Specific example: \n\nCH3CH(CH3)CH2CH(Cl)CH3\n\nOk, to me, I would name this molecule as 2-chloro-4-methylpentane because I'm thinking that halogens are higher in priority than methyls, but I'm being told it's actually 4-chloro-2-methylpentane.\nWhy is that?
138742
Please see my sources.  I couldn't have said it better.
89028
methyl group - the univalent radical CH3- derived from methane
84200
As the preceeding poster stated, "pentachloro" is not a chemical compound;  it could certainly be *part* of the name of a chemical compound, though, in which case it would simply mean that there are 5 (penta) chlorine atoms in the structure or a substructure of the molecule.  \n\nFor instance, pentachloroethane ( a toxic compopund) would have chemical formula C2 H Cl5.  Pentachlorobenzene would be C6 Cl5 H.  Pentachlorophenol (the nasty pesticide "lindane", which ) has formula C6 Cl5 OH, and so on.
175860
It's pentanal, because it's a terminal carbonyl. It's certainly not pentanol, because otherwise you'd have a carbon missing 2 bonds, or a carbon missing 1 bond and an oxygen missing 1.
379683
1.) Phosphorus trichloride is PCl3. Since it's created from the elements, the unbalanced equation is: P + Cl2 (g) -> PCl3\n\n2.) Carbon tetrachloride is CCl4 (tetra = four). For the equation, you need to take into account the hydrogen that's removed from the methane. CH4 (g) + Cl2 (g) -> CCl4 (l) + H2 (g)\n\nAfter showing you 1 and 2, you should be able to do 3, as long as you know that sulfide means Sulfur. Thus, Lead Sulfide is PbS.
521424
I believe it would be:\n\nNe\nCH3Br\nCH3CN\nCH3OH\n\nthe -OH is the strongest of these because of hydrogen bonding
5323
Since propane has three carbon atoms the isomers could be\n\nCl(CH2)-CH2-CH3\nor\nCH3-CH(Cl)-CH3\nThe isomerism in it is structural or position isomerism.\n\n\nThe reaction is as under\n             SN2  \nC3H9 + Cl2 --------------> monochloropropane\n\nThe mechanism is Nucleophilic Substitution. It is usually SN1 that is Unimolecular substitution.
362447
1. M1X1 (no subscripts), because alkali metals are +1 and halogens -1.\n2. M2X1 (subscript 2 for the metal), because group 6a elements are -2 in binary compounds\n3. MX2, because alkaline earths are +2.\n4. MX (subscripts each 1) because alkaline earths are +2 and group 6a elements -2.\n5. MX3, because group 3a metals are mostly +3. An important exception is thallium, which is either +3 or +1. So thallium chloride could be TlCl or TlCl3.
345138
Of the choices you listed above, then Alkane is the answer.  There are other compounds which fit this description, for example the cycloalkanes which is a sub-group of alkanes.\n\nThese are also called aliphatic compounds and saturated hydrocarbons.
353846
The reaction in question is:\n\nPCl5(g) <-> PCl3(g) + Cl2(g)\n\nfor which we can write an equilibrium constant, the value of which we are told is 11.5 (this really should have units of atm):\n\nK = 11.5 atm = (p_PCl3)*(p_Cl2)/(p_PCl5)\n\nwhere p_xxx stand for the partial pressure of the species "xxx" at equilibrium.\n\nThe molecular mass of PCl5 is 208.24 gm/mol, so 3.243 gm of PCl5 contains 3.243/208.24 = 1.557*10^-2 moles of this compound.\n\nWe will assume all the gases in this problem obey the ideal gas law.  Using this law, we can calculate what the pressure of 1.577*10^-2 moles of PCl5 would be if it were sealed in a  438 ml (= 0.438 liter) container and heated to 600K.\n\nWe have that pV = nRT\n\nwhere p is the pressure\nV is the volume\nn is the number of moles\nT is the absolute temperature\nR is the universal gas constant = 8.2057*10^-2 liter*atm/(mol*K)\n\np = nRT/V\n\nPlugging in the values appropriate to this problem, we have:\n\np = [1.577*10^-2 mol * 8.2057*10^-2 liter*atm/(mol*K) * 600 K]/(0.438 liter)\n\np = 1.75 atm (only report 3 significant figures because the volume was only given with this many).  This is the answer for part (a).\n\nFor the remaining parts of this problem, let's define "x" as the degree of dissociation of PCl5 at equilibrium.  x can range from 0 (no dissociation) to 1 (completely dissociated).\n\nx = 1 - (p_PCl5 at equilibrium)/(p_PCl5 before dissociation)\n\n(p_PCl5 at equilibrium) = (1-x)*(p_PCl5 before dissociation)\n\n\nWe know from the stoichiometry of the reaction, that each mole of PCl5 that dissociates produces one mole each of Cl2 and PCl3 gas.  Because we're assuming all the gases behave ideally, this means that the partial pressures of Cl2 and PCl3 will each increase by the amount that the partial pressure of PCl5 decreases.  We also know that the reaction produces Cl2 and PCl3 in a 1:1 molar ratio that their partial pressure must be equal at equilibrium.  This means that:\n\n(p_PCl5 before dissociation) = (p_PCl5 at equilibrium) + \n(p_PCl3 at equilibrium)\n\nand\n\n(p_PCl5 before dissociation) = (p_PCl5 at equilibrium) + \n(p_Cl2 at equilibrium)\n\nThese can be rewritten in terms of the degree of dissociation of PCl5 as:\n\n(p_PCl3 at equilibrium) = x*(p_PCl5 before dissociation)\nand\n(p_Cl2 at equilibrium) = x*(p_PCl5 before dissociation)\n\nFrom above, we also have that:\n\n(p_PCl5 at equilibrium) = (1-x)*(p_PCl5 before dissociation)\n\n\nGoing back to the equilibrium constant, we have that:\n\n11.5 atm = (p_PCl3)*(p_Cl2)/(p_PCl5), \n\nwhere all the pressures refer to the equilibrium partial pressures.  Plugging in the three expressions above for the partial pressures expressed in terms of the degree of dissociation and the initial pressure of PCl5, we have that:\n\n11.5 atm = x*(p_PCl5 before dissociation)* x*(p_PCl5 before dissociation)/(1-x)*(p_PCl5 before dissociation)\n\n11.5 atm = x^2 * (p_PCl5 before dissociation)/(1-x)\n\nWe already calculated that pPCl5 before dissociation would be 1.75 atm, so:\n\n\n11.5 = x^2 * 1.75/(1-x)\n\n11.5 - 11.5x = 1.75*x^2\n\n1.75*x^2 + 11.5*x - 11.5 = 0\n\nUsing the quadratic formula to solve for x, we find that:\n\nx = 0.882 or x = -7.453.  Only the positive root has a physical meaning, so x = 0.882 (i.e., the PCl5 is 88.2% dissociated.)  This is the answer to part (d).\n\nWe can now plug this value of x back into the three equations above to calculate the partial pressures of all the gas species:\n\n(p_PCl3 at equilibrium) = 0.882*1.75 atm\n(p_PCl3 at equilibrium) = 1.54 atm\n\n(p_Cl2 at equilibrium) = 0.882*1.75 atm\n(p_Cl2 at equilibrium) = 1.54 atm\n\n(p_PCl5 at equilibrium) = (1-0.882)*1.75 atm\n(p_PCl5 at equilibrium) = 0.207 atm\n\n\nThe total pressure is given by the sum of the partial pressures:\n\np_total = 1.54 atm + 1.54 atm + 0.207 atm = 3.29 atm.
253609
Hmm, kinda hard to help you here, since we can't see all your options for the first two.  The ellipses (. . .) indicate that you have more in your message than is being displayed.  Perhaps if you retyped them, it would be easier for us to help.\n\nHere's what I *can* help with:\n\nHalogens:Reactive as Noble Gas:STABLE\n(the halogens are very reactive gases, while the noble gases are extremely stable, almost to the point of being inert)\n\nOxygen:molecule as chloride:ION\nOxygen gas is a molecule, while chloride is the ionic form of chlorine\n\nRatio:subscript as oxidation number:CHARGE\nthe ratio of components in a mixture is denoted by the subscripts, as the oxidation number of an atom is denoted by its ionic charge
416052
1\na) H2C = CH - CH2 - CH2 - CH = CH2\n ( 1,6-hexadiene )\n\nb) H2FC - CHF - CFH2\n ( 1,2,3-Trifluoropropane )\n\nc) H3C - CH2 - CH2 - CH2OH\n  ( 1-Butanol )\n\nd) H3C - CH2 - CH2 - CHO\n  ( Butanal )\n\n2.\na) Usually FeBr3 is used in addition to Br2 to brominate benzene, but for this:\n\nProduct: Bromobenzene\n\nb) \nProduct: Ethyl butanoate\n\nc) Dunno this one.\n\nd) \nProduct: 1,2-Dichloropropane
165803
(a)2-Chloro-2-methylpentane\n(b)2,3-dihydroxy-2-methylpentane\n(c)Probably 1-Ethyl-2-methylpropanesulfonic acid also can be called 2-(i-propyl)propane sulfonic acid\n(d)2,3-Epoxy-2-methylpentane (you form an epoxide of the alkene)\n(e) You'll probably get a mixture of 2,3-dibromo-2-methylpentane and 3-bromo-2-methyl-pentan-2-ol\n(f) You'll get two products.  Isopropanol and n-propanol.
109723
just a comment: the correct spelling i think is 3-methyl-aminobenzene. i not so sure but it seems like a benzene ring with a CH3 and an NH2 spaced one C-H bond apart in the ring.\n\ntry this site http://chemexper.com \nthis might help.
1073
Carbon tetrachloride.  You could also call it tetrachloromethane.
380262
Perhaps there is also a H atom at C*\n\nThen final product is\n\n(CH3)2CH-CH=CH-CH2-CH(CH3)2\n\nA dehydration reaction occurs.
249264
a.  1-Bromobutane will undergo a faster reaction than 1-chlorobutane, as bromide is a better leaving group than chloride, due to being a weaker base.\nb.  1-Bromobutane will undergo a faster reaction than 2-bromobutane, as primary halides undergo SN2 reactions faster than secondary halides, due to steric considerations.