In the gas-phase, homolytic bond-dissociation rxns such as: HCl-->.H(g)+.Cl(g) are much less endo, and therefore much easier to achieve than the corresponding gas-phase heterolytic bond-dissociation (HCl(g)-->H^+^(g)+ Cl^-^(g))  Why so? (Hint: This is a "coulombic" issue)\n\nH^+^  means positive ion\nCl^-^ means negative ion
216332
The difference between \n\n1) HCl-->.H(g)+.Cl(g)   and\n\n2) HCl(g)-->H^+^(g)+ Cl^-^(g)\n\nis that in equation 2, you have two charged species.  We know that a positively charged species likes to interact with a negetively charged species (coulombic effect), thus making them much harder to dissociate.  This is what your hint was trying to address.  \n\nIn the gas phase, there also do not exist any solvents or counterions to stabalize the charged products in reaction 2.  The only reason that NaCl dissociates in water is due to the dipole moment of water which helps stables the Na^+^ abd CL^-^ ions.  In the gas phase, these don't exist so you would be hard pressed to acheive equation 2.
144252
Here is the answer.. Just remember periodi table...\nElements on left most side are electro positive and elements on right most are electro negative. (The last row is nobel gases..so leave it)\nNow Cl- belongs to right most side... its more electro-ve\nRb, Ca are on left side.... they are electro positive(They will give  electron with out much force)\nSi, S are in middle-- their properties are in between...\n\nThats it!
60342
The borohydride will dispense a Hydride (H minus) to the carbonyl site causing the following transformation: R-COOH ---> R-CH2O- \n\nAdding the HCl will protonate the oxygen leaving you with: R-CH2OH
415545
Not impossible but Hco3- is considered a very weak acid with resilience towards losing another Hydrogen Ion that's a good explanation for Hydrogen Carbonate formation instead of Carbonate
65765
Positive and Negative ends of a molecule depend upon the electornegativity of the atoms of the molecule. Lets take a simple example of water here. H2O. \n\nOxygen is water is bound to 2 Hydrogen atoms. Oxygen is highly electronegative. So, it pulls the electrons in the O - H bond towards itself partially. So, this creates a partial negative charge on it and also a partial positive charge on Hydrongen.\n\nElectronegativity actually means "the ability of an atom to pull a lone pair of electrons towards itself". Electronegativity depends on the number of electrons present in the outermost shell. \nEx: Sodium has 1 electron in the outermost shell. So, its tenedecy to release this electron and get to a stabler configuration is more than gaining electrons further. But, in case of Chlorine, there are 7 electrons. It needs just 1 more electron to attain stable configuration. So, Chlorine tends to pull electrons towards it and therefore Chlorine is highly electronegative. \n\nNow, looking at the molecules u have given...\nNitrogen is highly electronegative. It has 3 unpaired electrons in its outermost shell, but it wud still go ahead and grab a few more to get to a fully filled state. So, in NH3, N would be partially negative and H would be partially positive. \n\nSo, is the case in HCN too. Cyanide ion (CN-) is a highly electronegative ion and therefore acts as the negative end. \n\n\nHope this information helps.
424877
when bonding takes place by sharing of electrons then it is said to be covalent bond. now it can be possible in two ways.\n\nfor eg if two hydrogen are combining to form a bond then they both will share one electon each and the force applied from there nucleus will be same so the elecron pair will remain nearly equal with both the hydrogen.so you can say that it remains exactly at centre from both the nucleus\n\nbut if you consider HCl. here also both hydrogen and chlorine are going to share one elctron each when bond formation takes place. but as electronegativity of chlorine is quite high(ie its attraction from nucleus is quite high) compared to that of hydrogen so eventhough they both have shared one electron each but still electron pair will remain towards chlorine.\n\ni hope this will help you out
445535
(D) is the right answer because O is more electronegative than N and N is more electronegative than S. So H-O bond is more polar than H-N bond and H-N bond is more polar than H-S bond.
531787
HCl_aq + NaOH_aq -> NaCl_aq + H2O_l\n(H+)_aq + (Cl-)_aq + (Na+)_aq + (OH-)_aq -> (Na+)_aq + (Cl-)_aq + H2O_l\n(H+)_aq + (OH-)_aq -> H2O_l\n\nFor all the other equations, the (Cl-), (K+), (NO3-), (Na+) are spectator ions, and the n.i.e is the same:\n(H+)_aq + (OH-)_aq -> H2O_l
231813
Hydrogen peroxide give hydrolysis of complexes, since the hydrogens on water ligands are VERY acidic the ligands become hydroxides and the complex usually ends as a metal oxide. Ammonia acts as a lewis base with the complex accepting the electron density and most likely kicking off some chlorides. HCl just storms the complex with chloride ions. And The metal oxygen bonding is much stronger than the metal chloride bond (mainly due to differences in pi backbonding) so the chlorides shouldnt do much to the original complex. \n\n\nAs for the exact molecular formula of what should happen, try asking god. For the answer that your teacher wants try reading a book.
349315
This deals with Bronsted-Lowry (B-L) acids and bases, and conjugate-base pairs \n\nB-L acids: donate H+ ions: STRONG ACIDS dissociate into ions\nH20 + HCl --> Cl- + H+  (H+ is the same as hydronium [H3O+])\nso it can be written as H20 + HCl --> Cl- + H3O+ \nThis looks like it donates a H+ ion, but it dissociates because it is a STRONG acid.  \n\nwk acid: HA= wk acid; A- = conjugate base\nHA + H2O ---> H3O+ + A-\n\nB-L bases: accept H+ ions: STRONG bases dissociate\nNaOH + H2O --> H2O + Na+ + OH-\n\nwk base: B = wk base; HB = conjugate acid\nB- + H2O --> HB + OH-\n\nThat's all you need to know to write all eqns.
334041
HCrO4- ---------------> H+   +  (CrO4)2-\nThis is an equilibrium problem, assume that x moles of HCrO4- dissociate which means that x moles of H+ and   (CrO4)2- are produced.\n\nAt equilibrium, \n[HCrO4-] = 0.250M-x\n[H+] = x\n[(CrO4)2-] = x\n\nKa = [H+] * [(CrO4)2-]/ [HCrO4-]\n3.0*10^-7 = x*x/(0.250-x)\n3.0*10^-7 = x^2/(0.250-x)\nFor these kind of problems you have to assume that\n0.250-x = 0.250\nTherefore, \n3.0*10^-7 = x^2/0.250\nx^2 = 7.5*10^(-8)\nx = 2.74*10^(-4)\n\n[H+] = [(CrO4)2-] = x = 2.74*10^(-4)\n\nPercent ionization = x/[HCrO4-] * 100\nPercent ionization = 2.74*10^(-4)/0.250 * 100 = 0.1095%\nor approximately 0.11%
457164
I guess that you know the simplified structure of amino acids: they contain a  carboxyl group, an amino group and a R group.\n\nProton of the carboxyl group tend to be transferred into amino group which leads into the formation of a "salt" and over 90% of the amino acids are in the form of this peculiar ion, having both positive (NH3+) and negative (COO-) charge. This results an ion which is not electrically charged towards its environment.\n\nReal ionisation of aa happens when OH- or H+ acts with it, resulting R-CHNH2COO- or R-CHNH3COOH+, respectively. aa's are thus amphoteric and react with hydroxyls or protons only in the -CHNH2COOH group. This causes no modifications of the R- group. \n\nIf the subject of your study are proteins, other things can happen. But observing only the chemical reactions of amino acids, no modification of R- group happens, I guess.\n\nI
4420
SO3+2 clearly is an error. It would demand a +8 charge of the sulfur atom (or a stable positive sulfur ion with an isolated peroxide ligand), which is impossible.\n\nA superscript "2-" would indicate a sulfite ion, whereas SO3 is sulfur trioxide, a neutral molecule forming a gas at room temperature.
322655
it must be polar because the structure is tetrahedral with a lone pair at one corner of the tetrahedron and Cl atoms at the  rest three corners. so the individual As-Cl bond moments are not cancelled bcoz of the asymmetry of the molecule and the net polarity cant be zero.
340858
This is kinda hard to draw on here...\n......_\nK--Br|\n......_\n\nThe lines around Br represent 2 electrons, and the -- represents a bond...ignore the periods
542695
coefficients denote how many ions you have, while subscripts denote how many of that element is in the molecule.  here, it should be a subscript, because you have 2 Cl-1, not one molecule of Cl(2)-1.
68419
When a salt derived from a strong acid and a weak base dissolves in water, the solution becomes acidic. For example, consider the process\n \nNH4Cl(s) ----H2O ---> NH4+(aq) + Cl-(aq)\n\nThe Cl- ion has no affinity for H+. The ammonium ion NH4+ is  a weak conjugate acid of the base NH4+(aq) and ionizes as follows \n\nNH4+(aq) + H2O(l) <=> NH3(aq) + H3O+(aq)\n\nor simply  \n\nNH4+(aq) <=> NH3(aq)  + H+(aq)\n\nThe equilibrium constant ( or ionization constant) for this process is given by \n\n\n\nKa = Kw / kb  =   1.0*10^-14 / 1.8*10^-5 = 5.6*10^-10\n\n\n      x^2 / .10 =  5.6*10^-10\n\n           x^2  =   5.6*10^-11\n\n             x  =    7.5*10^-6 M  =  [H+]\n\npH  =  -log [H+]\n    \n    =  5.12\n\n     my answer is  C , NH4Cl