Experiment 4 He N-Bottle P: Fill & Download for Free

Though normally in mathematics we like to abstract and generalize in order to increase the scope of subject matter covered by the proposed solution, please allow me to address a question I see frequently pop up on Quora, do you need to know math (whatever that means) in order to be a computer programmer (whatever that means) - and the likes, and craft an answer that hopefully exposes the idea that being mathematically literate only helps.We may claim that this is yet another geometric progression but, instead, let us do a transformation or a rearrangement of sort and switch to a different numeric base - a useful problem-solving approach that often bares fruit whether we seek an optimal pan scales weighing strategy to identify a not fair coin, an optimal number of lab mice needed to find one poisoned bottle of wine from a set of 1,000 or if we want to encode the number of a chest with a treasure in it, by strategically turning just one, and letting our accomplice decode that number.An exact power of two, name it [math]p[/math][math][/math], has this nice property that in base two, in a [math]k[/math]-sized word for definitiveness, [math]p[/math][math][/math] is represented by exactly [math]k-1[/math] zeros and a single and the only [math]1[/math]. For example, for [math]k = [/math][math]4[/math][math][/math]:[math]2^0 = 0001 \tag*{}[/math][math]2^1 = 0010 \tag*{}[/math][math]2^2 = 0100 \tag*{}[/math][math]2^3 = 1000 \tag*{}[/math]Since each exact power of two is unique and each such number has a unique representation in base two, it follows that when we construct a sum of the above numbers and compute it the old-fashioned way, by adding the digits vertically, column-by-column then we will have a square matrix of sort whose not only rows but also columns represent an exact power of two:[math]+ \left\{\begin{array}{cccc}0&0&0&1\\ 0&0&1&0\\ 0&1&0&0\\ 1&0&0&0\\ \hline 1&1&1&1 \end{array}\right. \tag*{}[/math]which guarantees the absence of carries and which produces a result in the form of a row filled with nothing but [math]k\;[/math] [math]1[/math]s.Another transformation: what will happen if we add [math]0001[/math] to the above result?In the old-fashioned way still the right-most [math]1[/math] of the addend will trigger a tidal and unstoppable wave of carries that will generate exactly [math]k[/math] zeros and a single and the only [math]1[/math] in a new, [math](k+1)[/math]-st position of a new [math](k+1)[/math]-sized word:[math]+ \left\{\begin{array}{ccccc}0&1&1&1&1\\ 0&0&0&0&1\\ \hline 1&0&0&0&0 \end{array}\right. \tag*{}[/math]What property does the above result have?It (is a set that) consists of exactly [math]k[/math] zeros and a single [math]1[/math] - and, thus, constitutes an exact power of two:[math]10000_2 = 2^4_{10} \tag*{}[/math]which makes it, pretty much, an answer to our problem:[math]2^0+2^1+2^2+2^3 = 2^4 - 1 \tag*{}[/math]Intuitively: the next exact power of two (short of [math]1[/math]) has all the information built into it about the sum of all the previous consecutive exact powers of two.Our sample analysis was for [math]k = [/math][math]4[/math][math][/math] but in your case [math]k = 13[/math] (almost). Hence, the magnitude [math]S[/math] of the requested sum shy of [math]2^0=1[/math] is:[math]S = 2^{12+1} - 1 - 1 = 8190 \tag*{}[/math]Say, we want to verify our theoretic claim and see the binary representation of a reasonably arbitrary number ourselves by writing a compact version of C code to convert an unsigned integer no larger than what 64 bits ([math]k=64[/math]) can represent from base ten to base two.Effectively we have to probe all the (relevant) bits of such a number in its base-two form. This is where mathematics comes in. We recall elementary number theoretic Division Algorithm Theorem according to which the residues of the input integer [math]\bmod 2[/math] can never be larger than [math]2[/math]. In fact, these residues can only be either [math]0[/math] or [math]1[/math].How convenient. These residues in base-two dwell in the right-most bit of the word that represents the input number [math]n[/math] - and are, effectively, returned by the modulus operator %:residue = n % 2; How do we convert the binary image of [math]0[/math] or [math]1[/math] into the corresponding textual form?We leverage the Division Algorithm … again!Construct a two-byte array Bits with a `0` in its first position and a ‘1’ - in the second so that the actual binary residue can be used to find the correct textual image by letting the pointer arithmetic do all the heavy lifting:bin[ here ] = Bits[ residue ]; and copy the appropriate character into the output string bin at the appropriate location, here. We shall populate bin in the right-to-left fashion by starting at the last possible character and moving leftward, via here--, as the conversion progresses.OK, the modulus operator lets us find out what the right-most binary digit is. And the rest?We will use the right (as opposed to left) bit shift operator >> to move all the bits of an input number by one position over (to the right) - effectively pushing the next interesting bit into the right-most position where the modulus operator can act on it again:n >>= 1; and we will do so in a loop until we whittle the input number down to zero by grinding all, and only, the relevant bits with the modulus operator.Here’s the meaty, and a rather compact, code*:while ( n > 0 ) {  residue = n % 2;  here--;  *here = Bits[ residue ];  n >>= 1; } Save the contents of the code shown below in a file dectobin.c and build it like so:gcc -g -m64 -Wall -o dectobin dectobin.c and execute it like so:./dectobin 4096 1000000000000 or like so:./dectobin 4095 111111111111 Experiment away as you see fit.What famous problems that are connected to our usual suspect here are you aware of?When I was a kid I was introduced to the story of a king who, in return for the invention of the game of chess on an 8x8 board, asked the author to suggest his own reward. The story goes that the inventor asked the king to put one grain of rice on the chess board’s first square, twice that - on the second, twice that - on the third and so on and then - the wise guy had some stones - he asked for all the grains of rice thus obtained.It follows that there should be [math]2^{63}[/math] grains of rice on the chess board’s last square and that the total number of the grains of rice would be our sum for [math]k = 64[/math]:[math]2^0+2^1+2^2+\ldots +2^{63} = 2^{64} - 1 \tag*{}[/math]which happens to be the limitation (the hard upper bound) of our simple 64-bit converter:[math]18, 446, 744, 073, 709, 551, 615 \tag*{}[/math]What would the binary representation of the above number be? Try it out.Another puzzle connected to our sum is the Towers of Hanoi. Here is a GitHub link to the Java source code of that puzzle - the curious travellers can examine it and the brave souls may even build it and see if they can find the optimal number of moves, which happens to be:[math]2^n-1 \tag*{}[/math]that solves the puzzle. Good luck.*The (as is) code in its entirety:#include <stdio.h> #include <stdlib.h> #include <errno.h>  static char Bits[] = { '0', '1' };  extern int main( int argc, char* argv[] ) {  unsigned long n;  char* here;  size_t residue;  char bin[ 64 + 1 ];   if ( argc < 2 )  {  return 1;  }   errno = 0;  n = strtoul( argv[ 1 ], NULL, 10 );  if ( errno || n == 0 )  {  return 1;  }   bin[ sizeof( bin ) - 1 ] = '\0';  here = &bin[ sizeof( bin ) - 1 ];  while ( n > 0 )  {  residue = n % 2;  here--;  *here = Bits[ residue ];  n >>= 1;  }   printf( "%s\n", here );   return 0; } 

I worked at a full service gas station in Lexington, Kentucky, during high school (1989–1992 for reference). One of my best friends worked with me. We were good workers, but were definitely a couple of smart asses. There are so many things I’ll never forget. Here are a few of my favorite memories.A family pulls in but not up to the pumps. The dad is obviously pissed off. His wife has a map, and two kids are in the back seat. My buddy and I go outside to see what they want. The dad, trying to remain calm asks, “Can you show us on the map how to get to Circle 4? ” That was the road that went around Lexington. My buddy points to the main street in front of the station (which also happened to be Main Street) and says “just go back out there and take it four more lights down. You can go north from there or the fifth light down will put you southbound.” The dad says, “Thanks, but can you show us on the map?” My buddy says, “Well sir, you’re in Lexington.” The dad says, “Yeah, I know.” My buddy just dead pan says, “Well, that’s a map of Louisville.” I snorted. The looks on the mom’s face was priceless, and the dad said something like “Thanks. That’s what you get when a woman navigates,” or something else sexist like that.The station had regular customers who were on account. One was the neighborhood Baptist church. Normally it was the minister’s wife who came through for gas. Not to make a judgment here, but she always had a bit of an attitude. After all, she was a minister’s wife! So, the movie Boyz N the Hood had recently come out. One of the characters was named Tre. So, my buddy Kev (short for Kevin. I’m just going to start using Kev instead of buddy) go out to fill her car and wash the windows. While we’re standing there we’re discussing the movie. I’m asking Kev if Tre is spelled Tray or Tres. We’re going back and forth when Mrs Minister butts in and says “It’s spelled M E A D O W T H O R P E!” She’d obviously been eavesdropping and thought we were talking about the name of the church. Whatever, lady, please sign the receipt.Germans really know how to engineer a car. A feature on BMWs and Mercedes (and probably Audi, Porsche and VW) is the locking gas cap cover/door. It’s tied in to the door locks. So, if the passenger doors are locked, so is the gas cap cover/door. This is an important detail for two stories. Both were generated by Mercedes 190 drivers. The first, the wife of a local attorney pulls in (mom of a classmate, that’s how I knew). She tells me (not ask, no please) to fill it up. I go back and the gas door is locked. So, I politely ask her to unlock the doors so I can get to the fuel cap. She says, “You’re just doing it wrong. I’ll show you.” So, she gets out of the car (unlocking the doors in the process) and comes back and opens the fuel door. “See!” She exclaimed. I pointed out that she had unlocked the doors to get out of the car, but she was having none of it.This guy pulls in driving a brand-new Mercedes 190. It was a beautiful looking car. Dark paint, but not black. He asks me to fill it up but then says there’s a problem with the fuel door and he’ll come open it. So, he comes around and started prying on the door with his car key! There were deep scratches all around the door. Deep down to the primer, heavy-duty scratches. He gets the door open and says he’s got an appointment scheduled to get it fixed. I ask him if he was trying to get it opened with the doors locked. He said yeah. I pointed out the bayoneted that locks the fuel door and explained how it worked. I even closed the fuel door and opened it again, then closed it and had him lock the doors. Of course, the fuel door wouldn’t open. Realization dawned on his face. He said he felt stupid. I’m not sure how the dealership handled it, as he never came back.This is one of my favorites. It was Christmas Eve, things were slow, but there were three of us working. Me, Kev, and Ed. We’re located at a four-way intersection. There is a slight uphill grade headed into town. Our station was on the uphill side. We see this car struggle to get through the intersection and then pull over and come to a stop out front. The driver opens the hood and starts looking around. Ed was the manager and told us to watch out for this guy. Sure enough, the guy comes and asks if he can borrow some tools. Ed says no, they’re locked up and only the mechanic has the key (mostly true). The guy goes back to his car. He comes back a little while later and offers $20 to anyone who will come out and hold the battery cable wires to the battery. $20 was a lot for a high school kid back then. But, Ed says no, too dangerous. The guy leaves. Ed explains that that’s a good way to get electrocuted. The guy comes back a little while later. This time he zeroed in on me asking for help. “Can you come hold these cables, please?” I say, “Sorry, I really can’t.” He follows up with “Where’s your Christmas spirit?” So, I tell him, “I’m Jewish.” He gets all huffy then leaves. Finally he gets his car started. He pulls into our lot, past the pumps and office. As he’s passing the office, he flips us all the bird. Then his car died. Ed just stood up and laughed about as loud as he could. The car coasted off the lot. Surprisingly, the guy came back. He acted all sheepish and said, “Where’d that asshole with the broken down car go? ‘chuckle’. Can you guys please help me push my car off the road?” So we did. The place was closed on Christmas, but by the 26th the car was gone.This is probably my favorite story from my time there. Here’s some background. I’m not sure how all this breaks down. It was an independent Sunoco station. I’m not sure if it was a franchise because we were also a Firestone dealer. The name of the place was John Foltz Firestone. This is important to the story. Our mechanic and John Foltz had been friends for a long time. John said once that he hired the guy because he would be unemployable anywhere else. The mechanic was really good, but he had some mental issues. He had the kind of issues that required a paper grocery bag full of pills to control. That’s not an exaggeration. He had a grocery bag with about 15 different prescription pill bottles in it. Now, he was mostly harmless, but he did have a temper. He swore like a sailor all the time, but he was a nice guy. Foltz didn’t let him interact with customers much because this guy lacked a filter. Also, it was a good idea to stay out of the bays when he was working on a car. It wasn’t uncommon for a tool to be hurled at the wall if a bolt or screw was causing a problem. So, the guy normally left at 5 p.m., two hours before we closed up. One night early in my tenure he was asked to close up. I’d not been given keys to the place yet. The place had flood lights out front that needed to be turned on, and signage that needed to be turned off, plus the pumps and whatnot. As we’re getting ready to close up I said to the mechanic “Hey, don’t we need to open the fuse box and shut off the signs and stuff?” He just looked at me and said “Fuck Foltz and his fucking fuse box!” I was terrified, but looking back, that was an excellent example of alliteration and that’s why I like that story.I will never forget this one, ever. John Foltz is the one boss that I have the most respect for. He was a decent sort, borderline mentor to the guys working for him, patient, calm, reasonable. He was everything you would want in a boss. If you wanted to learn something, he’d show you. You needed advice, he was your guy. He’d butt in sometimes and offer guidance when needed too, just a great human being all around. Working there was one of my best experiences ever. Sometimes I really miss that job. Anyway, this happened mid 1993. I was home from my first year at school and came back to work for the summer. John was Catholic. I’d grown up going to Catholic church too and was even an altar boy for a while. He’d talk about the local priests and their sermons. I knew who some of them were. I don’t know this for a fact, but I think that instead of giving at the collection plate on Sunday, he just gave the priests in town free gas. Maybe he donated too, but it doesn’t matter. So, in the early 1990s pedophile priests were being outed left and right. One day the head priest at his church was suspended or something because some people had come forward with abuse allegations. I was sitting in the office when this guy pulls in. I recognized him from when he would fill in for Mass at my church (even before all this happened, I didn’t like the guy). John was doing an oil change or something in one of the bays and didn’t see him. Normally he went to talk to priests when they pulled in. I go out to start filling the guy’s car up. It was some old boat (an Oldsmobile I think) with the gas cap under the license plate. This priest gets out of his car to watch the process. He had some little yippie dog as well. I was just starting on washing the windows when John comes out, removes the gas nozzle from the car and hangs it up. He gets right in this priest’s face and says, “Don’t you ever come back here.” It was the most menacing thing I’d ever heard come out of John’s mouth. The priest just got back in his car and left. I don’t know if there was more to the story, but it put John in a foul mood. He asked me to finish the oil change and then lock up the place, then he left. Not that it matters now, but if anyone is interested the priest was Rev. Leonard B. Nienaber. Here are a few articles. Looks like he served some time and the Church had to pay the victims. So at least there’s that.Lexington Priest Charged with Sexual Abuse, by Tom Loftus, Courier-Journal, April 8, 1993Kentucky Diocese Settles With VictimsThere are more stories, but they’re the “you had to be there” type.

This is a very interesting question. I will try to explain this in detail.I did my Ph.D. in the lab of Prof M. Van Beylen. He was a student of Michael Szwarc - the pioneer in Living Anionic Polymerization.When you study living anionic polymerization your solvents need to be super dry, consequently the group of Prof Van Beylen used a lot of sodium and potassium metals. Concomitantly with it, they also had lots of sodium and potassium waste. We collected this metal waste in 2.5 L glass bottles covered with paraffin oil.The traditional way to dispose the alkali metal waste on lab scale is to suspend it in n-hexane followed by dropwise addition of isopropanol. This works fine if you have a few grams of metal, but not if you more than 2 kg. (Unless you have a few days to spare.) So our - enjoyable, nevertheless slightly unsafe - way of sodium and potassium disposal was a bit different. We would glide our 2.5 L bottles in the small river nearby the lab. Due to the turbulence of the water, the bottle would eventually flip and fill up with water. This would potentially give an explosion or at least a nice fire in the middle of the river leaving only some alkaline residue.I have to admit that this would not be acceptable anymore, but in our slightly juvenile opinion it was the fastest way to dispose it in a reasonable safe way. Although they told me that on one occasion the bottle drifted nearby the river side and started a small fire in the bushes. Obviously this event caused some extra amusement. The fact that we sometimes disposed the waste after a lab party in the middle of the night adds to the equation of pure nostalgia.So far the introduction to the fact that alkali metals fires are fascinating and one of the reason of this post.Why do alkali metals explode in contact with water?This seems a simple question. I guess most chemists would just state: “That is obvious: the reaction of alkali metals with water is very exothermic and you generate hydrogen. Due to the heat the hydrogen is ignited, which causes the explosion.”But if we look into detail it is not that straightforward.Let's check what happens if we would drop a pea-size piece of metal in water.Lithium just sizzles on top of the water (see Table 1 for densities). It does not explode.Sodium also floats on top of the water, it will ignite with a yellow orange flame most of the time however it does not explode.Potassium is not denser than water, it does react very violently and will ignite with a violet flame, it will often explode in contact with water.Rubidium sinks, reacts very violently and will explode.Caesium will explode on impact with the water.TABLE 1: Physical properties of the alkali metals + NaK alloy (data from Alkali metal - Wikipedia)[math]\small{\begin{array}{ccccc} \hline M & \text {Melting point/°C} & \text {Boiling point/°C} & \text {density/g mL}^{-1} \\\hline Li & 180.5 & 1342 & 0.534 \\ Na & 97.7 & 883 & 0.968 \\ K & 63.4 & 759 & 0.856 \\ Rb & 39.3 & 688 & 1.532 \\ Cs & 28.4 & 671 & 1.873\\ NaK & -12.6 & 785 & 0.866 \\ \hline \end{array}}[/math]Just to give you an idea of the complexity I will start with the reaction of the alkali metals with pure oxygen.Lithium gives Lithium oxide:[math]4Li(s) + O_2(g) \to 2 Li_2O(s)\tag{1}[/math]Sodium will be rather similar:[math]4Na(s) + O_2(g) \to 2 Na_2O(s)\tag{2}[/math]but it does react further to the peroxide [math]O_2^{2-}[/math]:[math]2Na_2O(s) + O_2(g) \to 2Na_2O_2(s)\tag{3}[/math]the peroxide will be the main product.Starting from potassium it will react in a 1:1 mole ratio to form superoxides: salts with the dioxide ion; [math]O_2^-[/math] (a charged ionic species with a single unpaired electron and a net negative charge of −1)[math]\begin{align*} K(s) + O_2(g) & \to KO_2(s) \\ Rb(s) + O_2(g) & \to RbO_2(s) \\ Cs(s) + O_2(g) & \to CsO_2(s) \end{align*}\tag{4}[/math]Superoxides are rather exotic, but they do have applications in for instance O₂ generators.The large superoxide [math]O_2^-[/math] anion can only be stabilized by the bigger cations with smaller ion charge densities. The Lithium cation is just too small and has a too high charge density.Actually the reaction with oxygen is one of the reasons you should keep the metal stored in an inert solvent that doesn't dissolve oxygen.But we are drifting away from the question. We were interested in reaction with water: all alkali metals react the same. They form the metal hydroxide and hydrogen gas (and heat.)[math]2M(s) + 2 H_2O(aq) \to 2 MOH(aq) + H_2(g)\tag{5}[/math]The traditional assumption is that the heat will ignite the hydrogen gas causing the explosion or more correct an implosion (reaction of hydrogen and oxygen to form water)[math]\begin{equation} O_2(g) + 2H_2(g) \to 2H_2O \tag{6} \label{oxy}\end{equation}[/math]But the heat generation is not a correct argument: if we look at the thermodynamics of the reaction (See Table 2) all alkali metals actually generate almost the same amount of heat per mole. If we would recalculate per gram we notice that Lithium generates much more heat per gram. Hence the argument that the generation of heat is responsible for the explosion can’t be correct.TABLE 2 Heat Release Based on the Reaction (based on data from ref. 1)[math]\textbf{(7)}\quad\ M(s) + H_2O(aq,\infty ) \to MOH(aq) + \frac{1}{2} H2(g)[/math][math]\small{\begin{array}{ccc} \hline M & \Delta H \text{/kJ mol}^{-1} & \Delta H \text{/kJ g}^{-1} \\\hline Li & -508.5 & -73.3 \\ Na & -470.1 & -20.4 \\ K & -482.4 & -12.3 \\ Rb & -481.2 & -5.6 \\ Cs & -488.3 & -3.7 \\ \hline \end{array}}[/math]The reason why all alkali metals are so close to each other might need some extra explanation. In order to have some qualitative data we can look at the Hess cycle of the Cation formation.SCHEME 1: Hess Cycle of the Cation formation:You can imagine 3 steps:The atomization energy is the energy required to vaporize gaseous atoms of the metal. As you can see from the table Li requires the most energy. The metallic bond in Li is the shortest (and strongest.)The first ionization energy is the energy required to form an ion. Since the valance electron is further from the nucleus in Cs, it will require less energy to get rid of it.The hydration enthalpy the energy that is formed to hydrate the ion. In the case of Cs the ions are larger in size there is less interaction with the lone pair of the water molecules.TABLE 3: Estimate of Enthalpy of formation of alkali metals with water. All data in kJ/mol. (data from ref [1] except hydration enthalpy from [2])[math]\small{\begin{array}{ccccc} \hline M & \text {Atomistion energy} & \text {1st Ionisation Energy} & \text {Hydration Enthalpy} & \\\hline Li & 159.4 & 526.4 & -519 \\ Na & 107.4 & 502.0 & -409 \\ K & 89.2 & 425.0 & -322 \\ Rb & 80.9 & 409.2 & -293 & \\ Cs & 76.1 & 381.9 & -264 \\ \hline \end{array}}[/math]In all 3 items there is clear trend from Li to Cs. The problem is that the change rate of the separate items is different, so - when combined - the trend is gone.For the smart people that do the actual math of the last table. It indeed does not correspond to the data of Table 2. But please keep in mind that the data was provided to give you an idea about the trends. (The enthalpy data of Table 2 was gathered from one source, if you would check others you might find other values.) I also left out the contribution of water (visualized by the infinity sign), and only included the enthalpy value of OH-formation. But - if you want - you can add it, which I did in table 4). You could also calculate [math]\Delta G[/math] using standard electrode potential or [math]E^°[/math]:[math]\small{\begin{align*} 2M(s) & \rightleftharpoons 2M(aq) + 2e^- \quad & E_{M}^° \\ H_2O(l) + 2 e^- & \rightleftharpoons 2 OH^- (aq) + H_2(g) \quad & E_{H_2O}^° = 0.83\\ \hline 2M(s) + 2H_2O(l) & \rightleftharpoons 2M^+(aq) + 2 OH^- + H_2(g) &E_{cell}^°=E_H^°-E_M^°\end{align*}}[/math][math]\Delta G_{cell}^° = −n \,F \, E_{cell}^° [/math]TABLE 4: Estimate of Gibbs Free energy and Enthalpy of the reaction[math]M(s) + H_2O(aq) \to MOH(aq) + \frac{1}{2} H_2(g)\tag{8}[/math][math]\small{\begin{array}{cccc} \hline M & E_M^°/V & \Delta G \text{/kJ mol}^{-1} & \Delta H \text{/kJ mol}^{-1} \\\hline Li & -2.84 & -194 & -222.7 \\ Na & -2.71 & -181 & -184.3 \\ K & -2.931 & -203 & -196.5 \\ Rb & -2.98 & -207 & -195.3 \\ Cs & -3.026 & -212 & -202.4 \\ \hline \end{array}}[/math]Thermodynamic data don’t tell you much of the speed of the reaction. Referring to the data of Table 3 you can make a fair estimation of the activation energy of the reaction by making the sum of the first two items (until formation of the ionized cation, but before hydration) . It turns out that activation energy is much higher for Lithium. This is the reason why Lithium reacts slower in comparison with the other alkali metals. But it remains puzzling why it eventually does not explode.It might relate to the solubility of the resulting hydroxides; Lithium hydroxide is less soluble in comparison with the other metals. (About a factor 10, see Table 5) So the the LiOH might form a film on the metal surface, which could explain an inhibition or at least a slow down of the reaction. Another factor is that Lithium has a fairly high melting point (see Table 1) in fact higher than the boiling temperature of water. The other metals will melt during the course of the reaction. But Lithium remains solid: the surrounding water act as a buffer and takes up the heat in the form of steam. The other metals will melt and might give small droplets which does give a much higher contact surface.TABLE 5: Maximum solubility of Alkali hydroxides in g/L near room temperature (data from Wikipedia)[math]\small{\begin{array}{ccc} \hline MOH & Solubility /g L^{-1} & temp /°C \\ \hline LiOH & 128 & 20 \\ NaOH & 1110 & 20 \\ KOH & 1210 & 25 \\ RbOH & 1000 & 15 \\ CsOH & 3000 & 30 \\ \hline \end{array}}[/math]But even considering all the above factors there are some other issues with the theory that hydrogen causes the explosion. If we would repeat our experiments again in an inert gas like argon it turns out that starting from Potassium it will explode sometimes. Without oxygen reaction [math]\textbf{\ref{oxy}}[/math] is not possible.One of the options would be a Steam explosion, but that seems highly unlikely. Due to the heterogeneous nature of the reaction the water and metal need to come into contact. You form hydrogen gas and heat at the surface of the metal, but once the reactions starts the hydrogen gas and generated steam should act as a barrier to prevent further reaction. (The Leidenfrost effect)In 2014 Mason and Colleagues [3] used a high-speed camera (~100 µs resolution) to film the explosion. They actually used Sodium-Potassium alloy, which is a liquid at room temperature (see Table 1) . The reason why they did this was two-fold: they could control the size of the droplet reproducibly by using a syringe, but the main reason was to have a clean metal surface. So free from oxides and other impurities.Using the high speed camera they could produce these wonderful pictures:Figure 1 A NaK-drop hitting the surface of water from above and below (with water as comparison) from ref [3]As you can see from the picture the reaction is virtually instantaneous. At about 0.3 to 0.4 ms you see some spikes coming out from the metal. These spikes were crucial in order to understand the mechanism.The spikes are the reason why alkali metals explode in water: a closer look at the pictures revealed that they are dendritic. (smaller spikes appear as secondary spikes etc.) In a matter of a few ms this causes a very high surface area, which can react with the water molecules creating the enormous amount of heat and hydrogen i.e. an explosion.They did some extra experiments with molten Aluminium (which did not cause the spikes) and also dropped a NaK-drop into liquid Ammonia. (Figure 2)Figure 2: A NaK-drop hitting the surface of liquid ammonia. From ref [3]So in liquid ammonia the spikes were clearly visible, but no subsequent explosion occurred. Ammonia is capable of solvating alkali metals (forming a blue solution), but the reaction to hydrogen is too slow. (Sodium in Ammonia is stable for days, although eventually it does form sodium amide)[math]M(s) + NH_3(l) \to \frac{1}{2}\ H_2(g) + M^+ + NH_2^-\tag{9}[/math]But obviously one question remains: what causes the spikes. The authors suggest that in contact with water the metal forms solvated cations(visible as the blue area in the above picture) almost instantaneous hence a fast formation of positive charges on the metal area. These identical charges repell each other due to electrostatic repulsion. This causes what is called a coulomb explosion. They did some ab initio calculation to proof their theory. In addition they could completely stop the explosion mechanism by adding small amounts of surface active species like hexanol into the water.In a follow up article ref [4] they published a very nice footage of this last experiment: link to movie (mp4)I urge you to watch it first before you read on. (It's 20 seconds long)The authors describe what is going on:After gently placing a drop of a sodium/potassium alloy on water under an inert atmosphere production of solvated electrons becomes so massive that their characteristic blue color can be observed with a naked eye. The exoergic reaction leading to formation of hydrogen and hydroxide eventually heats the alkali metal drop such that it becomes glowing red (at this stage to be around 600°C) and part of the metal evaporates. After the chemistry is essentially over. When the smoke clears, the drop cools and suddenly becomes completely transparent. It is consisting of molten hydroxide temporarily stabilizes on water via the Leidenfrost effect, bursting spectacularly after it cools sufficiently.Now, watch it again :-)This figure appeared on the cover of Angewande ChemieFIGURE 3: A small water drop on an alkali metal drop. With the blue, electron-rich region and white hydroxide as the final product of the reaction from ref [5]So to conclude: the explosion of Alkali metals in water is far more complex as it seems.Edit: thanks to the comment of User-11782962439727910095 I realized that Mason is a YouTuber called Tunderf00t. So I had a look into some of his videos. He explains the journey in this rather long (30 min), but very interesting behind the scenes video: (He ends with a question that he explains in his follow up article as explained above.)[1] The NBS tables of chemical thermodynamic properties selected values for inorganic and C1 C2 organic substance in SI units, Wagman DD, Evans WH, Parker VB, Schumm RH, Halow I,Bailey SM, et al., J Phys Chem Ref Data 1982; 11(2): 3–392.[2] Ionic hydration enthalpies, Derek W. Smith, J. Chem. Educ., 1977, 54 (9), 540[3] Coulomb explosion during the early stages of the reaction of alkali metals with ​water P. E. Mason, F. Uhlig, V. Vaněk, T. Buttersack, S. Bauerecker , P. Jungwirth, Nature Chemistry 2015, 7, 250[4] A Non‐Exploding Alkali Metal Drop on Water: From Blue Solvated Electrons to Bursting Molten Hydroxide P. E. Mason, T. Buttersack, S. Bauerecker, P. Jungwirth, Angew. Chem. Int. Ed. 2016, 55, 13019.[5] Inside Cover: A Non‐Exploding Alkali Metal Drop on Water: From Blue Solvated Electrons to Bursting Molten Hydroxide P. E. Mason, T. Buttersack, S. Bauerecker, P. Jungwirth, Angew. Chem. Int. Ed. 2016, 55, 12916.