You have made an error in the final integration: \begin{align} t &= \int_{[\ce{A}]_0}^{[\ce{A}]_t}\frac{\mathrm d[\ce{A}]}{[\ce{A}]_0 - 2[\ce{A}]} \\ &= \frac{-1}{2} {\ln|{A_{0}-2[A]}|_{A_0}... View answer Accepted answer 3 votes Each oxygen atom is connected to a hydrogen atom(which develops -1 charge on oxygen and +1 on hydrogen) and to another \textbf{oxygen} atom which contributes no charge to both(0 and 0). Similar for ... View answer 2 votes My suggestion is to stop trying to predict the product by rules like “rich becomes richer” or “proton goes to carbon having greater number of hydrogens attached to it”, etc. for Markovnikov addition ... View answer 2 votes " compound produces a solid" indicates that a reaction has occured.(hence no 3° amine group(according to this)). Further this is insoluble in base so there is a 2° amine group. "This solid dissolves ... View answer 2 votes The point is, we are measuring \textbf{change} in temperature. Not the temperature itself. Celsius scale has 100 divisions from 0 °C to 100 °C. Kelvin scale also has 100 divisions from 273 K to 373 ... View answer 2 votes In such type of questions, we can't directly determine Oxidation state by the normal algebra method. First you have to draw the structure. 1.In a covalent bond, the more electronegative element gets ... View answer 1 votes\ce{H-C#C-H ->[Cu2Cl2][NH4Cl] H-C#C- Cu+ }$$This then attacks \ce{H-C#C-H} to give \ce{H-C#C-CH=CH2} Note: This occurs only for terminal alkynes and only in acidic medium(NH4Cl here). ... View answer 1 votes Dipole moment is what matters in substituted biphenyls. The dipole moment in the first case is from left ring to right ring due to -OH groups predominant +M effect to the ring.(the alkene being an ... View answer Accepted answer 1 votes For your reaction, \Delta n_g = 3-(1+1)=1 (assuming all A,B,C are gases). K_p = \frac{[C]^3}{[A][B]} (\frac{P_{total}}{\Sigma n})^{\Delta n_g}, where \Sigma n is the total number of moles of ... View answer 1 votes I will first try to answer your second question. There's absolutely no mistake you made! As you assumed first step is RDS(rate determining step), You would expect K=K_{f1}. Continuing from where ... View answer 1 votes In this case, the absolute configuration remains same (as R) for \mathrm{S_N1} and changes (into S) for \mathrm{S_N2}. This is because the priority order of non substituted groups (wrt \ce{C2}) ... View answer Accepted answer 1 votes -NHCOCH3 is a slightly activating group due to the lone pair of N which shows +M effect. Hence it is Ortho para directing. View answer 1 votes Maybe these links would help: US3046204A - Method for making diamond tools - Google Patents https://patents.google.com › patent The Basics: Electroplating & Vacuum Brazing - Diamond ... https://... View answer Accepted answer 0 votes To make your problem simpler(equivalent of your problem) just balance this reaction: \ce{I2 -> I^{-} + IO3^{-}} in basic medium. While balancing, the \ce{OH-} on the LHS correspond to the ... View answer 0 votes K_p=K_c(RT)^{\Delta n_g}(try to prove this yourself) where \Delta n_g is number of gaseous products -Number of gaseous reactants. In your question, \Delta n_g=0 so K_p=K_c=2.43 View answer Accepted answer 0 votes In aprotic or very small polar solvents like chlorobenzene, solvated ions cannot form so there is no solvent effect. It follows a 'similar' order as that in gaseous phase. Coming to the question, on ... View answer 0 votes Singlet carbene is sp2 hybridised whereas triplet carbene is sp Hybridised. More s character, more nuclear force, more Electro negativity so more acidic hydrogen and hence less basicity. So singlet ... View answer -1 votes In the first case hydrogenation occurs(anti addition) whereas in the second case, acid base neutralization occurs. So they aren't identical. According to Wikipedia,$$\ce{2 Na + 2 NH3 -> 2 NaNH2 + ...

Q is pentan-1-ol and not pentan-3-ol as on dehydration, of pental-1-ol, only one product (alkene) is formed. Whereas on dehydration of pentan-3-ol, we get two alkenes, as the alkene can show ...