bromate nonahydrate formula?
87280
MBrO3-9H2O where M can be Na,K,Li, other +1 cations\n\nIf M is other than these you have to adjust subscripts
927283
NaBH4 + 2H2O → NaBO2 + 4H2\n\nthe moles of NaBH4 3*10^-3 * 1 = 3*10^-3 \nthe moles of H2 : 4*3*10^-3 = 12*10^-3\nthe volume of H2 : 12*10^-3 * 22.4 = 0.2688L\n\n2HCl + Zn => ZnCl2 + H2\n\nthe moles of Zinc : 1/65 moles\nthe moles of H2 : 1/65 moles\nthe volume of H2 : 1/65 * 22.4 = 0.3446 L
564798
Ba(NO3)2 + Na2CO3-->BaCO3 +2NaNO3\n\nto do a double displacement follow this formula\n\nAB + CD --> AD + CB
474531
CH3COOH + NAOH =.CH3COONA +H2O
574078
This might help :)\nhttp://www.chemicalforums.com/index.php?board=8;action=display;threadid=3263
889156
The answer is BrF3.\n\nand the balance equation is \n\nBr2  +  3F2  --->  2BrF3\n\nhow? get the moles of each using this formula:\nn = PV/RT\n\nyou'll get:\nmol Br2 = 0.041mol\nmol F2 = 0.123 mol\nmol pdt = 0.082 mol\n\nbase from this, you'll found that ratio of Br:F is 1:3. Hence, the product is BrF3. you'll also found that ratio of Br2:F2:pdt is 1:3:2 which also agrees with our balance equation above.
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
606155
NaOH + HCl ---> NaCl + H20
921837
NaHCO3 --> Na+ + HCO3-\nCa(NO3)2 --> Ca(2+) + 2NO3-\nAl2(CO3)3 --> 2Al(+3) + 3CO3 (-2)
980626
ChemDraw from CambridgeSoft Corporation may be help you. Look at www.cambridgesoft.com
130958
The equation is correct, if I perceive it right. You have 3 ethoxy groups on the boron atom, which does correspond to that formula. It looks like what might happen if you reacted borane (BH3), possibly tethered in a THF complex, with ethanol (EtOH) to get the product and H2 gas, but a lot of heat too. Such a substituted Borane would be rather reactive, as a Lewis acid.\n\nB(OC2H5)3 + 6O2 ---> H3B03 + 6CO2 + 6H2\n\nis my best guess
771931
Starting material:\nCH3CH2CHBrCH2Br ?\n\nProduct wanted:\nCH3CH2CH2CH=CHCH3 ?  Cis or Trans?\n\nIf the starting material is 1,2 Dibromobutane, then treating it with base (eg. NaNH2) will give:\n\n CH3CH2CHBr–CH2Br   +   base   ——>  CH3CH2CH=CHBr   +   base   ——>  CH3CH2C≡CH\n\nIf this is done with excess NaNH2 in liquid ammonia or dry diethyl ether, then the sodium salt will be produced:\n\n(CH3CH2C≡C)- Na+\n\n(CH3CH2C≡C)- Na+ + BrCH2CH3  ——>  CH3CH2C≡CCH2CH3 \n\nCH3CH2C≡CCH2CH3 when heated gives the more stable compound CH3CH2CH2C≡CCH3  (this is 2-hexyne).  The path will be different from here depending on if you want a cis or trans product.  A controlled catalytic addition (Lindlar catalyst) of H2 will give the Cis form.  Na in liquid ammonia will give the Trans.\n\nThere might be an easier way, but this path is a classic.
777907
Ninhydrin(Triketohydrindane hydrate)----C9H6O4\nLugol(Iodine Potassium-Iodide)-------  KI3 \nBiuret(Allophanamide)-----NH2CONHCONH2\n\nI cant find a chemical formula for benedict. But i found a equation.\n\nBenedict's reagent (also called Benedict's solution or Benedict's test) is a reagent used as a test for the presence of monosaccharides (such as glucose and fructose); it is also used for some disaccharides like maltose, or more generally for the presence of aldehydes, in a solution.\n\nBenedict's reagent contains blue copper(II) sulphate (CuSO4)· 5H2O which is reduced to red copper(I) oxide by aldehydes, thus oxidizing the aldehydes to carboxylic acids. The copper oxide is insoluble in water and so precipitates. The color of the final solution ranges from green to brick red depending on how many of the copper(II) ions are present. Benedict's test cannot be used to detect the presence of aromatic aldehydes.
921488
2ClO4(-) --->  2ClO4(-)\n2H(+) + 2OH(-) ---> 2H2O\n\nwhere sign in brackets are superscripts.
556705
Ba(OH)2, MgBr2, CH3COOH, NH4OH\n\nFirst two are strong electrolytes, second two are weak (do not completely dissociate).
233816
None.  The sell chemicals with alkyl groups.  In CH3Cl, CH3 is the alkyl.  They don't exist on their own.  Start with\n\nhttp://www.sigmaaldrich.com/Local/SA_Splash.html
921966
Br2 is extremely soluble in CCl4 - much more soluble than in water.\nLook up "partition".