March 25

What are the IUPAC nomenclature rules for naming group 13 hydrides, that ditinguishes bonding motifs for structural isomers? For instance, there are two isomers for closo-diborane(4). Both have a boron-boron bond, however, the ground state isomer, has two hydrogen bridges. Plasmic Physics (talk) 00:55, 25 March 2013 (UTC)[reply]

Most discussions about interstellar travel start out with something like "at Voyager's current speed, it would take 73,000 years to reach the nearest star". So what's wrong with taking 73,000 years, or even a million? What can possibly break a spacecraft in 73,000 years, since there's no wind, rain, bacteria, or vandals, and very few micrometeroids compared to inside the Solar System? --140.180.249.152 (talk) 05:21, 25 March 2013 (UTC)[reply]

What's wrong with it is that we will either be extinct or will have picked it up and put it in a museum long before it reaches anything. In either case, it won't help us make contact with aliens. StuRat (talk) 05:51, 25 March 2013 (UTC)[reply]

We most recently discussed the topic of the longevity and eventual fate of Voyager's material components in February 2011, and there are links to earlier discussions. Nimur (talk) 05:53, 25 March 2013 (UTC)[reply]

StuRat: You're right, but I was really asking whether human technology could stay functional after that long in space.

Nimur: so according to that discussion, this is actually a feasible way of reaching another star? Assuming the spacecraft stays dormant until it gets near enough to power its solar panels, that discussion seems to say nothing bad will happen to it. --140.180.249.152 (talk) 06:23, 25 March 2013 (UTC)[reply]

It is extremely extremely unlikely that anything that involves electronics or mechanisms made by man would still function after that length of time. On a time scale of thousands of years, semiconductor alloys are unstable, and most metal alloys are unstable. Semiconductor alloys of course includes solar panels. Man has considerable expertise making high reliability electronics for military, space, and telecommunications applications. Even so, not all failure mechanisms are understood, and such components have been found to have rapidly increasing failure rates (both catastrophic failures and drift outside specifications) after only 40 to 60 years. Really, the only way to build something that will still be functional after 73,000 years is to build it with the very best available practice, test it until it fails, figure out why, and build another corrected for the discovered failure mode. Repeat as necesary until you've tested a batch contiunously for 73,000 years.

Electronic and mechanical systems suffer from random failure in addition to identifiable failure mechanisms, such as Coffin-Manson fatigue failures. The only way to address this is to build in self-repairing or redundant systems. For a service life of 73,0000 years, the simplest functionality is likey need so much redundancy and self correction that it will be physically enormous in volume and mass.

And then it could, and probaly would, still fail due to crashing into something unexpected, or damaged by interstellar dust.

Wickwack 120.145.136.89 (talk) 07:12, 25 March 2013 (UTC)[reply]

Once it gets past the point where the solar wind is significant, it will`have a problem. It will still be outgassing, but the gasses won't be blown away - they will form a very thin atmosphere around the spacecraft held there by gravity.. When material outgass, atomic oxygen is a commonly found gas, and it is very corrosive. It will be very thin, but it will also have a very long time to work on the spacecraft. --Guy Macon (talk) 09:35, 25 March 2013 (UTC)[reply]

All sorts of things can happen to materials they are a mixture or touch each other or are small, which all happen in semiconductors, over such long periods, but it would mostly be at 4°K so that should slow things down considerably. An interesting problem would be having a computer working at that temperature for that length of time and waking up properly when it got within a few thousand million miles of the destination. Dmcq (talk) 14:29, 25 March 2013 (UTC)[reply]

That's an interesting question, what temperature would it really reach? After all the interstellar gas can have quite a high temperature. Dmcq (talk) 14:43, 25 March 2013 (UTC)[reply]

Designing for a minimum temperature of 4 K would certainly rule out conventional solid state electronics. The mimimum permitted storage temperature for military spec parts is generally -55 C (218 K) [Ref Fairchild 54/74 databooks, National Semi databooks, etc). Most commercially used alloys have phase changes (more than one solid phase) at low K temperatures, and this aspect and its consequences is poorly understood, as there is virtually no commercial incentive to research it. Most likely a designer would go for some sort of long life nuclear isotope power source to supply both electrical power and heat, but that is another field that does not have sufficient practical knowlege. Wickwack 120.145.9.126 (talk) 14:58, 25 March 2013 (UTC)[reply]

All deep space probes need RADIOISOTOPIC heater units to keep the electronics warm. There long half-life mean they will keep the spacecrft warm for many more decades- long after the radioisotopic thermoelectric generator fails to deliver sufficient power to keep the space craft alive, Also, gravity is the week force, therefore, the atomic recoil from any oxygen out-gassing would send these gases far away from the space craft – no solar wind required. The whole assemblage would need to be no more than a gnats whisker from absolute zero for these atoms not to achieve escape velocity from a few hundred weight of ali and plutonium. If any of you want to contribute to some artificial contrivance that may still be working in 10,000 years time... then I refer you to: [1] But I'm putting my money on a simple egg timer. However, if the Eloi have given up eating eggs by then, even that wont be any use to them, (Stonehenge still works though but it would be difficult to launch into space). --Aspro (talk) 23:53, 25 March 2013 (UTC)[reply]

Warm for many decades eh? The OP wanted 7,300 decades or better. Re size you forgot random failure and the need for redundancy and/or self repair - see above. Egg timers need consistent gravity. If some nutters want to bury a clock programmed never to repeat its' chimes for 10,000 years, well good luck to them - it sounds like fun. But all human endeavour involves human error and oversight. It won't be until 10,000 years is up before we find out what their human erors were. My guess is <<< 10,000 years before it fails. This is why the onboard computers of US sourced military airplanes continued to be manufactured for decades with magnetic core memory (technolgy of the 1950's and 1960's) and not the various far more compact semiconductor memories used commercially since the 1970's. They have lots of experience with magnetic core memories and can trust them. Nobody has decades of reliability engineering experience with semiconductor memories. Wickwack 58.170.130.186 (talk) 01:18, 26 March 2013 (UTC)[reply]

Simply substitute Plutonium-238 for Plutonium-239. That has a half-life of 24,100 years which has the warmth of freshly drawn cows milk. Only 8 kilograms will still one with more than a kilogram over this time period. Magnetic core memory's were used in military applications simple because they had high resistance to Electromagnetic pulse. So, providing the probe doesn't stray into a nuclear war between the Klingons et.al. That should not be an insurmountable problem. An egg timer could be housed in a centrifuge but I was not suggesting this for a deep space time-piece. Just pointing out that such a silicon- sodium- aluminum- boron oxide device would last a very, very long time. Transistor radios came out in the late fifties. Your own ears can lay witness that some still work to day. So we already have decades of experience of semi-conductors. Once the dopants are locked in to the substrate there is little reason (so far) to suggest that one day the might suddenly decide pick up their bags and emigrate to some where else. If science was left to the pseudo-skeptics, Homo-sapien would still be at the stage of trying to rub two boy-scouts together to create fire. My money however, is still on the egg timer.--Aspro (talk) 17:22, 26 March 2013 (UTC)[reply]

RTGs ALREADY use ²³⁸Pu. See radioisotope thermoelectric generator. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 15:32, 27 March 2013 (UTC)[reply]

Yes. That's why is suggested 239!--Aspro (talk) 01:18, 29 March 2013 (UTC)[reply]

You write about things you know nothing about. Having worked as an electronics engineer, I can tell you that the mean time between failure for 1950's and 1960's transistors is such that most radios had at least one component failure after only a few years. Radio servicemen used to make a good living out of fixing them. Assuming failures occur randomly, and no repairs are made, and after 5 years half of radios still work, then after another 5 years one quarter will still work, after a total of 15 years, one eighth will still work, and so on. If one million were sold, after 50 years you'd still have 976 radios still working - more than enough to keep the collectors and nostalga nuts happy. But if you only built one radio to the exact same quality, after 50 years, you'd have only approx 0.001 probability of it working 50 years later. In real life it's more extreme than that. You can buy all sorts of old radio parts on ebay. But what you don't find on ebay is the specialised germanium convertor transistors (these convert the incoming radio frequency to a standard internal frequency for amplification). The reason is they fail after about 40 years or so due to an internal whisker growth phenomenon not anticipated by the manufacturers. They have normal relaibility until the whiskers grow long enough to cause a short, then, between about 40 and 50 years, they all fail. A somewhat different situation applies to tube radios. Many collectors claim, incorrectly, that the tubes made in the 1930's are the most long lasting and reliable - becaus they have a 1930's radio that still works. In fact, any radio tech that worked back then can tell you that each radio sold was a never ending stream of profit from repairs and replacement tubes. Back then what caused tubes to fail was incompletely understood. So most got made with failure modes built in. The ones that still work today are the ones, that by random fluke, got built without the failure modes. It was not possible back then to predict which were the good ones and which weren't. (The situation was markedly improved just before and during World War 2).

It is misleading to say that we have decades of experience in semiconductors - because semiconductors have gone through several drastic changes in technoloy and manufacturing technique. Memory and computer technology in the 1970's was based on TTL. In teh 1980, it was NMOS, in the 1990's CMOS took over.

I began in electronics in the mid 1950's when transistors were new. By then what caused tubes to fail was very well understood. We were all told that the new trasistors would be by far the most reliable parts in the radio/stereo/tv and far better than tubes because nearly all the tube failure modes did not apply. It didn't work out that way though. Those old 1950's and 1960's transistors failed pretty often. If you wanted high reliability in the late 1950's you built with tubes. Transistors had their own new failure mechanisms that engineers had to learn about. — Preceding unsigned comment added by 124.178.132.17 (talk) 04:21, 27 March 2013 (UTC)[reply]

All you are saying is that you had contact with cheap consumer electronics only!.Aspro (talk) 01:08, 29 March 2013 (UTC)[reply]

Actually my electronics career has included the professional high quality electronics field. But that is irrelevant. You brought up consumer radios as a supposedly long life product. The points I made here are 1) just because some samples of a mass produced product are still around and still work does not mean that the average failure rate was very good, and I ran with your example to explain it. This is an application of statistics and applies whether the goods are cheap and nasty or not. 2) Predicting the reliability of new technology is fraught with serious problems, and there will be new failure modes that need to be discovered. This applies regardless of whether the goods are cheap consumer grade or not. We were told, for both consumer and professional applications, that transistors would be dramatically more reliable than tubes. And so they were, but only after 20 years of manufacturing experience revealed all the new unanticipated failure modes. I still have a 1969 Motorola (Motorola was a major manufacturer of transistors) technical catalog that specifically says that for high reliability applications eg military, telecomms, germanium alloy power transistors should be selected, because the factory engineers had over 14 years experience making them, and how to make them so as to avoid failure was well understood. They were carefull to say that while they made their silicon mesa transistors (which eventually rendered germanium obsolete due to better performance) acording to best in industry standards, they had only 5 years experience and so had no right yet to claim the reliability was as good as their germanium product. Actually, I can if you like quote specific examples where cheap consumer grade parts turned out more reliable than the carefuly made professional equivalent. As always, human error in the factory was the main problem. You cannot predict human error. Wickwack 124.182.155.158 (talk) 11:08, 29 March 2013 (UTC)[reply]

I'll say this one more time: any human endeavour involves human error. The only was to ensure that a mechanism or electronic circuit will functionally last 73,000 years is to accumulate 73,000 years of testing, in order to detect the errors.

Wickwack 04:04, 27 March 2013 (UTC) — Preceding unsigned comment added by 124.178.132.17 (talk)

Why do the electronics need to stay warm? Absolutely nothing needs to be operational during the interstellar cruise phase. When the probe arrives at the target star, its solar panels can wake up the electronics simply by providing power. Also, what's the timescale for outgassing--will it still be outgassing decades after launch? --140.180.254.209 (talk) 00:11, 26 March 2013 (UTC)[reply]

I told you, above. Electronic parts are rated for a minimum permitted storage temperature (-55 C for military spec parts, various higher temeratures for commercial parts. If the parts are cooled below the allowed minimum temperature, they will be damaged (due to differential contraction of internal structures and many other mechanisms). When they are powered up, the minimum allowed temperature is higher. While it should be possible to design special parts to go lower than -55 C, designing for a low as the 4 K of deep space is just not on - a whole new technology would have to be invented. The problem is especially accute for solar panels, due to their large area, forgetting for the moment that atleast one panel will need to be exposed for the whole trip, to be blasted for 73,000 years worth of interstellar dust and micro meteorites. Wickwack 58.170.130.186 (talk) 01:04, 26 March 2013 (UTC)[reply]

I'd be principally concerned about being able to get there accurately enough for the star's light to power it up. Voyager deviated from its path simply because of thermal radiation and even a tiny deviation would be enough for failure. Dmcq (talk) 16:02, 26 March 2013 (UTC)[reply]

Something that no one has brought up yet: OK, so the deep interstellar probe arrives at is distant destination. What is the point if it can't send any information back? Voyager sends back info at about 160 bit/sec. That is not because the technology is now old but due the bandwidth-distance issue. The further the probe is away from the receiver, the slower the data has to be sent. A modern fast laser link would have its signal lost in the light noise from cosmic star light scatter. However, stars don't radiate that much in the radio frequency that Voyager uses, so, at present we are stuck with that spectrum. So, such a craft trying to send it data back to Earth would have to reduce its data rate to a very, very low speed. Since the first probe went to Venus the Arecibo telescope has radar mapped the surface from here on Earth (no probe needed). So its more than possible/probable that long before 73,000 years has passed this hypothetical probe would be an obsolete white elephant. --Aspro (talk) 21:53, 26 March 2013 (UTC)[reply]

Thanks to everyone who responded! The engineering discussion was fascinating. I searched up tin whiskers, and my mind was blown at the fact that such things can pop up from tin, and that we still don't understand why. --140.180.254.209 (talk) 07:14, 27 March 2013 (UTC)[reply]

If people work on electronics working on Venus at 500°C [2] the 4K long term storage environment will also be under evaluation. --Stone (talk) 09:51, 27 March 2013 (UTC)[reply]

Don't count on practical results any time soon though. Phase diagrams of metal and semiconductor alloys are simple at high temperatures, and complex at low temperatures. 500 C in absolute temperature terms is less than a 2:1 improvement over standard electronics. To go down to 4 K is a 55:1 improvement. Wickwack 121.215.47.30 (talk) 11:22, 27 March 2013 (UTC)[reply]

I cannot understand the concepts of dark matter and dark energy and their percentages of 22% and 74% respectively. I think there is no seperate matter nor energy. I think they are inter-related. So matter and energy in universe should be in the ratio 1:1. Even matter cooled to absoluate zero temperature should have its nuclear energy. Why there is a different ratio?--G.Kiruthikan (talk) 06:24, 25 March 2013 (UTC)[reply]

You CAN separate matter and energy. A rock is matter. Light and heat are energy. It's true that you can transform matter to energy and vice versa--for example, you can annihilate matter with antimatter and create photons, or smash protons together and create a shower of particles from their kinetic energy (which is what happens at the Large Hadron Collider). But that doesn't mean matter and energy are the same thing.

Now, the next question might be: so what's dark energy? The most precise not-too-controversial answer is that it's something which 1) has negative pressure (meaning it makes the universe expand faster rather than trying to contract it), and 2) behaves like the energy of empty space. In other words, its energy density stays constant as the universe expands, so if the universe gets 2 times bigger in volume, there's 2 times as much dark energy. Are there more informative explanations of dark energy? Yes. Do any of them work? No. --140.180.249.152 (talk) 07:05, 25 March 2013 (UTC)[reply]

Now, how in the name of God is this supposed to tie in with the law of conservation of mass-energy, which it directly contradicts??? 24.23.196.85 (talk) 07:57, 25 March 2013 (UTC)[reply]

Don't be confused because both words start with "m". Mass and energy are the same thing. Matter is not equivalent to mass. Mass is a property of matter, just as properties like volume, density, color, etc. are. --Jayron32 12:58, 25 March 2013 (UTC)[reply]

"Dark matter" and "dark energy" are just names, and not very good ones. There is no general definition of the term "matter" in physics. To cosmologists it means "nonrelativistic particles", which have the property that they dilute by a factor of x³ when the universe expands by a linear factor of x (or a volume factor of x³). "Radiation" is relativistic particles, which dilute by a factor of x⁴ instead. The "dark energy" dilutes by a factor of x⁰, which is not true of "energy" in general as cosmologists use that term. "Dark matter" is also inconsistently named because it can be relativistic, unlike cosmologists' normal matter. Relativistic dark matter is "hot dark matter" and nonrelativistic dark matter is "cold dark matter". So you shouldn't try too hard to make sense of the individual words in these compound terms. -- BenRG (talk) 16:22, 25 March 2013 (UTC)[reply]

(ec) Maybe I should have been clearer in my original answer. The fundamental distinction between matter and dark energy, as used in cosmology, is that the former's density decreases as the universe's radius cubed whereas the latter's density stays constant. So if the universe's radius gets 2 times as big, its volume is 8 times as big, meaning the density of matter is 1/8 the original. If you want, you can forget the distinction between matter and energy and consider all matter to be its equivalent rest mass energy. In that case, rest mass energy density dilutes as r^3 whereas dark energy density stays constant, which is why they have different observational consequences. --140.180.254.209 (talk) 16:28, 25 March 2013 (UTC)[reply]

Still not possible in the context of conservation of mass-energy -- this essentially means that as the universe expands, dark matter/dark energy is constantly being created from nothing at all! 24.23.196.85 (talk) 00:19, 26 March 2013 (UTC)[reply]

Energy conservation in general relativity is a difficult topic, one which I think it's safe to say that no one really understands. The problem is that energy is clearly carried by gravitational waves in practice (see Orbital decay from gravitational radiation), but you can't assign energy to a gravitational wave without breaking the equivalence principle. The only way to get an energy-conservation law that respects the equivalence principle is to define it on the boundary of spacetime, which is not where the energy seems to actually be. This seems to be related to the difficulty of formulating a sensible theory of quantum gravity, since quantum mechanics absolutely requires energy conservation. And you may know that recent research in quantum gravity has suggested that it might be naturally defined on the boundary of spacetime (gravitational holography and AdS/CFT). So, yes, the dark energy is created from "nothing at all", and the exact meaning of that isn't understood. It may actually show that the universe is finite in size and has a finite maximum entropy, since when the cosmological constant is positive (and only then), you end up with a cosmological event horizon at far future times, which only allows you to interact with a finite volume of the universe, and at even later times, everything inside decays away and you just have the horizon and Hawking radiation from it, which has a finite entropy that can never be exceeded. Classically, there's a huge exponentially inflating universe outside that you can't see, but it's not clear that that's true or even can consistently be true in quantum gravity.

In short, I don't know. -- BenRG (talk) 02:26, 26 March 2013 (UTC)[reply]

(ec, again) Good observation. Energy is not conserved in General Relativity, and dark energy, being the energy of empty space, really IS being created from nothing at all. In fact, matter is the only component of the universe that obeys energy conservation. The energy density of radiation goes down as 1/x^4, where x is the linear size of the universe. But since the universe's volume is proportional to x^3, total energy goes down as 1/x: the energy in radiation decreases as the Universe expands!

This can be explained by looking at Noether's theorem, which says that energy conservation is a result of the time-invariance of the laws of physics. In other words, if a law predicts the same result no matter what time it is, the law conserves energy. Immediately we see that the universe is not time invariant. There was an unambiguous past where the universe was smaller, hotter, and denser, and an unambiguous future where the universe will be bigger, colder, and sparser. For more information, see here, or here for a more technical treatment. --140.180.254.209 (talk) 02:39, 26 March 2013 (UTC)[reply]

This argument from Noether's theorem, which I've seen many times, makes no sense. Yes, the solution of GR used in cosmology breaks time-translation symmetry. It also breaks Lorentz symmetry. But no one would say that GR breaks Lorentz symmetry—it's just certain solutions that do. Likewise, GR does not break time-translation invariance; it's certain solutions that do.

Laws associated with a certain symmetry don't automatically fail in a solution that breaks that symmetry. Energy conservation is not violated in the presence of a time-varying electromagnetic field, even though that field has a geometrical interpretation as a gauge field. Depending on how you stated the law, it could be violated, but it doesn't have to be: you can save energy conservation in a way that also respects locality and the gauge symmetry, by assigning an energy to the field in a certain way.

The fact that you can't do this with the gravitational field in GR, not just in cosmological solutions but in general, does not follow from Noether's theorem as far as I can tell.

It's also weird how willing Sean Carroll is to discard the principle of energy conservation as though it were an optional part of physics. Presumably he wouldn't say that general relativity violates the laws of thermodynamics, even though energy conservation is one of those laws. And when he says everyone agrees on the physics, he's ignoring the quantization problem at the very least. Even at the classical level, although everyone may agree on the axioms, that doesn't mean anyone understands everything that's entailed by those axioms. -- BenRG (talk) 20:41, 26 March 2013 (UTC)[reply]

In order to find the tension in simple pendulum, we equal the net force in the centripetal direction (T-mg*cos(angle)) to mv**2/r. By doing so, we presume that the result of the combined forces (tension and gravity) is circular motion. Is there a way to find the tension without this assumption? Thanks. 109.160.152.227 (talk) 07:39, 25 March 2013 (UTC)[reply]

Since the pendulum is constrained to circular motion, what is wrong with assuming it? It will not be uniform circular motion, as it approximates sinusoidal velocity, but you only need the tension at maximum, which when the pendulum is centred. So you are really only asuming uniform circular motion at the centre position. Wickwack 121.215.146.244 (talk) 10:32, 25 March 2013 (UTC)[reply]

We know it is constrained to circular motion by experiment. Is there a theoretical way to prove it? 109.160.152.227 (talk) 11:13, 25 March 2013 (UTC)[reply]

We know it is constrained to circular motion by the definitions of a circle and of a pendulum. The rod constrains the mass to points a specific distance from the pivot. A circle is the set of all points equidistant from a specific point. 38.111.64.107 (talk) 11:53, 25 March 2013 (UTC)[reply]

This is a perfect case for the use of the Lagrangian mechanics formulation, allowing to solve for the motion by analyzing conservation of energy, instead of the Newtonian style representation of forces that are proportional to acceleration. The motion is constrained (by tension) but we don't know the magnitude of the tension force. We do know that energy is conserved. So, set up the Lagrangian; instead of setting the equation to zero, set it equal to the unknown constraining force, and solve for that as a residual. This is a standard homework problem in an elementary mechanics class. Nimur (talk) 15:57, 25 March 2013 (UTC)[reply]

This method described by Nimur is the same as the principle of least action where you take into account the constraint using a Lagrange multiplier. The Lagrange multiplier is then the tension in the rod. Count Iblis (talk) 17:04, 25 March 2013 (UTC)[reply]

You can model the pendulum arm (or any rigid body undergoing elastic deformation) as a spring with a very large spring constant. Then the pendulum bob is in principle free to move in two dimensions, but in practice won't move far from the circular region where the arm has its preferred length. The tension in the arm is given by Hooke's law, and is a function of the dynamical variables. You can derive the original (explicitly constrained) solution by taking the spring constant to infinity. -- BenRG (talk) 00:43, 26 March 2013 (UTC)[reply]

Thank you all. 109.160.152.227 (talk) 16:52, 26 March 2013 (UTC)[reply]

Is there a (chemical) reaction that can reverse copper acetate back to copper?Curb Chain (talk) 08:22, 25 March 2013 (UTC)[reply]

Sure, find an element that is higher on the Reactivity series than copper is, or (for a more advanced answer) one with a lower standard reduction potential than copper. What you are basically looking for is a metal that can act as a sacrificial anode for copper. That shouldn't be too hard, since copper is a fairly unreactive metal, so most other metals would work well for this purpose. --Jayron32 12:56, 25 March 2013 (UTC)[reply]

Reduction with zinc. Plasmic Physics (talk) 03:55, 26 March 2013 (UTC)[reply]

Can you give me a chemical formula?Curb Chain (talk) 09:14, 26 March 2013 (UTC)[reply]

And what reaction would reduce zinc acetate(?) back to zinc?Curb Chain (talk) 09:39, 26 March 2013 (UTC)[reply]

Reduce it with magnesium. Any metal more active than magnesium will react with the aqueous solvent the produce the hydroxide and gaseous hydrogen. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 15:36, 27 March 2013 (UTC)[reply]

Acetate + H
₂O is a solvent?Curb Chain (talk) 00:55, 28 March 2013 (UTC)[reply]

One problem you may be having is thinking that the acetate would be involved in the reaction. In most cases (especially in aqueous solution) it probably wouldn't. The reaction you're dealing with is Cu⁺+X(s) -> Cu(s) + X⁺, or equivalent, where X is an element higher on the aforementioned reactivity series. If you're attempting to reduce zinc instead of copper, it's the same reaction, but with a more active metal. - So how do you reduce a metal high on the list? Typically electrolysis (in a non-aqueous solution) is used. For example, the Castner–Kellner process to make metallic sodium or potassium, or the Hall–Héroult process to make aluminium. Note you can also use electricity to directly reduce copper and zinc (see electroplating). -- 71.35.109.213 (talk) 16:02, 26 March 2013 (UTC)[reply]

Has any manufacturer of electric cars ever thought that for tanking an electric car, the easiest way is to change the battery? That would require building a network, that would need to keep a stock and track who's got what, but what's the alternative? Waiting several hours seems unacceptable for most drivers. OsmanRF34 (talk) 11:19, 25 March 2013 (UTC)[reply]

Our electric car and charging station articles mention this as an alternative to recharging. Better Place was one of the first providers of battery swap services. Gandalf61 (talk) 11:35, 25 March 2013 (UTC)[reply]

And you avoid much of the bookkeeping by making the battery not part of the car, but a lease (preferably from the maker, who will also recycle and recondition them). --Stephan Schulz (talk) 11:40, 25 March 2013 (UTC)[reply]

And if some turkey comes to me offering shares in a battery swap company targeting electric cars, I'll show him the door right smartly, for these reasons:

• The service life of a battery is roughly halved for each 10 C increase in temperature. If the company has swap stations in the southern areas of Australia (22 C typ), commercial pressure will make it charge ($ charge not electric charge) on the basis of a battery life of many years. Then some drivers will go on a tour up to the top end, where the temperature is around 40 C. The batteries will come back stuffed in 1 year.
• The batteries have to be charged (electric this time, not $) - presumably from the grid. Now, modern automotive diesels operate at about 40% efficiency. Power stations can do a bit better, up to 45% efficiency. Losses in the grid and local electricity distribution bring it back to about 40%. Add in the battery charge efficiency, always less than 100%, and the business is on the wrong side of marginal.

At the very least, the electronic controls need to be swapped out with the battery, otherwise the battery swap company is at risk from cars with faulty controls damaging the battery.

Then there's the safety aspects of cars with a ton of battery on board - that will make every collision like hitting a big block of cement at 2x speed, with the added danger of electrical fire etc.

Wickwack 124.182.153.240 (talk) 13:09, 25 March 2013 (UTC)[reply]

I don't agree with any of those concerns.

• The issues of battery life versus temperature are the same issues that all sorts of companies have with leasing all sorts of equipment - and they somehow manage. When you rent a car, they can't tell whether you're going to rev the thing to redline continually, brake like crazy, scrub the tires and so forth...but somehow they manage to stay in business. Same deal with the batteries. It can be managed.
• The batteries have to be charged - yes, but when you talk of the efficiency of diesel engines, you're assuming that the world continues to use fossil fuel to generate electricity. It's not about efficiency - it's about global warming. If your electricity can come from wind, solar or nuclear - then the CO2 footprint of an electric car is close to zero...vastly less than for a diesel. So simple fuel efficiency isn't the issue. Plus this comment is off-topic - we're talking about recharge mechanisms - not whether electric cars are a good or a bad idea.
• Safety aspects are also a ridiculous concern. The Prius has a bunch of batteries in the trunk, so do any number of electric vehicles. Sure batteries weigh a lot - but electric motors are small and light. Compare that to a tank of highly inflammable liquid and the huge engine with all of it's cooling equipment!

The real problem with battery-swap stations is the initial cost. A gas station on a US freeway has half a dozen pumps and usually about half of them seem to be occupied, and it takes a minute or two to gas up and get going - so as a VERY rough estimate, I'd bet they have around five customers per minute...on the average. Of course gas powered cars have a range of around 400 to 500 miles - and electric cars are about 100 - so you need to swap batteries (say) 4 times more often than that. If each of them needs a complete battery swap and a battery takes (say) three hours to recharge - then they need to keep at least 3x60x5x4=3,600 battery packs in stock and on charge at all times to keep up with "average" demand. Probably they should double that to allow for peak demand (Nobody wants to show up at a gas station that has no recharged batteries in stock!), so let's say you need 7,000 battery packs at every garage, continuously on charge. According to Nissan Leaf, the cost of a battery pack for that car is $18,000...so every gas station has to have about 126 million dollars worth of batteries in stock and on-charge at all times! There are about 170,000 gas stations in the US - so upwards of $20 trillion dollars worth of batteries have to be purchased, stored and maintained in order to have a viable battery-swap system in place.

SteveBaker (talk) 16:00, 25 March 2013 (UTC)[reply]

I think a realistic range is more like 200 miles - see Tesla Model S. Also, at least when I was in the US, often 90 % of pumps were empty. So there is a factor of about 10 in your calculation. Fuel stations may be more occupied in high-population areas, but there people drive, on average, shorter distances, so they can refuel at home overnight. Also, of course, this infrastructure does not need to be there in one step. Especially with modern electronics and online services, its much easier to dynamically direct cars to stations that have batteries, and batteries to stations that don't. This is only slightly less convenient than having three fuel stations on every crossing. What I see more as a practical problem is a lack of standardisation of batteries, and the additional constraints on the car construction to make the battery easily replaceable. This may shake out over time, but it might take a while. --Stephan Schulz (talk) 16:18, 25 March 2013 (UTC)[reply]

I stand by my numbers:

• I don't think you're correct about the average occupancy of pumps in US gas stations...but in any case, the number of batteries they'd need to stock would have to match the peak customer demand - not the off-peak demand. If the peak customer demand were any less than the number of pumps out front...why would they have built that number of gas pumps? Whatever calculation they use to decide how many pumps to get - that calculation applies identically to the number of batteries they'd estimate they'd need. So if we're being careful...we should say that the number of batteries a station would have to keep in stock would have to be equal the number of gas pumps they have multiplied by the peak number of customers per hour per pump multiplied by the number of hours it takes to recharge a battery.

• Yes, there are lots of gas stations close to each other at many intersections - but that's how much refuelling capacity we'd be replacing. Sure, you could halve the number of gas stations - but each one would get twice the amount of business - so the number of batteries that would have to be stored and recharged overall is exactly the same.

• The 100 mile electric vehicle range that I used is a very good estimate. The claimed 200 mile range of the Tesla is *very* atypical. With the 85kWh, 200 mile battery pack, the Tesla is a high end super-car and costs $95,000(!) - the slightly more affordable 40kWh version has only a 120 mile range and still costs $59,000. The Nissan Leaf is a more realistic kind of car that the majority of people are more likely to own - and according to the EPA, it has a 75 mile range...so even worse than I estimated.

• The idea of finding an alternative (nearby) gas station in the event that the one you want is out of batteries right now could be very bad for consumers. If you have to replace your batteries every 75 miles (the Nissan Leaf number) then having to drive 10 miles out of your way to find another battery swap station starts to look VERY unattractive! On my recent drive from Texas, through New Mexico and into Arizona, we were getting kinda low on gas and my GPS told me that the nearest filling station was 18 miles away and the second nearest was close to 30 miles away. With a gas-powered car, that's OK - 18 miles is well within my "reserve" and the odds of that gas station being out of gas is almost zero. But for an electric vehicle, there had better be a charged battery at that first station because if you have to drive 30 miles to get to the second closest one, then by the time you get back onto the freeway, you'll only have 45 miles of juice left and it'll be time to turn around and drive back again! With shorter vehicle ranges, you need MORE filling stations - not less!

• You might be correct about not needing battery-replacement-stations in built-up areas - but I have to guess that the battery-replacement companies are going to be strongly resistant to you using their batteries and recharging them yourself. If you're going to spend trillions of dollars on these replacement battery stations, you *REALLY* want people to pay to use them! So I'd expect we'd find some kind of computer-interlock thingy built into the battery that would stop you from recharging your Exxon battery yourself. This is potentially fixable by legislation...but it's tricky.

I agree that battery standardization is a tough problem. Honestly, I think hydrogen powered vehicles are more likely to be in our future. SteveBaker (talk) 20:38, 25 March 2013 (UTC)[reply]

I think we may be talking about different things. I talk about a reasonable system that can be constructed with the best known technology - I assume that we can get Tesla quality down to the masses. If you want to drive through large empty spaces, you don't want a Nissan Leaf. But you don't want a Smart for that today either. As for finding batteries: You assume your current workflow (drive to fuel stations until you hit one that has fuel). I'm thinking about a navigation system that knows where to find available batteries at any given time, and that will automatically guide you to the most convenient location on demand, and might even call ahead to ensure they hold one for you. Also, there is no reason why you cannot recharge the car from a reasonable electrical outlet in pinch. Sure, it takes more time, but it won't let you get totally stranded. "Fuel" stations will need high-powered electricity for recharging anyways, so you can get a 100 amp line to the recharger easily. Similarly for the business model. You can simply lease the batteries for a monthly (or mileage-based) fee, with electricity provided for free. Then whenever you reload at home, the batterie company saves a charge (but the actual electricity is small change compared to other costs, anyways). Something similar works e.g. for cell phone plans, so why wouldn't it work for batteries? --Stephan Schulz (talk) 23:58, 25 March 2013 (UTC)[reply]

Regarding Steve Baker's first set of comments:-

• Relavence of temperature dependence: The service life of batteries roughly halves for each 10 C increase in temperature. The service life of an internal combustion engine is also temperature dependent, but not any noticeable degree. Most wear occurs during warmup, whcih is shorter in hot climates. Noboby goes around saying how much they are dissapointed their car engine lasted only a year or two because they live in a hot climate. So your argument here is a nonsense.
• Claimed zero CO2 footprint for electric cars. It never ceases to amaze that people have such a head in the sand approach. The reality is that almost all electric power is generated from either fossil fuels or nuclear, and nuclear for lots of reasons is a solution only for countries with very large populations and/or a need to hide production of weapons grade material. Things will remain this way for the forseable future. Why do you think photo-voltaic panels are so expensive, when not subsidised by stupid governments? It's a lot to do with that they have to be made with about the same order of electric energy consumed in the factory as they will produce in their service life. Having the CO2 emitted by a power station in China, where the panels are made, instead of from each car, IS NOT a solution to the World's pollution and greenhouse issues.
• Safety aspects: You must be joking. How many fuel tank fires/esplosions actually occur? Personally, the only ones I know about occurred in Ford Pinto cars about 30 - 40 years ago, and it is understood these cars had an easily corrected design flaw, and there were only a handful of Pinto fires anyway. People who think there is not a safety aspect of a added huge mass are also "head in sand" types. And batteries also cause electro-chemical fires when physically damaged. You need to take into account that while battery cars are very much in the minority, and are still new (which means not driven by less carefull drivers) they won't show up in accident statistics. If battery cars were most or all the cars on the roads, it will be a very different story. This is similar to the experience of our local bus company, who have have been slowly converting their fleet to LPG fuel on government subsidy (LPG being considered a little more environmentally friendly than diesel, and something we have a lot of in this country). They now have one-third of their fleet on LPG, about 500 busses. At first, everything seemed good, but with 500 busses on the road every day they now have about 1 bus fire per 3 months, usually totally destroying the bus. This is just not acceptable. They've been lucky that passengers were able bodied, few, and were able to leave each bus in time. Fires in their fleet of ~1400 diesel busses were about 1 each 10 to 15 years, with none totally destroyed.

Wickwack 58.170.130.186 (talk) 23:45, 25 March 2013 (UTC)[reply]

If the batteries are provided by the car maker as a lease, or by the fuel station network as a lease, different battery lifetimes are irrelevant. Whoever operates the system simply must average cost over all driving conditions. Not trivial, but something that businesses do all the time. And the time that solar panels used about as much energy as they provided are long in the past. Current generation photovoltaics are very much energy-positive. As are other alternative energy sources, like wind, hydro, solar-thermic, and so on. Denmark produces 20%-30% of electricity with wind turbines. Norway produces 99% from hydropower (but lends some of its capacity to Denmark for storage). Germany has about 20% of electricity covered by renewables. --Stephan Schulz (talk) 00:08, 26 March 2013 (UTC)[reply]

As I have shown, batteries are different from any other technology in that their life halves for each 10 C rise in temperature. The asme battery that lasts 5 years in the southern extremity of West Australia will last only one year at the top end, usage patterns being the same. Averaging over all conditions works in most fields because they don't have this sort of variation. Most countries don't have rivers suitable for hydro power to any extent. Germany has a lot of nuclear power - they can afford it with their big population, but it has become a political liability.

Wind has its own problems. It doesn't blow with total relaibility, so power companies using it have to have spinning reserve covering 100% of the wind gneration capacity. Our local power company has some large wind power farms as the Government made them do it. The spinning reserve is diesel, as their base load coal fired power stations cannot take sudden large changes in load. Big diesel engines cannot tolerate running long periods at idle, so they have to have a base load on the diesels at all times, at less efficiency in CO2 footprint than their base load coal fired stations. That considerably destroys the CO2 advantage of wind power.

You see a pattern in all this? The pattern is that all these alternative energy sources (wind, photo-voltaic, even nuclear) have to be government forced, and almost always government subsidised. Why is that? Because they are not commercially viable. Why are they not commercially viable? Because they don't much reduce fossil fuel power generation, or evironmental impact, they just shift it somewhere out of sight. It suits China at the moment to sell us solar panels made with their cheap labour and their cheap coal-fired electricity running of coal mined without much regard to OH&S. It will be interesting to see what happens 10 years from now - As their standard of living rises up to Western Standards, and their labour ceases thereby to be cheap, and their Government gets on top of their bad health and environmental issues, you might find they won't sell us cheap panels any more. Wickwack 58.170.130.186 (talk) 00:51, 26 March 2013 (UTC)[reply]

Your first point is not about technical, but about business questions. Businesses regularly deal with a factor of 5 in cost. There are many McDonalds in downtown Manhattan. The rent there is a lot more than 5 times higher than the cost in, say, Seward, Nebraska. What's more, McDonalds is maintaining a number of restaurants in Paris, with high rents and nobody who want to eat there. It's part of their strategy to offer (nearly) the same experience for a similar price all over the world. If you talk about the number of McDonalds restaurants, or mainstream cars, differences simply average out. Your second point has some valid arguments, but it very much overstates the case. Even today, there are ways to handle the variability of wind. It's not free, as many people assumed, but then there is no such thing as a free lunch. You need better electricity networks, you may need high-voltage direct current to connect larger areas so that local effects can, again, average out. You can use pumped-storage hydroelectricity in some areas, you can use molten salt storage with concentrated solar power as Desertec plans to do, and so on. There are some technical problems, there is a price, but there is nothing we cannot do on a technological level even today. Yes, alternative energy currently works best with government subsidies. But which form of energy does not? Nuclear has beed largely developed out of the public purse. Who do you think builds the roads tankers drive on distributing fuel over the country? And so on. And we haven't even started on externalised costs. --Stephan Schulz (talk) 06:30, 26 March 2013 (UTC)[reply]

Yes, building rental in cities will be 5 times that of small towns, but what fraction of the amortised cost per burger is rent? My guess is that labour is a very much greater cost, even though they seem to employ mostly low wage teenagers. One MacDonald's shop looks the same as another, and the floor area seems unrelated to trading volume, but their staff levels do seem in proportion to the trading volume - up to 20 or so in busy stores at busy times, only 3 or so in quiet stores (what they call a front-of-house person aged about 15, a back-of-house person also about 15, and a duty manager, all of 17! Good training though - I've learnt from experience that if you hire an ex-MacDonald's person, you get a hard working customer pleasing worker ) A battery lease company will have as its' major cost the cost of the batteries. As I said before hydro power is only for those countries fortunate in having suitable rivers and gorges to dam up. Most don't. The same applies to hydro storage, which is well established (many decades) but only in those locations suitable. You can't be serious about regarding highway costs as a subsidy to fossil fuel generation, as all sorts of other traffic uses the same roads in far greater volume. Things like molten salt storage are just good ideas, not proven established technology. One of our local universities (Murdoch) put a lot of time and money researching phase change storage (a la glauber's salt and similar), but they were stumped by the problem of supercooling. Supercooling can be virtually eliminated by adding in an evenly mixed dispersion of some non-reactive substance, but it tends to settle out a little bit each cycle, so after a finite number of storage/release cycles, the dispersant is all in the bottom, and the salt will no longer change phase. Wickwack 121.221.236.225 (talk) 07:53, 26 March 2013 (UTC)[reply]

A few comments:

1) Price of batteries should come way down when they are made in the quantities envisioned here.

2) Electric cars probably aren't suitable for Australia, what with the high temperatures and long distances. At most, they could be used as "city commuter vehicles", where each is limited to one city, with no infrastructure for recharging them between cities.

3) Recharging will likely occur during off-peak hours, where electricity prices are lower. StuRat (talk) 02:51, 26 March 2013 (UTC)[reply]

Re (1): I doubt it, as batteries are already made in large quantities in automated plants, and the price pretty much reflects the materials consumed.

Re (2): You got that right. And a high proportion of people buy cars suited for that weeked away, or visiting Aunt Joan in another city. Not entirely rational, as 99% of the time they are just commuting, but that's what they do.

Re (3): True. Most Australian power companies do not offer off-peak concession rates (the ones that do have turned it around - they charge a premium for on-peak use), but that is merely a policy decision. If a customer uses more than 50 MW.hour per year, deals get negotiated. I negotiated 6 cents/kW.hour for my employer for 10 PM to 5 AM consumption, several rates ending up at 10 cents/kW.hour for peak hour consumption, whereas the same power company charges a flat 12 cents/kW.hour to homeowners. Wickwack 58.164.230.22 (talk) 03:36, 26 March 2013 (UTC)[reply]

2) Gasoline/electric hybrids might be more suitable for Aussieland, providing electric power while in cities, and gasoline to bridge the distances between them. StuRat (talk) 23:38, 26 March 2013 (UTC)[reply]

Indeed. The local taxi industry is trialling hybrids. Yep, you've guessed it - there's a government subsidy to taxi owners to buy them - not a good sign. It will be interesting to see if the taxi industry continues with them when the subsidy and trial officially ends. In taxi use, there is a small advantage: as taxis make a lot of short trips, dynamic braking (slowing and braking the car by using the electric motor as a generator to recharge the battery) results in extended range without starting the engine. People accept a higher noise level and the odd shudder in a taxi than they will accept in their own car, so the engine can be a more efficient diesel instead of a gas engine. Wickwack 124.178.132.17 (talk) 01:43, 27 March 2013 (UTC)[reply]

Wickwack - you seem obsessed with the subsidies provided to renewable energy sources. Given that coal miners and gas producers also receive subsidies, what's the real balance point here? HiLo48 (talk) 04:41, 27 March 2013 (UTC)[reply]

I guess I can seem obsessed with subsidies - but that is because subsidies prop up things that are not commercially viable. And if its not commercially viable, it's probably not a good idea, for the reasons I have given.

I thank you for your question. My view is that our Govt's decison to subsidise coal mining ranks amongst the most bizare, stupid, and disgusting decisions they have ever made, and they have made quite a few stupid decisions. It's stupid and bizare because of all the forms of energy, coal is the dirtiest, most polluting, and most greenhouse worsening. If the Austrlain Govt was serious about reducing greenhouse emissions, they would shut coal mining, and coal fired power stations, down, not subsidise them. Trouble is, that would put a lot of people out of work, and they include folk with political clout. It's stupid because the Govt are imposing a "carbon tax" allegedly to induce us to reduce our energy consumption. But in case that works, they subsidise/compensate the coal industry. This is no different to saying the people eat too many chocolate bars, its making the peasants fat, so we'll put a tax on chocolate, so the price goes up. Ooops, the chocolate industry say that they'll loose jobs as people will eat something else, we better prop them up with a subsidy. There's a simple rule about subsidies - they mean something is wrong.

Generally, subsidies prop up industries that are nor comercially viable. The coal industry IS viable, even with the carbon tax (I haven't turned the lights off at home because the carbon tax put electricity charges up - I just carry on and grumble) - it does not need a subsidy. — Preceding unsigned comment added by 121.215.47.30 (talk) 06:39, 27 March 2013 (UTC)[reply]

There should not subsidies for solar power, and there most certainly should not be subsidies for the coal industry either. If something is accepted as being bad, legislate to reduce or prevent its' use. People will find an alternative.

When it was realised that freon ruined the ozone layer, we didn't put a tax on it to discourage use, and then subsidise the freon industry to prevent job loss. If we had, we'd still be using freon. We simply said, 30 years ago, the freon factories can get stuffed, it's now banned. We should do almost the same thing with coal. It's use is polluting, and it makes the greenhouse effect worse. We should just have a law progessively reducing the quantity that is allowed to be used, and let coal miners find something else to do. Or, if you don't accept climate change, just leave it as it is. Lots of people loose their job - and find another. Been there myself.

The gas producers have claimed, rightly, that while gas is a greenhouse promoter when burnt, it is not quite as bad as coal, and it burns clean. Therefore, given that we are economically tied to using hydrocarbon fuels anyway, we should try to use more gas and less coal. This is a whole subject on its own, so I'll leave it for now.

Some subsidies are ok. The Govt has subsidised indigenous radio stations and the indigenous (black people) music industry. This has had a significant factor in reducing alchoholism and violence in remote areas by providing folk with something to do, and get employment. I think it is a good thing.

Wickwack 121.215.47.30 (talk) 05:53, 27 March 2013 (UTC)[reply]

This is off topic and belongs at Humanities. But I should at least briefly point out that politics is about more than markets - military and economic war need to be considered. It is conceivable that if Australia lets its domestic coal production wither, it could be cut off from its normal energy imports, whether by military action or through the connivance of some cartel that demands the prime minister signs documents in recognition that Master Blaster runs Bartertown. Of course, I doubt that, and have great suspicion of the coal companies; my point is that this is valid political discussion that someone over at Humanities might give you a lot of insight about. Wnt (talk) 14:31, 28 March 2013 (UTC)[reply]

You are entirely correct in saying that politics is about more than markets. However, it is well known here that the coal mining industry and its unions are an effective lobby group. If coal has a strategic importance, and it may well do so, it's best left in the ground in peace time. Once we've dug it up and consumed and sold it, it's gone. If it's in the ground, we can resume digging it up again at any time - it's not rocket science. Our Govt cannot be concerned with any strategic value of coal reserves, as we export considerable tonnage to China. Australia's power generation history is cyclic, changing from building new power stations to run on coal, then oil, then back again a couple of times, as the price of oil went up and down, and new pollution control requirements made coal slightly less economic. They even converted some existing power stations from one to the other and back again. Australia is well blessed with natural gas, which can, and is, used to run power stations. But due to political messing about, only a fraction. It should be most. It's quite easy to build power stations to be dual fuelled, so they can burn oil when that's cheap, and burn gas when that is cheap. Wickwack 124.178.52.113 (talk) 00:21, 29 March 2013 (UTC)[reply]

Hi. I am in the completion stages of writing a chemistry lab report, and so this may be the first time that I'm asking for help on an assignment problem on the Reference Desk. I will show an attempt to complete the question myself in order to demonstrate which part I am stuck on. All logs are in base 10.

Part 1

I am asked to calculate the theoretical pH after adding 25.0 mL of 0.120 M ammonia with 12.5 mL of 0.120 mL HCl. The pK_b of ammonia is given as 4.75. Here are the steps I follow to get the pH of this mixture.

Given that the pK_b = 4.75, K_b = 10^-4.75 = 1.78 x 10^-5. I then use the equation that in solution K_b = [OH^-]² / [NH₃], and so [OH^-] = 1.46 x 10^-3 M. The original pH of ammonia solution without any addition of HCl is equal to 14 - pOH, where pOH = -log(1.46 x 10^-3), so pH = 11.2. So far, so good.

Here's where the problem arises. I take the [OH^-] concentration in total solution as the original molarity multiplied by a 25/37.5 ratio, giving [OH^-] = 9.73 x 10^-4. I then assume 100% dissociation for HCl, so that [H₃O⁺] = [0.120 M] x 12.5/37.5 = 0.04 M. I then subtract one from the other, so effectively the new [H₃O⁺] is equal to 0.039 M. Taking pH = -log(0.039) = 1.41, that is the value I get for pH.

Here's the real dilemma. According to my problem sheet, the combination of these two solutions occurs at the half-neutralization point, which creates a partial conversion of NH₃ into NH₄⁺. Do I have to take into account the conjugate acid, NH₄⁺? I have read that at the half-neutralization point, pH = pK_a, and pOH = pK_b. That means pOH = 4.75, and so pH = 9.25. This makes sense from a "half-neutralization" standpoint, but it contradicts the previous answer. Uh oh! Where is my calculation error?

Part 2

As if the previous problem didn't make me sound stupid enough, here's the next issue. I analyze a titration curve, finding the equivalence point of a reaction between aqueous acetic acid (weak) and sodium hydroxide (strong) to be pH~8.3. I use the volume of titrant NaOH added, rather than the pH value, to calculate the concentration of unknown NaOH. Here's method one.

K_a of acetic acid is 10^-4.76 = 1.7378 x 10^-5 = [H₃O⁺]² / [HA]. Since [HA] = 0.1056 M originally and is diluted to a 1/7 solution (25 mL original acid plus 150 mL deionized water), effective [HA] = 0.01509 M. So, I get [H₃O⁺] = 5.12 x 10^-4, which I assume to be equivalent to [OH^-] in the solution at the equivalence point. (Hmm...have I forgotten to account for pH>7?) At this equivalence point, I find that 22.6 mL of NaOH solution at initial unknown concentration is added. I use the concentration-volume formula, with the product of [OH^-] concentration with the total volume 0.1976 L divided by the NaOH titrant volume, to find that the stock solution of NaOH has a concentration of 4.48 x 10^-3 M, which is quite low.

The second method I use takes the same approach, except this time I assume no dilution took place, and I use 0.1056 M as the original acetic acid concentration, so [OH^-] becomes 1.35 x 10^-3 M, and the total volume of mixture is 0.0476 L. This time I get [NaOH] = 2.84 x 10^-3 M, within an order of magnitude of the first calculation.

The final method I use takes the formula K_a = x² / ([CH₃COOH] - x), where "x" is the original [NaOH] concentration. For this, and using the original [CH₃COOH] = 0.1056 M, I get x = 1.346 x 10^-3 M. I take the average of the three values to be [NaOH], but I still doubt that I did this part correctly.

Thus, I have demonstrated above that I have attempted to solve the questions without answers that make sense. Please enlighten me as to my erroneous methods used in calculation. Thanks. ~AH1 ^(discuss!) 13:55, 25 March 2013 (UTC)[reply]

By your own admittance this is a Homework Question. We can not pass your exam for you.--Aspro (talk) 16:04, 25 March 2013 (UTC)[reply]

From the Notes, "We don't do your homework for you, though we’ll help you past the stuck point." I am not asking you to solve the problem for me, but to point to any relevant articles or concepts that may help. I have specifically gone through my thinking process, highlighting areas of difficulty, as it's always been done for homework questions, although this is my first time trying to get help. Have the rules now changed? I am now trying to get past the stuck point, by solving "Part 1" in a different manner, but would still appreciate any kind of guidance. ~AH1 ^(discuss!) 16:28, 25 March 2013 (UTC)[reply]

• For #1, it is even easier than that. At the 1/2 neutralization point, pH of the mixture = pKa of the acid form of your conjugate pair (this is a consequence of the Henderson-Hasselbach equation). So, if you know the pKa of the ammonium ion (and you do if you know the pKb of ammonia, since pKa+pKb = 14), then you know the predicted pH at the half-neutralization point. --Jayron32 16:30, 25 March 2013 (UTC)[reply]

• But I already demonstrated (in paragraph 5) that I was capable of applying this knowledge to finding the pH at the specific half-equivalence point. Thank you for confirming that for me. Though how do you find the pH for other ratios, such as a small amount of strong acid, at the full equivalence point, or with a large amount of strong acid...do I use the equation in paragraph 4, a variant of the equation I gave in paragraph 9, or neither? Thanks again. ~AH1 ^(discuss!) 16:39, 25 March 2013 (UTC)[reply]

Edit: Wait...let me try solving it using pKa. ~AH1 ^(discuss!) 16:42, 25 March 2013 (UTC)[reply]

OK, so I just got the same pH value, 9.25. I'm now trying to verify my general method by somehow setting up the equation so I have a surplus of basic ions, which is proving rather difficult. How do I go about doing this? I'm going to try it first with the simplest example. ~AH1 ^(discuss!) 16:52, 25 March 2013 (UTC)[reply]

AH. This, by the way, is THE classic acid-base titration problem. Just about every general chemistry class uses it. You've got several ways to solve these problems, depending on where you are at in your titration:

• If you are calculating the pH of the solution before adding ANY strong acid, then you are just calculating the pH of a pure weak base, and this is a simple Ka problem. If you are calculating the pH at any point between the first drop of strong acid and the equivalence point (including the 1/2 equivalence point, as noted above, which is a special case) you use the Henderson Hasselbach equation. You need to do a bit of stoichiometry to figure out how much at any one point is in the acid form, and how much is in the base form, but that is very easy as the moles of acid = the moles of acid added, and the moles of base = the moles of base initially - moles of acid. Just plug those numbers into the HH equation and solve for pH. If you are calculating the pH the exact equivalence point, that's the pH of a weak acid; keep in mind that the moles of NH4+ are going to be the same as the moles of the NH3 at the start of the reaction, BUT the volume is now the TOTAL volume (you've diluted it some). Otherwise, you calculate this the same way you calculate the initial pH, but for the acid form and not the base form. If you go beyond the equivalence point, you now have an excess of the strong acid. You're just calculating the pH of a strong acid, which is just the moles of strong acid/total volume, and then take the -log of that. --Jayron32 21:53, 25 March 2013 (UTC)[reply]

I think I've finally figured it out. For the general method, if 3.00 mL of acid is added to the ammonia solution, then the pH is 10.1. Am I right? Thanks, everyone. ~AH1 ^(discuss!) 17:05, 25 March 2013 (UTC)[reply]

Side note: In order to solve my problem, I needed to represent NH₃ as NH₄⁺OH^-. ~AH1 ^(discuss!) 17:08, 25 March 2013 (UTC)[reply]

Not important, the stoichiometry is the same either way. --Jayron32 02:50, 26 March 2013 (UTC)[reply]

Here is a research question I am trying to answer for my project. Do ocean floor borrow pits (these are created by dredging) have hypoxic or anoxic conditions? Do they encourage the surrounding area to be hypoxic or anoxic?--anon — Preceding unsigned comment added by 99.146.124.35 (talk) 23:27, 25 March 2013 (UTC)[reply]

I don't see why they should, unless the dredging somehow increases biochemical oxygen demand. Of course I'm not an expert on marine biology, so anyone with more definitive info is welcome to contribute. 24.23.196.85 (talk) 00:05, 26 March 2013 (UTC)[reply]

Not an easy question to answer. Dredging for the purposes of deepening a channel where there is flowing water means that the water would only become slightly hypoxic as it flowed by (until the organic mater had decomposed). But if it is a blind trench or pit below the depth where the wave action above would cause mixing (I.e., well below sixty odd feet), then it could become truly anoxic and a death trap for any animal venturing into it..Aspro (talk) 00:29, 26 March 2013 (UTC)[reply]

The sea floor is largely a uniform 4 degrees C (or slightly lower) and fully saturated with oxygen unless there are other specific local and recent factors involved as Aspro and IP 24 mention. [3] Dredging itself is mechanical and would not deplete the oxygen nor would it result in permanent deoxygenation. μηδείς (talk) 01:16, 27 March 2013 (UTC)[reply]

Yes, it is not the dredging process itself but oxidation of freshly exposed hydrocarbons in the deeper sediment that causes the water to lose free oxygen. Carbon dioxide (and possibly sulphur dioxide) laden water is heavy and it does not want to diffuse upwards. Deep wells and mine shafts are higher in these gases, there have also been fatalities when people have gone down into poorly ventilated legs of oil platforms. Once, critters start to succumb to these traps, their bodies too, will help deplete the free oxygen and so possibly maintain a permanent anaerobic environment – if there is no mechanical means of mixing with the higher waters. The OP needs to define the 'ocean floor borrow pits' with enough precision for us to gauge the actual conditions. Out of interest: what is the brief your project? Aspro (talk) 13:46, 27 March 2013 (UTC)[reply]

An Ocean floor borrow pit is a hole created in the ocean by the removal of sand from the ocean floor. What do you mean by "enough precision"? These holes can be any width and any depth, though it is possible that deeper (or wider)holes would be more favorable to hypoxic conditions. --anon — Preceding unsigned comment added by 99.146.124.35 (talk) 23:27, 25 March 2013 (UTC)[reply]

I am trying to answer this general question for my honors program thesis (60 pages of writing) "is coastal armoring and beach nourishment justifiable?". I am studying Earth Science and Political science, so I am looking at this question from many angles. One of these is how beach nourishment effects marine species.--anon — Preceding unsigned comment added by 99.146.124.35 (talk) 23:27, 25 March 2013 (UTC) — Preceding unsigned comment added by 149.152.23.48 (talk) [reply]

Well, dredging is one thing, excavation is another. Dredging is going to leave relatively shallow tracks whose sides will collapse in on them assuming a coarse-grained substrate near the shoreline. There may be some immediate local decay and deoxygenation, but this should rectify itself very quickly. Immediate destruction of reefs, clam beds, and other physical disruption of animal life, not the creation of long-tern anoxic environments will be the damage done. It's not possible to give a generalized prediction, since local conditions will vary greatly. μηδείς (talk) 10:13, 28 March 2013 (UTC)[reply]

Ok, so you are aware that there are pro's and con's to beach nourishment. The beach front hotel owners and business might find it 'justifiable' in order to provide them with a living in the same way that a farmer may want to drain marsh in order to grow more crops but in doing deprive the local wild life their habitat. Yet traditional ecology of an eroding beach will change anyway. Shallow water dredging (which is cheaper) supplying such a beach, may (I think) bring it back to something close to what it was before. Together, with a similar ecology (although maybe not exact due to any different mineralogy of the new deposits. Coastal areas increase tide hight (because the water gets shallower). Therefore, it think that the increased oxygen demand will be redly met by the volume of water in movement. So hypoxia would not be a long term problem in my view. Anyway, the resort owners would not want their reseeding golden beaches replaced with dredged material full of mud and stuff. I'm personal not against Armoring because it often provides extra niches for other life. As a thought: it might be worth googling for companies that do this sort of beach reconstruction and emailing them. In this enlighten (?) day-and-age they are probably required by law (in most of US and Europe) to consult with ecologists before starting such work. They may have a lot of information about this. A thesis that shows that the author has gone to this trouble would -I think- have more impact on the examiners. If the reclaimers like you, you may even persuade them to let you go aboard one of their dredgers and take some photographs. These little opportunities are what often comes from doing research – if you don't ask you don't get. P.S. If you do get any photos please upload them to Wikimedia Commons... we need them.--Aspro (talk) 18:25, 28 March 2013 (UTC)[reply]

Robert E. Loveland, retired, of Rutgers University would be an excellent go-to if you can contact him. He was the head of their coastal marine biology research in the 90's, but last I heard they were reorganizing and shutting down some of his projects. If he's still able to be contacted he was big on studies of the Jersey coastline especially in regard to its effect on horseshoe crabs.