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Aluminium is a metal thus formed of metallic bonds. The positive Al cations are regularly arranged in a lattice and surrounded by a sea of free/delocalised negatively charged electrons (therefore metals can conduct electricity). The negative charge of electrons and positive charge of cations have strong attraction forces which require a lot of heat energy to ...


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One solution would be to reduce the amount of solvent (under vacuum for instance), then filter out the precipitated NaCl at high temperature (at higher temperatures NaOH has as much as >10x the water solubility of NaCl). The common ion effect will play in your favour, as the high water solubility of NaOH at higher temperatures will help you reach NaOH ...


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In this case, you have to know that $\ce{Cl2}$ and $\ce{NaOH}$ react at low temperatures according to $$\ce{Cl2 + 2 NaOH -> NaCl + NaClO}$$ and that at high temperature, the hypochlorite $\ce{NaClO}$ gets disproportionated according to $$\ce{3 NaClO -> 2 NaCl + NaClO3}$$ There is no way of predicting it. It must be known. So b) is the good answer. Note ...


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@Jon Custer makes a good point in his comment so lets look beyond the electronic shell configuration of V and Fe to help understand the different in melting points. Consider the following: Pauling’s Electronegativity Fe > V 1.8 vs 1.6 Vander walls radius Fe < V 0.126 nm vs 0.134 nm And consistent with Fe having a higher electronegativity the first ...


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Without getting into the chemistry which has been so clearly discussed by @MaxW, from my personal experience in "ebonizing woods" [particularly Oak and Black Walnut due to their tannin content] the light yellow solution you get as soon as the iron has dissolved, works the best. I am assuming that the red solution obtained after allowing the iron ...


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Permanganate oxidizes iodide to iodine $\ce{I2}$ at all pH values. In acidic conditions, the reaction is finished here. But in basic solution, the iodine $\ce{I2}$ is transformed into iodide and iodate according to $$\ce{3I2 + 6 OH^- -> 5 I- + IO3^- + 3 H2O}$$The mixture iodide + iodate reacts the other way as soon as the solution becomes acidic : $$\ce{...


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At room temperature, about $75$% of $\ce{H2}$ is $\ce{o-H2}$ and $25$% is $\ce{p-H2}$. As the temperature drops, the relative amounts of $\ce{p-H2}$ increases. The two forms are in a temperature-dependent equilibrium $\ce{o-H2<=> p-H2}$, $\Delta H = \pu{-1.66 kJ/mol}$. Ref. M. Eagleson, Concise Encyclopedia Chemistry, Walter de Gruyter Berlin, New York ...


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According to (1) and (2), the values in your table would be as follows (all values in Å): $$\begin{array}{c c c} & \ce{PF4Cl} & \ce{PF3Cl2} \\ \ce{P-F_{eq}} & 1.541 & 1.546 \\ \ce{P-Cl_{eq}} & 1.999 & 2.004 \\ \end{array}$$ we have $x < y$ and $a < b$, which means that (b) is the correct answer According to Bent's rule: atomic ...


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To complete your literature survey, include this publication cited at least 11 times so far: «Dichlorotrifluorophosphorane (PCl2F3): molecular structure by gas-phase electron diffraction and quadratic force field» by French et al. in Inorg. Chem. 1985, 24, 2774–2777, doi 10.1021/ic00212a014. Despite the paywall, ACS's «In lieu of an abstract, this is the ...


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For an isobaric process you can integrate the expression $dS = dq/T = C_p dT/T$ to compute the entropy change at another temperature. Since the heat capacities are assumed constant you can proceed as follows: $$\begin{align}\Delta S_{\pu{380 K}}^\circ &= \Delta S_{\pu{298.15 K}}^\circ + \sum_i \nu_i\int_{298.15 K}^\pu{380 K} \frac{C_{p,i}}{T}dT\\&= \...


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PQ Corporation sells many compositions of sodium and potassium silicates. Concentrations of potassium silicate range up to 40% in water. The product literature includes pH, viscosity, SDS and just about every measurable property you could ask for. The reason such data are not in academic literature or book form is probably that they are considered lower-...


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First of all: Reduction is just a fancy way of saying "gain of electrons". The posed question can thus be reformulated as: What happens to water, if it gains more electrons? In order to answer this, we should have a look at the oxidation numbers of water: Since oxygen is more electronegative, it has an oxidation number of -II, while the two ...


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Reduction processes as posed here, “rearranges” the atoms upon taking up two electrons. One way you may have been able to “derive” or take a good guess is to consider the electrolysis of water, breaking water down into $\ce{H2}$ and $\ce{O2}$. In this case there has to be both an Oxidation and a Reduction. In the hydrolysis case, the $\ce{H2}$ is the ...


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I get that this makes sense, but I have no idea how to derive it myself. Any help would be appreciated! Good question because such questions were pursued by prominent physical chemists and Nobel laureates (Ostwald/ Nernst and Hoffmann). The reduction of water can be tested visually and the apparatus to demonstrate it indeed named after Hoffman. Basically, ...


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Potassium silicate is synthesized by melting $\ce{KOH}$ above $400$°C, and adding some solid $\ce{SiO2}$. Its composition depends on the amount of silica used. The dissolution process does not look so. But it is a slow chemical reaction with a rather viscous liquid. The result can be cooled down and gives a solid substance practically insoluble in water at ...


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The NCERT textbook is partially wrong for elements after sodium. The flame colors and wavelengths do not match. Keep in mind that for alkali metals, the Bunsen burner flame shows multiple wavelengths for potassium, rubidium and cesium. It is very easy to excite their outermost electron in such a low temperature flame to higher energy levels. My point is ...


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You're absolutely right: the visual appearance of the potassium spectrum is a moderate purplish or lilac, but it has a very strong peak in near IR, which we don't see (but would likely show as very bright viewed with a digital camera, particularly with internal IR filter removed). See the spectra in this paper. BTW, a digital camera clearly shows the light ...


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To clarify: $\mathrm{p}K_\mathrm{a1}: \ce{NH3^{+}-R-COOH <=> NH3^{+}-R-COO- + H+}$ $\mathrm{p}K_\mathrm{a2}: \ce{NH3^{+}-R-COO- <=> NH2-R-COO- + H+}$ For aminoacids, the the pH calculation is the same as for acidic salts of diprotic weak acids $\ce{MHA}$. Unfortunately for you, $\mathrm{pH}=\dfrac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{...


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You are confused about chemical equations. The stoichiometric coefficient (the numbers before a compound) do not indicate the mass of the compound, they indicate the amount of substance. For example $$\ce{CaCO3 -> CaO + CO2}.$$ This equation means that $\pu{1 mol}$ of calcium carbonate will decompose to give $\pu{1 mol}$ of $\ce{CaO}$ and $\pu{1 mol}$ of $...


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You are mentioning $0.500$ g $\ce{CaCO3}$. Whatever the use of this substance later on, it contains $\ce{0.500 g/(100 g/mol CaCO3) = 5.00·10^{-3} mol CaCO3}$. Of course, if this sample reacts with an acid later on, it will produce $\ce{5.00·10^{-3}}$ mol $\ce{CO2}$. And these $\ce{5.00·10^{-3}}$ mol $\ce{CO2}$ occupies a volume $V$ = $\ce{5.00·10^{-3}} RT/p$...


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We consider mixing of $\ce{H2SO4(aq)}$ and $\ce{Ca(OH)2}$ in molar ratio 2:1. Molecular equation version: $$\ce{2 H2SO4 + Ca(OH)2 -> Ca(HSO4)2(aq) + 2 H2O}$$ Eventually if $[\ce{Ca^2+}][\ce{SO4^2-}] = [\ce{Ca^2+}] \cdot K_\mathrm{a2} \cdot \frac {[\ce{HSO4-}]}{[\ce{H3O+}]}> K_{\mathrm{sp,}\ce{ CaSO4}}$ ($\mathrm{p}K_\mathrm{a2}=1.99$): $$\ce{Ca(HSO4)2(...


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For spontaneous reaction: $\Delta G \lt 0$ Since $\Delta G = -nFE$, it is safe to say that for spontaneous reaction: $E \gt 0$. Thus, to answer this question, you must have access to the electrochemical series, the one like Guest posted in his/her answer. Let's look at the first reaction: $$\ce{Ni + Sn^2+ -> Ni^2+ + Sn} \tag1$$ The two relevant half-...


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I am not sure where your solutions come from but 4.42a isn't correct. The products are calcium sulfate and water, and not, as Poutnik points out, $\ce{HSO4-}$. So the molecular formula should finish like this: $\ce{Ca(OH)2(aq) + H2SO4(aq) -> CaSO4(aq) + 2H2O(l)}$ Calcium sulfate is actually pretty insoluble so would normally appear as a precipitate but ...


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The rate of a reaction, or how fast reactants become products, depends on the collisions of particles. This is because when a particle collides, the reaction occurs (provided that it has enough KE). So, if we increase the amount of particles (the concentration), we will have more collisions and therefore the reaction will go by faster.


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Historically speaking, potassium derives from pot ash and now potash when -- if there were no geologic sources of $\ce{K2CO3}$ -- woods were burnt in a pot. The addition of water will dissolve $\ce{K2CO3}$ and other high soluble matter, but not the ash. According to wikipedia, the addition of amine may lower the amount of KCl, but basically then, as by ...


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The composition of the wood ashes varies with the nature and the origin of the wood. In the average, it is a mixture of $40$ % - $70$ % $\ce{CaCO3}$, $5$% - $10$% $\ce{MgCO3}$, $5$% - $10$% $\ce{K2CO3}$, $5$% - $10$% $\ce{Na2CO3}$, $2$% - $5$% $\ce{SiO2}$, $2$% - $5$% $\ce{Ca3(PO4)2}$, $0.5$% -$2$% $\ce{Al2O3}$, <$1$% $\ce{NaCl}$, <$1$% $\ce{KCl}$, ...


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It seems that by partial crystalization, it can be somewhat purified as the most soluble salt, but cannot be easily purified to high purity degree. It may be andvantageous to crystalized near water freezing point. Solubility in $\pu{g/100 mL}$ for $\pu{0^{\circ}C, 20^{\circ}C, 100 ^{\circ}C}$ $\ce{KCl}$:28, 34.1, 56.3 $\ce{K2CO3}$: 105, 111, 156 $\ce{NaCl}$:...


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HINT - The picture below shows the situation.


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Current IUPAC recommendations suggest following electronegativity ordering. The symbol for iron (a less electronegative element among the two) should be written first: $\ce{FeCo}.$ From the section IR-4.4 Sequence of citation of symbols in formulae [1, pp. 58–59]: IR-4.4.2 Ordering principles IR-4.4.2.1 Electronegativity If electronegativity is taken as the ...


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The molecular formula: $$\ce{ 2 K3PO4 (aq) + 3Ni(NO3)2 (aq) -> Ni3(PO4)2 (s) + 6KNO3 (aq) }$$ The complete ionic formula is: $$ \ce{6K+ (aq) + 2PO4^3- (aq) + 3Ni^2+ (aq) +2NO3- -> Ni3(PO4)2 (s) + 6K^+(aq) + 2NO3^-(aq) }$$ The net ionic formula is: $$ \ce{2PO4^3- (aq) + 3Ni^2+ (aq) -> Ni3(PO4)2 (s) }$$


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Distilled water unfit for human consumption should be identified on the label with giant red letters; otherwise it should be mineral-free and quite pure. Getting an accurate pH will be difficult since there will be such low ionic conductivity. Possible contaminants like low vapor pressure organic compounds? Like ethanol, methanol, benzene? Highly unlikely. ...


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The ones you want to look out for are calcium and magnesium. Either try finding a water that has lower content of those two ions or try finding a water filter that can soften the water. I grew up in an area with very hard (but potable) tap water and such a water filter was constantly used before boiling our tea water. It had the nice added side-effect that ...


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Your brand B mineral water is pretty hard water. Boil it for 5 minutes, cool, decant the water. Do your tea with that.


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Long time ago, I supervised an undergraduate on a summer tea project (for fun). In South Asia there is a natural strawberry colored tea made from slightly matured green tea. You boil the leaves in water with a small amount of sodium bicarbonate. Ice cool it, and boil again and aerate it with a ladle. The tea becomes reddish on aeration and after adding milk, ...


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In your simple model, you start off with: nitrogen: $\mathrm{3(sp^2)^1 + p^2}$ left oxygen: $\mathrm{(sp)^1 + sp^2 + p^1 + p^2}$ right oxygen: $\mathrm{(sp)^1 + sp^2 + p^1 + p^2}$ The σ bonds are easy. Use $\mathrm{sp(O) + sp^2(N)}$. That leaves you with: left oxygen: $\mathrm{sp^2 + p^2 + p^1 + (\sigma_\ce{O-N})^2}$ nitrogen: $\mathrm{p^2 + (sp^2)^1 + (\...


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The problem is we are assuming pairwise sharing of electrons. That may not be true. Think of diborane -- it has two hydride moieties that share their electrons in three-center bonds and thus contribute two electrons simultaneously to both boron centers. Thereby each boron atom has a full complement if eight valence electrons around it. The bridging ...


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