get the total energy of the aluminum. divide the grams of aluminum by its molar weight and then use Q=m*c*delta t where the Q's are the same, c for the water is 4.18 J/gC and then solve for the mass. good luck!
105926
Initial equation:\nAl(OH)3 + H2C03 --> Al2(CO3)3 + H2O\n\nOn the right side, there are 3 CO3 ions, so we must put a "3" in front of H2C03 on the left hand side of the equation. Similarly, there are 2 Al ions on the right hand side of the equation, so we must write "2" in front of Al(OH)3 on the left hand side of the equation.\n\n2Al(OH)3 + 3H2C03 --> Al2(CO3)3 + H20\n\nNow, we see that the number of "O"s from OH ions and the number of "H"s from the OH ions (in Al(OH)3) and H ions (in H2CO3) is NOT EQUAL to the H2O on the right.\n\nOK so number of "O"s from OH ions = 2 X 3 = 6\nNumber of "H"s from OH ions and from H2C03 = (2 X 3) + (3 X 2) = 12\n\nOne H2O molecule has 2 Hs and 1 O, so 6H20 molecules will have 12 Hs and 6 Os.\n\nThus your final equation will give\n\n2Al(OH)3 + 3H2CO3 --> Al2(C03)3 + 6H20\n\nI hope I made this clear enough. I suggest that you read the chapter again or ask your teacher about this.
890642
Al + Fe(NO3)3 ---> Al(NO3)3 + Fe.\n\n2K + 2H2O ---> 2KOH + H2\n\n2KOH + CuSO4 ---> Cu(OH)2 + K2SO4\n\n3NaOH + H3PO4 ---> 3H2O + Na3PO4\n\nBest?
707894
Cu 2+ ClO3 1-\n\nCu(ClO3)2
606075
Cu(s) + 4HNO3 (aq)--> Cu(NO3)2(aq) + 2H2O(l) +2NO2(g)