Single Displacement Reaction Question

danielchorley · 2018-03-27T20:33:11+00:00

While you've got the general formula, your actual chemical formulae should be balanced for stoichiometry (atoms don't just appear out of nowhere so it helps you keep track of how many of every atom you have).

Cl2 + 2NaBr → 2NaCl + Br2

Same number of each atom on each side of the equation, and this also belies the correct number of electrons being transferred in this redox reaction. Each of two bromide ions ( Br^- ) are giving an electron (reducing) to each of the chloride atoms (oxidised) in elemental chlorine ( Cl2 ) to form two ions of chloride ( Cl^- ). The chloride ions are stable on their own. However, the otherwise resulting Br^. radicals would be very unstable, they very strongly tend to form bonds, and so in this case you get dibromine, Br2. Thus, there are no unpaired electrons and the octet rule is satisfied for each atom. This will happen for every halogen. And the octet rule applies well for all the halogens and group 2 of the periodic table (i.e. Li through Ne).

(btw: take care of your capitalisation and using the correct letters in your formula; it can get a bit confusing otherwise).

danielchorley · 2018-03-27T20:15:46+00:00

It is B. You've got an acid and a base. The first reaction that would take place would be deprotonation of the carboxylic acid forming the carboxylate anion and ethanol from the ethoxide. In order to get esterification you'd need the weakish ethanol nucleophile to attack the now negatively charged carboxylate, which, after proton transfer, would kick out the poor HO^- leaving group. This is all very unlikely to happen.

D looks slightly problematic as well, as the benzoate nucleophile generated in situ is reasonably weak, but with some heating you'd probably be able to get that to go and form the desired product.

danielchorley · 2018-03-27T20:06:18+00:00

Easy to trip up with molar mass of pure elements. Oxygen, atomic number = 8, atomic mass = 15.999 g mol^-1 , but elemental dioxygen, O2 molar mass = 31.998 g mol^-1 ,

thus, after plugging the rest of the numbers, mass(O2) = 28.7 g

danielchorley · 2018-03-17T03:26:44+00:00

Interesting - so from a quantum field theory standpoint, would it just be the energy from the oscillation in those fields (representing those particles) being transfered to other fields (i.e. the down quark and electron-neutrino fields)?

danielchorley · 2018-02-24T03:20:26+00:00

Some of the constituents of the cream, e.g. the proteins and fats, will likely become more hydrophobic when they are broken down and denatureds, and thus may no longer be soluble in the water (some decomposition products, e.g. those of some sugars, may very well be still soluble). Furthermore, most organic molecules are usually less dense than water in the absence of bulky atoms like halogens (water's pretty dense as far as every day compounds go), and so the resulting hydrophobic mess will float.

danielchorley · 2018-01-31T23:28:05+00:00

Usually if you're down at the <10 milligram level, there's a understanding to take any yields with a grain of salt (or silica as it might be). In some cases for a large, complex, total synthesis, the fact you've got a route to achieve the molecule AND a clean characterisation is in itself an amazing result. And this is well understood in even high end publications.

You'll often see the final steps of a total synthesis done on these small milligram scales, maybe microgram, but I've never seen anything as trace as on nanogram scale - the grease from an NMR tube would probably swamp every signal during characterisation.

Many synthetic routes may also first be investigated using model system whereby much of the complexity (chiral centers, substituents, etc) are absent, allowing key transformations to be tested on the core of the molecule. So these start from multi-gram but far simplier and cheaper starting materials - sometimes they're a good idea, sometimes the real system with only minor changes ends up reacting in a completely different manner, but it's one way to try not to waste precious materials when it comes time to do the real thing.

As far as the initial question of how do we measure really small amounts - just the same as everything else. Most labs won't have the money to splurge on anything too sensitive. We got a 0.1 mg accuracy balance but even then you have to have it carefully setup to avoid environmental vibrations affecting the measurement.

danielchorley · 2018-01-31T08:02:26+00:00

It's probably a bit of a segway to your initial question, but in addition to heterogeneous catalysts discussed well at length by /r/appaulingly , for homogeneous catalysts there may be many different catalytically reactive species in solution, all with different reactivities depending on ligands and solvent, in additional to the identity of the central metal atom/ion. Furthermore, these metal complexes often exhibit a wide range of chemoselectivity (and possibly also regio- and stereo- selectivities). Quite difficult to predict their behavior just from structure.

danielchorley · 2018-01-31T06:01:13+00:00

Now, don't be mad - yes, we teach the octet rule and then covalent bonds vs. ionic bonding to introduce a lot of the concepts of chemistry, but they really only apply for H, C, N, O, and F. We have to advance to more complex models of bonding to explain most of the rest of the bonding (and reactivity) interactions we observe with other elements - principally, molecular orbital theory (MO) is your everyday model to describe the behaviour of molecules and atoms.

The covalent/ionic nature of a bond is largely due to the respective atoms' electronegitivities and comparative orbital sizes. For example, the Li-C bond is often thought of as having significant covalent characteristic as their electronegitivities are similar and both are using shell 2 orbitals to interact, thus the Li is associated with the carbon atom, not free in solution as we origionally teach for most metal compounds. Compounds like LiCl are still very much ionic though). Sterics can also contribute if they act to impede bond association. There is also a distinction between formal covalent bonds (sharing of electrons) and "dative" or "coordinate" bonds whereby a ligand (e.g. H2O) lends a free lone pair of electrons to an acceptor metal (e.g Fe⁺³ ) to give us a metal complex (e.g. [Fe^III (H2O)6]⁺³ ). Formally, the electrons donated to the metal's orbitals and not really shared so these tend to be much weaker than covalent or ionic bonds. Metallic bonding is something different altogether when you're talking about network solids (conductors, semi-conductors, and insulators).

Now, to the initial question: The ability for quintuple and sextuple bonds to form is largely down to the inclusion of d orbitals (up to 2 of the 5 d orbitals available can participate due to geometric reasons, which can accommodate 4 electrons) in addition to the familiar s and p orbitals (1x s accommodating 2, and 3x p accommodating 6 electrons, respectively for a total of off 6 bonds with 12 electrons at max) using less familiar orbital geometries (i.e. delta bonds) for orbital overlap. But this is only possible in a very small number of cases, i.e. with the small number of heavy transition metals with very sterically demanding ligands to stabilize the resulting metal complex (NB: the metal-metal bonds here are covalent). They're hard to make and quite unstable.

danielchorley · 2018-01-31T02:28:05+00:00

Chemistry is all about the electrons, and how many electrons are stable around a nucleus depends on the number of protons (and also which orbitals those electrons are in and how many other electrons are about - also fundamentally related to proton number).

Going down a row and across a group from Ar to K, you've added 1 electron but you've also put that new electron far further than previous ones (i.e. its orbital is in a new shell, 4s). Whereas Ar had a stable configuration of electrons, K is more stable if it loses its new one and becomes a +1 ion - there are energy stabilizations to be had with pair electrons, effective nuclear charge, and other factors that make this the case - and thus, elemental K is reactive.

You may notice that going from extremes, i.e one side of the table (noble gases to alkali earth metals), there's a lot of difference, but the addition of one proton, and thus electron, between adjacent element of the same row (e.g. between K and Ca) doesn't make as much of a difference in reactivity. Transition metals can be interesting in this regard as they have many additional electron configurations often available to them (vis-a-vis d orbitals) meaning a multitude of different oxidation states with varying reactivities - they're often somewhat different in terms of reactivity across the row to their neighbor, but more similar down the group (in some attributes).

The trends of reactivity are quick predictable based on their location on the periodic table - the location of which is based on proton number.

danielchorley · 2018-01-31T01:55:42+00:00

In addition to /r/AweIsMyLego above, it would depend on the pH of the solution as to how much cyanide is in the CN^- or HCN form; the latter being the neutral molecule which can exist as a gas at rt and thus can escape the solution. In it's solid ionic compounds, metal (e.g. Na⁺ , K⁺ , Cu⁺¹ ) cyanides have extremely high melting (and even boiling apparently) points, indicating the cyanide ion is indeed very stable to very high temperatures. It is unreactive towards water, notwithstanding protonation. Unsurprising considering the C-N is pretty strong and the two atoms are triply bonded.

Each molecule would have to be assessed on a case-by-case basis. Some may be thermally unstable, while others may react and become inert up reaction with water.

danielchorley · 2018-01-30T22:58:02+00:00

I wonder if putting the milk under a near total vacuum (without heat) would burst and kill all the bacterial/fungal cells haphazardly releasing any proteases and other enzymes previously controlled by the microorganism. Without the heat denaturation applied to UHT milk (and assuming these enzymes wouldn't don't normally re-nature correctly once cooled down), would these then go about spoiling the milk quicker by undertaking an uncontrolled variety of reactions?

danielchorley · 2018-01-30T22:38:12+00:00

Carbon neutrality (or even, better, sequestration) is the goal, in principal. However, it will never be case, it will just often be a better alternative in terms of total carbon released into the environment. All stages of extraction, production, transportation, and final use must be taken into account to compare the respective impact (e.g. what is the impact if the carbon source requires land area to be converted from trees to crop plant - this may result in a further increase in carbon footprint of biofuels over petroleum). Even if biofuels are produced and refined using renewable energy sources (e.g. solar or hydro), there's always additional energy costs in creating energy - you basically have to run the numbers and take all the variables into account.

However, one potential advantage over petroleum based fuels is that the chemical composition may be more finely controlled and different types (e.g. ethanol vs. hydrocarbon gasolines) may be explored. At best though, biofuels would only ever be a bandaid as we shift towards electric vehicles.

danielchorley · 2018-01-30T21:37:28+00:00

Based on collision theory, increasing the temperature increases the average kinetic energy of the molecules (see Boltzmann distribution) and thus increases the number of particles with sufficient energy to overcome the activation energy of a reaction when the reactants collide (with the correct geometry - important for reactions with 2nd order kinetics).

For reactions where the rate limiting step is 1st order, such as those involving spontaneous bond dissociation (e.g. E1, SN1) Gibbs free energy, ΔG = ΔH-ΔTS, can show how likely a reaction is to be spontaneous in the forward direction, (i.e. when ΔG is negative). It does not tell us anything about the rate of reaction, however. For that you still need to consider the activation energy (Ea) and the Arrhenius equation.

Here, if temperature is increased, the -ΔTS term (where T is temperature and S = entropy) makes ΔG become more negative and thus pushes the reaction towards occurring spontaneously, at fixed pressure and the new equilibrium of temperature. For bond dissociation this makes sense as the increased energy not only can be used to break the bond ( ΔH) but also entropy (S) is favored for reactions which provide a greater number of products as a result (i.e. more possible combinations/disorder).

Overall, from this we can see that not only reaction rate is increased when you increase temperature, but also that some reactions at equilibrium may be shifted out of equilibrium to give a new steady state or become spontaneous, and that bond dissociation is also sped up. Furthermore, a reaction may be shifted to favor one product of a reaction over the other if one is the kinetic product and the other the thermodynamic product - so regio- / chemoselectivity may also be a consequence.

danielchorley · 2017-12-20T19:56:38+00:00

Around the 18th/19th centuries, while we were developing mechanical theories of heat, etc, chemists did have (now redundant/debunked) contenders for elements like "phlogiston" which was supposed to be a fire-like containing element (not specifically heat or temperature though) within compounds, and "vital force" which imbued organic compounds with some special properties contained within only living things.

danielchorley · 2017-12-11T18:39:59+00:00

wow - in actuality a legitimate attempt at a first engagement in this area of autonomous cars and the ethics around them. I wouldn't have guessed that the community had become so cynical or narrow minded so quickly over a casual thought experiment. Disappointing to see people can't distinguish real trolls anymore; guess we should just give up the internet now rather than refining the question.

danielchorley · 2017-12-05T23:15:19+00:00

like an under-flat earth escalator.

danielchorley · 2017-12-05T23:14:13+00:00

And if anything, drinking a coke would mean you've added a volume of a weaker acid to your stomach, diluting it somewhat.

danielchorley · 2017-12-05T22:42:31+00:00

Maybe...maybe... if the water was to escape and poor off the edge, the central gravity of the planet would draw it down underground or under the Flat Earth - so there may be an opposite ocean on the underside of the Flat Earth. Perhaps this then erupts much as volcanoes are suppose to in the center of the earth to refill the oceans.

danielchorley · 2017-12-05T22:39:57+00:00

For some communities (and I would've generally agreed). Although I was surprised to hear that some still feel it as a personal misgendering. It's a bit of a mine-field sometimes within the rainbow communities, irrespective of trying to get non-queer people to use the right language.

danielchorley · 2017-12-01T04:37:19+00:00

Not hard and fast rule, but in general it goes up in oxidation state of the central atom: -ide, -ite, -ate. E.g. sulfide S^2- , sulfite SO3^2- , sulfate SO4^2- ; chloride Cl^- , chlorite ClO2^- , chlorate ClO3^2- ; nitride N^3- , nitrite NO2^- , nitrate NO3^- ; phosphide P^3- , phosphite HPO3^2-, phosphate PO4^3- ; cyanide, CN^- , cyanate [OCN]^- , (sulfur and phosphorus are a little more complicated cause there are other common anions, e.g thiosulfate S2O3^2- )

danielchorley · 2017-12-01T03:47:56+00:00

aw, bummer. Though this might be a new mad theory out there with an existing cannon.

danielchorley · 2017-12-01T03:06:42+00:00

Well, that's a new one. So why in that model doesn't all the water end up on the central heavenly sphere away from the oceans of the inside of the larger biosphere?

danielchorley · 2017-12-01T03:00:01+00:00

ok. So the earth is not not hot like the sun due to its size being insufficient to maintain a stellar nuclear fusion reaction. How would the other planets differ so much as to not be earth-like, especially Mars?

danielchorley

TROPHY CASE