1. What are hydrides? \n\n2. Write the formulas for the hydrides of Na, Ca, As, and S.
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Hydrides are compounds that have the anion H-.\n\nTherefore, the formulas are NaH and CaH2.
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NaOH + NaH2PO4 ===> Na2HPO4 + H2O\n\nH2S + 2KOH ===> K2S + 2H2O\n\n(NH4)2SO4 + 2KOH ===> 2NH3 + K2SO4 + 2H2O\n\nSO2 + Ca(OH)2 ===> CaSO3 + H2O\n\nIn the above equation, further oxidation in air could occur by\n\n2CaSO3 + O2 ===> 2CaSO4
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1. electron\n2. ionic\n3. molecules of compounds\n4. decomposition (not sure)
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a) Ag2S\nb) SnCl4\nc) Na3N\nd) SrI2\n\nDo us a favor and pay attention during class. That way you'd know how to do these.
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1.) hydrogen and bromine forming hydrogen bromide (chemical symbols: H (hydrogen) Br (Bromine)\n\nH2 + Br2 → 2HBr\n\n2.) fluorine and argon forming argon trifluoride (Fluorine- F and Argon- Ar)\n\n3 F2 + 2Ar → 2 ArF3\n\n3.) sulpher and oxygen forming sulphur dioxide (Sulphur- S and Oxygen- O)\nS + O2 → SO2\n\n4.) calcium and chlorine forming calcium chloride (Calcium- Ca Chlorine- Cl)\nCa + Cl2 → CaCl2
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There are a few methods, most of them require memorization, or a good chart.  For example, ionic compounds get named by adding the names of their ions, but molecular compounds get named by each individual element and a prefix of how many there are.  Then you have acids...  \n\nhere are some examples:\n\nNaCl - Sodium Chloride\nNa2CO3 - Soduum Carbonate\nNa3PO4 - Sodium Phosphate\n\nNotice that sodium is always sodium, not mono, di or tri.\n\nMolecular compounds are different, here are some examples:\n\nNF3- nitrogen trifluoride   \nNO - nitrogen monoxide OR nitric oxide (Ionic)\nNO2 - nitrogen dioxide  \nN2O - dinitrogen monoxide (laughing gas)\n\nI wont even go into acids, or Hydrocarbons\n\nI hope that gets you started anyway
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1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 9.270 g x (12 / 44) = 2.528 g\nhydrogen: 3.795 g x (2 / 18) = 0.422 g\noxygen: 3.797 - 2.528 - .422 = 0.850 g\n\n2) convert to moles.\ncarbon: 2.528 / 12 = 0.211 mol\nhydrogen: 0.422 / 1 = 0.422 mol\noxygen: 0.850 / 16 = 0.0531 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.211 / 0.0531 = 4\nhydrogen: 0.422  / 0.0531 = 8\noxygen: 0.0531 / 0.0531 = 1\n\nThe empirical formula of the substance is (C4H8O)n\nThe atomic masses add up to 72.1, which is also the molar mass of the compound. Therefore n=1 and we find C4H8O
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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. :)
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HCl + NH4OH → H2O + NH4Cl.\nThere would be heat given off, so you could verify that a reaction took place.\n\n2HCl + Ca(OH)2 → 2H2O + CaCl2\n\nedit: The steps for balanacing the reaction:\n1)We know the salt produced is CaCl2, so an acid with Cl is HCl and a base with Ca is Ca(OH)2. \n2) write the skeleton reaction:\nHCl + Ca(OH)2 → H2O + CaCl2\n3) there are 2 Cl's on the right side, but only 1 on the left. To balance this, place a 2 in front of HCl\n2HCl + Ca(OH)2 → H2O + CaCl2\n4) Now there are 4 H on the left and only 2 on the right, so, place a 2 in front of H2O.\n2HCl + Ca(OH)2 → 2H2O + CaCl2\n5) check with the other elements, each side has one Ca, and 2 O. It all works out ☺
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AB + CD ---> AD + CB\nlogic doesn't work here as Si doesn't react/bond with Ca or Phosphorus.and Ca(PO) should be Ca3(PO4)2\nand SiO should be in the form of SiO2. \nso unless you want hypothetical compounds SiO2 won't form with either of the two other reactants.\nbased on the forumlae from an SI data book, the above reaction would not balance.\ngood luck solving this one. :-)
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H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
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H2, or hydrogen gas.\n\nFor example, take magnesium metal and hydrochloric acid.\n\nMg + 2HCl -> H2 + MgCl2\n\nThis is a displacement reaction, in which a free element (Mg) breaks the bond between a compound (HCl) and then bonds to the ion with the opposite charge (Cl-) forming a new compound (MgCl2).  The ion left over (H+) loses its charge and becomes a pure element (H2).
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1.) do your own homework\n2.) this is the computer forum, not high-school chemistry.  Next time post to the right furom.
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These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
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2. 2 Na + 2 H2O --> H2 + 2 NaOH\n3. 2 HNO3 + Ca(OH)2 --> Ca(NO3)2 + H2O\n4. C5H12 +  8 O2 --> 6 H2O + 5 CO2
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1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 7.623 g x (12 / 44) = 2.079 g\nhydrogen: 4.161 g x (2 / 18) = 0.462 g\noxygen: 3.470 - 2.079 - .462 = 0.929 g\n\n2) convert to moles.\ncarbon: 2.079 / 12 = 0.173 mol\nhydrogen: 0.462 / 1 = 0.462 mol\noxygen: 0.929 / 16 = 0.0581 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.173 / 0.0581 = 3\nhydrogen: 0.462  / 0.0581 = 8\noxygen: 0.0581 / 0.0581 = 1\n\nThe empirical formula of the substance is (C3H8O)n\nThe atomic masses add up to 60.1, which is also the molar mass of the compound. Therefore n=1 and we find C3H8O
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1) Zinc Chloride\n2) Copper (II) Sulphate\n3) there is no such thing as NH4OH, but oh well..\n    Ammonium Hydroxide\n4) Iron (III) Sulphide\n5) Aluminium Phosphate\n6) Lead (II) Nitrate\n7) Sulphur hexafluoride\n8) Sodium Sulphate\n9) Mercury (II) oxide\n10) Potassium Clorate (V)\n11) Tin (II) Choride\n12) Iron (II) hydroxide\n13) Mercury (II) Sulphate\n14) Carbon Dioxide