9

It is rather: $$\ce{2 Al + 6 H2O + 2 OH- -> 2 [Al(OH)4]^- + 3 H2 ^}$$where $\ce{OH-}$ comes either from hydroxide dissociation, either from carbonate hydrolysis. $$\ce{CO3^2- + H2O <=> HCO3- + OH-}$$ The reaction with carbonate would gradually slow down as the carbonate/bicarbonate buffer will kick in. Initial $\mathrm{pH}$ is ( see notes below ...


8

While calcium sulfate is usually termed insoluble, it is not the ‘sitting at the bottom like a rock’ type insoluble; rather, it is the ‘there’s no practical way for me to get the two ions into the same solution without precipitation, but I’m still able to identify both ions in solution’ type insoluble. Those sentences don’t really help, so let’s look at ...


5

To answer this question, we have to assume that the given buffer solution is an ideal solution of which the total volume of it always equals to the summed volumes of sodium ethanoate ($\ce{NaOAc}$) and $\ce{HCl}$ solutions when they are added together. It should also be assumed that $\mathrm{p}K_\mathrm{a}$ of ethanoic acid (acetic acid; $\ce{HOAc}$) is ...


3

The answer is: it depends. Dissolution of a salt implies that the entropy gained exceeds the cost of breaking lattice interactions (the solution enthalpy, assuming it is positive). Electrostatic interactions compete with kT (thermal jostling). Under physiological conditions, long range interactions are strongly screened by intervening solvent molecules and ...


2

Once $\ce{Ba^2+}$ is eliminated as poorly soluble barium(II) sulfate, there is a solution of soluble copper(II) and zinc(II) sulfates. Considering this has been done by careful/gradual addition of sulfuric acid, and the fact that both remaining sulfates undergo hydrolysis, the solution is going to be slightly acidic — this is one reason for sodium hydroxide "...


1

Using the definition of $\mathrm{pH}$ (which is a log value) i.e. the difference of $\mathrm{pH}~1$ value is 10 folds difference in real terms. You have $\pu{100 ml}$ of $\ce{NaOH}$ @ $\mathrm{pH}~13.$ If you dilute by $10$ times, then the new $\mathrm{pH}$ will be $12,$ so you'll need to dilute it another $10$ times to reach $\mathrm{pH}~11.$ If you start ...


1

$\ce{CuSO4}$ exists as $\ce{[Cu(H2O)4]SO4}$ in solution. It is blue in colour due to the presence of $\ce{[Cu(H2O)4]^{2+}}$ ions. Now if aqueous $\ce{KF}$ is added, the solution turns green due to formation of complexes $\ce{[CuF4]^2-}$ and $\ce{[CuF6]^4-}$(Initially copper(II) fluoride($\ce{CuF2}$) is formed but it formed complexe in present of water). The ...


1

Find some nice instruction: http://nzetc.victoria.ac.nz/tm/scholarly/tei-Bio06Tuat03-t1-body-d2.html As Picric Acid stored under layer of water, we could know it's concentration in the water. It should be enough for most of the applications. Temperature ° C. grams picric acid/100 grams solution 0 | 0.67 10 | 0.80 20 | 1.10 30 | 1.38


1

If accuracy is not essential, weigh a small amount drained for a fixed time but still wet, and then weigh again dried, to find the ratio of wet-to-dry weight (carefully disposing the dried, sensitized, explosive without destroying the scale or personnel). Though this could give a rough idea of the actual dried weight, it would vary from batch to batch and ...


1

The old school method of generating ultra-pure water by distillation was to add a small amount of potassium permanganate in a slightly alkaline solution. All the organics were oxidized by the permanganate ion in boiling water. A small of $\ce{MnO2}$ so formed during boiling would take care of the acid(s). This water was good enough for electrochemical ...


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