how would hydroquinone react with silver bromide?
605802
FOR PHOTOGRAPHY- \n\nBleaching is an important phase in the recording of holograms on silver-halide materials, ensuring a high diffraction efficiency, and so important for holographic images.\n\nBleaching can be regarded as the reversed process of development. During the developing process, a silver ion is reduced to free silver and the developed film appears rather dark, whereas during the bleaching process, metallic silver is oxidised to silver ion. It is interesting to consider the chemical equation for development being reversible. \n\nFor example, the action of hydroquinone on silver bromide is\n\nC6H4(OH)2 + 2AgBr + 2OH C6H4O2 +2Ag +2Br +2HOH\n\nThe reversed action is the formation of hydroquinone and silver bromide caused by the oxidising agent quinone acting on metallic silver. In practice, the reversal (oxidation) can only take place in an acid environment. The development (reduction) only takes place in an alkaline solution.\n\nThere are various bleach solutions formulated on the basis of the different oxidisers.\n\nIn the mixing of the bleacher deionized water is used. This is very useful since ordinary drinking water contains salts and fluorine.
1121888
Well you got C3H3Br + CH3S,  I'd guess something like ethane and bromosulfide.. C2H6 + BrS..
223672
Could the bromine react with the iron to form FeBr3?  If I remember right, the Febr3 should make it easier to react with toluene (hence the iron catalyst).  I'm not sure about the math, but it seems like toluene was the excess reagent so FeBr3 shouldnt be leftover...\n\nDo you know if either of the distillation products are a bromotoluene?  From what you said, dibromotoluene could have come out.  Also, the methyl group has activating properties, so there will likely be more than one substitution, possibly even 3
725081
The answer is Cu + Cu(NO3)2 -> Cu + Cu(NO3)2\n\nThere is no reaction.
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
726370
There should be no reaction.  Provided there is enough water to dissolve the CuCl, the Compounds would separate into their Ion counterparts.\n\nSo, youd end up with Cu+, K+, and Cl- in solution
630717
All of the answers regarding the lack of oxygen to react with the silver are true, but another factor is that metals oxidize quicker when in the presence of an ionic solution, and the air on Mt. Everest has very few ions dissolved in it. So, then not only is there less reagents for the process of oxidizing silver, but there is also a lack of catalyst to speed the reaction up.
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You will get Silver Chromate as the precipitate.  It is red in color.\n2AgNO3(aqueous) + K2CrO4(aqueous) -> Ag2CrO4(solid) + 2KNO3(aqueous)
326922
Another easy reaction:\n\nMg(s) + HNO3 ===========> Mg(NO3)2 + H2 (g)\n\nMagnesium reacts rapidly with nitric acid to make magnesium nitrate and hydrogen gas
47432
3AgC2H3O2 + Na3PO4 -----> Ag3PO4 + 3NaC2H3O2
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
1018313
Silver is being oxidized, nitrogen from the nitrate is being reduced.\n\nOxidation half reaction:\nAg --> Ag+  +  e\n\nReduction half reaction:\nNO3- +  4H+ + 3e  -->  NO + 2H2O\n\nNet balanced and reduced equation:\n3Ag +  NO3- +  4H+ --> 3Ag+ + NO + 2H2O
215160
http://www-ec.njit.edu/~skawinsk/chem243/chap11/chap11.htm\nthis may help–near the bottom
138848
Here is one example ,\n\nAgNO3(aq) + NaCl(aq)-> AgCl(s)white precipitate + NaNO3(aq)
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easy!\n\nadd silver nitrate\n\nit will make silver bromide precipitate.\n\nthis is pale cream in colour.\n\nfurther confirmation is done by adding conc ammonia to it, and if it dissolves, it proves that the original solution had bromide ions.
471526
I suspect that the product is AlSO4.\n\nI'm thinking that the Al swaps with the Cu.  That would leave a free Cu+ which might displace the H from H20 and form either CuOH or CuO, either one of which would probably produce H2.  \n\nSo, I guess you would first get CuOH and AlOH, as intermediate reactions.\n\nI think the equation is:\n\n                          NaCl (goes over the arrow)\n6CuSO4 + 4Al + 12H2O ---------------> 2Al2(SO4)3 + 6Cu(OH)2 + 6H2
713038
the other product is zinc chloride.\nand the reaction is single replacement.