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About TWJian

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  1. paved88: You are welcome. Pleased to be of help. Cuthber: Ammonia deprotonates water to form NH4+ and OH-. Therefore: If "Most of the ammonia will exist as NH3 within the solution (Which is written as NH4OH rather then NH3+H20) Then "Since the concentration of OH- is low, ammonia is a weak base" Anything wrong with that? Unless you are quoting me out of context as: Since the concentration of OH- is low...<missing text>, most of the ammonia will exist as NH3 within the solution (Which is written as NH4OH rather then NH3+H20). Please don't reverse my premises to get a (substantially irrelevant as well as non-existent) contradiction. As for you: You seemed to have accepted my explanation that NH4OH is a shorthand way of writing NH3 + H2O (in order to show that as a base, it reacts similiarly to most metallic hydroxide). Then you claimed: If "There's a whole lot more ammonia there than OH so practically none of the NH3 is, in fact, present as NH4OH.) Then "Since practically no NH4OH is present, even as a dissociated species and even less is present as as such it's plain wrong to say that it is formed". You seemed to have missed the essence of my argument since NH4OH represents (NH3+H2O). So if there are a lot of NH3 present in water, the concentration of NH3 + H2O would be high, and therefore the same would go for NH4OH. Removing the word 'practically' would also help. Unless you want to suggest that theoretically, NH4OH (as the molecule, not the representation of NH3+H2O), would be formed, but the amount is so small as to be negligible. In that case you are shooting yourself in your foot since that would allude that NH4OH (as the molecule) actually would be formed, according to theory(which means it is consistent with true descriptions of reality, as of that moment, or else it would be a hypothesis instead). Care to explain what you really meant? Regardless, this is irrelevant either way as I have already conceded that NH4OH (as the molecule) does not form at all. I was merely using NH4OH as a shorthand way to allude that NH3+H2O behaves like a metallic hydroxide. "This being a scientific forum I think many people would agree with me that putting stuff that's plain wrong is unhelpful." True, but just because something is false, it doesn't mean that it is just plain wrong, especially if it is more easily understood and offers an excuse for what is happening, and even more so if it provides a gateway to the truer explanation (which is probably still false anyway but is better) Furthermore, scientific textbooks and journals should have higher standards of differentiating what is 'plain wrong' and 'wrong, but convenient explanation that is accepted by most of the community' I've seen aqueous ammonia being referred to as ammonia hydroxide in biology and even chemistry textbooks (university ones too), and even in some research papers, scientific indices, and journals. (http://www.cipav.org.co/lrrd/lrrd5/3/syria3.htm , http://potency.berkeley.edu/chempages/AMMONIUM%20HYDROXIDE.html) They usually explain briefly that it is just a representation of NH3 and H2O (or allude to such an explanation), which I also did. If I should not post such a 'common mistake' here, wouldn't those scientists be of greater error, as they are subject to a more stringent and heavier burden of truth than a "science forum"? Of course, if you want everybody to use the truest explanation all the time, regardless of its convenience, start by petitioning universities, publishers, etc. to stop putting in words like "ammonium hydroxide", "H+", stop using diagrams which picture delocalized electrons with double and single bonds (You can't even find such bonds since the electrons are all over the place, and hence delocalized)", and so on. And, "If you hadn't bothered to mention the (substantially irrelevant as well as non existent) NH4OH, not only would you have avoided your second verbose post, but your first one would have been shorter." Yep. Shouldn't have done that due to the can of worms opened up. Hopefully, they are stuffed into the can by now. "If you want, I will debate the fact that the reaction NH3 + H+ can be studied in the gas phase..." No thanks, this thread is getting off-topic enough at this rate. "In that instance you can decide whether or not NH3 is a strong base and you don't need to involve NH4OH at all.." NH3 is a weak base. Period. "so the assertion "Since the concentration of OH- is low, ammonia is a weak base. " isn't generally true either but I doubt anyone is listening." If it is such a big mistake that it is just "plain wrong", then why isn't anybody listening? Of course, if this is on the debate forums or the questioner is a college/university student, then I would use the longer and 'truer' explanation. As things are, NH4OH is suitable enough for a 10th grade chemistry class. Calling a spade a spade means I'm going to use the most convenient and suitable explanation for a particular condition. I am not going to call it a "yellow spade", "plastic spade", or a "digging implement with a handle made out of a matrix of sclerenchyma cells (read: wood) and a curved trapezoidal wedge made out of a composite of austenite, bainite, martensite, cementite, ledeburite, pearlite, and spheroidite (read:steel) despite them being 'better' and 'truer' explanations unless the situation calls for it. End argument. (If possible, I would rather not respond to your subsequent post (if you reply), as this is getting off-topic and paved88 has already mentioned that his research is already finished.) Thank you.
  2. Fine.. maybe I should have said when ammonia dissolves in water, it is practically identical to NH4OH, but technically, it is not. Then I would have to explain what I meant... Ammonia deprotonates a small amount of water to form NH4+ and OH-. This reaction is reversible, and NH3, H20, NH4+, OH- exists in equilibrium (although there's no actual equilibrium, but a shorthand way of describing how they behave, which otherwise would have taken a wall of text to describe. The chemical behaviour of NH3 and H2O together resembles that of NaOH, KOH and other soluble bases, and hence most scientists and engineers would also refer to a solution of ammonia as NH4OH, ammonium hydroxide. This is technically incorrect, but is done for convenience. (For the same reasons, most people write H+ instead, and benzene and acetic acid are drawn with double bonds, VSEPR theory is used to predict molecule shapes, etc. etc.) Chemically, though, ammonia in solution is treated as NH4OH, even though it is technically incorrect as it is impossible to isolate NH4OH, and hence, is assumed to not exist. This is reflected in the calculations for the disassociation of ammonia in solution, which is treated as NH4OH. Kb (ammonia)=1.79 x 10^-5, meaning that the amount of moles of OH- is squareroot(1.79*10^-5) per mole of NH4OH, and hence per mole of NH3 present in the solution. (even this is technically incorrect, since I am ignoring the change of the amount of NH4OH, which is rather negligible anyway. A more accurate solution would be arrange it as x^2+1.79(10^-5)x-1..79*10^-5 and solve for x using [-b+squareroot(b^2-4ac)]/2a (Note + instead of plus minus since the value must be positive). Even then, it is still technically incorrect since I'm ignoring a whole bunch of other factors, such as shielding, change of temperature, and so on.) Most of the ammonia will exist as NH3 within the solution (Which is written as NH4OH rather then NH3+H20). Since the concentration of OH- is low, ammonia is a weak base. However, both ammonia and the NH4+ ion are quite reactive, (delta G for most reactions tend to be really negative, and it also forms a lot of organic compounds and inorganic complexes) so it tends to react with other chemicals even if it is a weak base. Should NH4+ be responsible for the reaction, it will shift the disassociation of ammonium hydroxide (note that saying disassociation of ammonia is even more incorrect) towards the product side, yielding a constant supply of NH4 to take part in the reaction. This also has a side effect of increasing the pH of the solution. In fact, the strength of an acid or base does not affect its reactivity(rate of reaction), unless it involves acid-base reactions which generally yield a salt and water. Hydrofluoric acid for example, has a disassociation constant, k(a) of 7.2*10-4. Quite a weak acid, but you wouldn't want to go about dripping it on your skin unless you have a (quite painful) death wish (several drops could kill if left untreated). Same goes for formic acid, though it's significantly less dangerous. Yours verbosely, TWJian (See the wall of text now? Granted, most of my posts are long, but explaining every single thing like this would have raised the verbosity level by an order of magnitude or more)
  3. Got it. Thanks. Apparently I was doing too many parallel axis questions and was mind was swimming with them. Managed to find three ways to solve (a), though two of them are quite similar. The obvious solution 9and apparently shortest) is I(A)= 0.2kg (0+0.5^2+0.5^2+2*0.5^2)m^2 = 0.2 kg*m^2 As hinted by you. 2) Let d be the distance between two vertices. Since the vertices of the square trace a circle with radius, r= squareroot(2*(d/2)^2), use the formula: I(A)= I(CoM)+mh^2 =mr^2+ mh^2, h=r I(A) = 2mr^2=0.8kg*2*2*(0.25^2)m^2 =0.2 kg*m^2 3) Similiar to above, but proves that I(CoM) of a square with point masses at vertices is similiar to a circular band of negligible thickness. I(A)= I(CoM)+mh^2 = 4m(particle)*(distance of particle from CoM)^2+mh^2 = 4*0.2kg*(2*0.25^2)m^2 + 0.8kg(2*0.25^3)m^2 Note: same as above = 0.2kg*m^2 Therefore, the rotational inertia of a square with the only point masses at the vertices over an axis passing through one of the axis perpendicular to the plane the square is drawn in is: I= Total mass*Length of square^2 And for rotation with the axis passing through the CoM: I=Total mass*Length of square^2*(1/2) For a circular band rotating with an axis passing through it's fringes (or a particle at distance r from the center): I=2mr^2 Now for (b) Apparently I thought the blasted square has to rotate counter-clockwise despite being able to rotate clockwise... D C A B to A D B C change of height, delta h=h(i)-h(f)=0.5m (Initial gravitational potential energy)+(Initial rotational kinetic energy)= (Final gravitional potential energy)+(Final rotational kinetic energy) Since initial rotational velocity=0, mgh(i)=mgh(f)+0.5I(rotational velocity)^2 rotational speed=squareroot[2mg(delta height)/I] (speed is positive since it is the magnitude of velocity (in this case)), despite velocity being negative due to clockwise rotation.
  4. Assuming the 'correct' way is to solve as the force needed: 80g*n=500g Solve for n, and you get the force needed to counteract the weight of the car (the force exacted by gravity) Which is the force you need to apply to make the car effectively weightless. This just means the car will be just barely touching the ground (seems more like it's floating) with F(normal)=0. Problem is, the car is not going anywhere at all, and you are not actually lifting the car. The question asked "How many people will it take to lift the car 0,2 m off the ground?" Try to solve for the 0.2 m part with Newton's 2nd Law. (You cannot) Even if you put in more force to accelerate the car(initial acceleration), the time required for the car to arrive at z=0.2 m is not given, and you can't solve it at all. Unless of course, you want to launch the car as a projectile and set the maximum vertical displacement as 0.2m... but.. Isn't assuming the whole question is a work-energy question far simpler? I am going to apply Occam's razor to this problem, and hence I'm going to use the law of conservation of energy to solve this, unless there is a better method of computing the amount of people needed to lift the car (a given mass) by a given height. (a given displacement). Newton's second law cannot do it, and therefore the Law of conservation of energy should be used, sans the availability of a better solution.
  5. paved88, Let me guess.. you wanted to study the effects of acidic and alkaline solutions upon the solubility of a Jolly Rancher right? That means it is about the effects of different concentrations of H3O+ and OH-, therefore you would try to avoid other ions which might act with the Jolly Rancher... Avoid oxidizers at all cost. Bleach is sodium hypochlorite, which, although alkaline, could oxidize sugars into ketones, aldehydes, and carboxyllic acids (which are mostly soluble but a pain in the neck to isolate in this experiment), especially if the solution is acidic. (bleach actually has added HCl to become neutral, and the resulting oxidation would yield more HCl, increasing acidity and.. well, you get the point) If you are not sure what chemicals are oxidizers, avoid any acid with an O in there, including nitric acid. Sulfuric acid, incidentally, would dehydrate sugar instead, yielding insoluble carbpn if sufficient in high enough concentrations. Avoid chemicals with -ate or -ite unless you know what you are doing. They contain oxygen. Sprite contains carbonic acid. Shouldn't matter much.. but I'm going to drop carbon in case an organic reaction occurs. Vinegar. An organic compound. Which means you need to know organic chemistry due to the ensuing reaction between ketones, carboxyllic acids, etc. Drop this. Ammonia is NH3, but if it is dissolved in water, it becomes NH4OH. This is important because of the ammonium ion, NH4+. Don't use ammonia since it tends to form complexes (complex coordination ions) with transition metals and undergoes complex reactions with organic compounds, forming amines and god knows what. You already have some organic compounds in the Jolly rancher. Try not to introduce more organic compounds unless you want to do organic chemistry. There are however, some safe reagents which you could use which other ions will probably not react with sugars... As hermanntrude said, the standard chemicals would be HCl and NaOH. (Hydrochloric acid/Some drain cleaners/murcuric acid and Sodium Hydroxide/Lye) You could also use KOH, potassium hydroxide but it's more expensive and harder to find. This is because Cl- won't react with the sugars (unless you have oxidizers present) and Na+ compounds are all soluble. (so it will still remain as Na+ as long as you have water) Are you performing an experiment by yourself using household chemicals, and not using laboratory chemicals? If possible, ask for some lab-grade chemicals to help your research.
  6. What exactly(chemically) is a Jolly Rancher? If the solute in question does not react(chemically) with H3O+ or OH-, then the pH levels should not matter. If it does, and the solubility of the product is greater then the original solute (if the product is less soluble then the following arguments would also be reversed) 1) If the solute reacts with H3O+, then increasing the concentration of H3O+ (solution is more acidic) would increase the rate of the forward reaction, yielding more of the more soluble product, which corresponds to an increase in solubility. Increasing the concentration of OH- would decrease the concentration of H3O+ (solution becomes more alkaline) due to the autoionization potential of water (or whatever polar solvent you are using) 2)If the solute reacts with OH-, then increasing the concentration of OH-(solution becomes more alkaline) would increase the solubility, and vice versa. (et cetera, et cetera as in (1)) This holds true even if the original reactant is not soluble (and hence, not a solute) since a soluble product would certainly be more soluble then a non-solute. Keep in mind, however, that the solute could also react with other ions present in the solution due to the acids or bases you use, as well as molecules and ions present if you are not using water as your solvent. I am assuming it doesn't due to the information you provided. I can't presume that the solute reacts with anything else since the 'anything else' is unknown, and hence I ignored it. I could be wrong. A list of of all major reactants would be useful.
  7. "They can all lift 80 kg from ground level to 1 m above ground level this way." Doesn't the appearance of force (through 80kg*g) and displacement suggest that this is a work-energy problem? 80*9.81N of force and 1 meter of vertical displacement mean each person can do 80*9.81J of work, or 80*9.81n J of work, where n is the number of people doing the lifting. Lifting the car would increase the gravitational potential energy of the car, and therefore you need to input work to achieve that. Therefore, n(80kg*9.81m/s^2)= 500kg*9.81 m/s^2*0.2m Cancel out g or 9.81m/s^2 since it appears at both sides of the equation. (I would have removed it in the first place before I started calculating, but it seems clearer this way) (Note: J=kg*m^2/s^2) Solve for n. Note that other information are irrelevant unless you want to determine if you can conceivably fit in the n amount of people needed to lift the car. (i.e.: Is it possible to have enough people to lift the car vertically by 0.2m?)
  8. (Please delete the other duplicate thread. Sorry. Refreshed by accident) Recently I came across a confusingly-worded question (at least to me) in Physics. Semantics might matter, so I'm putting the entire question down ad verbatim: Four particles, each of mass 0.20 kg, are placed at the vertices of a square with sides of length 0.5m. The particles are connected by rods of negligible mass. This rigid body can rotate in a vertical plane about a horizontal axis A that passes through one of the particles. The body is released from rest with rod AB horizontal. Find: a) the rotational inertia of the body about axis A b) the angular speed of the body about axis A when rod AB swings through the vertical position I'm assuming that the square is arranged like this, on an xz plane: X X A B With the rotational axis at A and pointing out(or in, doesn't matter( of the page. The square would then rotate like a wheel. and the CoM would have a mass of 0.8kg at the center of the square. To find I, let I=I(CoM) +mh^2 (Parallel axis theorem, where h is the perpendicular distance between the axis through the center of mass (CoM) and A) h is therefore squareroot(2*0.25^2) right? Now how do I calculate I(CoM)? Assume it's a square and use I(com)=1/12*m(a^2+b^2), with a=b and I(com)=1/12*0.80(2x0.25^2)? Doesn't seem right as the mass are concentrated at the four vertices instead of being evenly distributed throughout the square...So... Assume it's a hoop with negligible thickness since the movement of the vertices appear to trace a circle? (am I even right?) Use the equation I(com)=mR^2, with r=h (from mh^2) and I(com)=0.80 squareroot(2*0.25^2) ? This interpretation(of the rotation) is even worse for (b). I'm assuming that energy is conserved and therefore: [mgh + 0.5*I(angular velocity)^2]initial= [mgh+0.5*I(angular velocity)^2]final Since AB is horizontal and ends up as vertical, There is no net change of h and therefore the gravitational potential energy is the same on both sides. With angular velocity(initial)=0 (since the square is released from rest), wouldn't angular velocity(final) be zero as well? That is, assuming the square and even rotate around A without any provided torque in the first place, since the CoM would definitely have to move upwards first before moving down to it's original position, thereby nessacating an input of energy due to increased mgh. So, how does the blasted square even move (except by falling down in translation motion if A is not locked down) in the first place? Obviously I am wrong. But where? How should I answer this question properly? Thanks for any help.
  9. I know that infinity is indefinite, and thus could be regarded as not being a number. With that, I realize that attempting to perform any arithmetic operations on infinity(other then 1/infinity, but that technically isn't an arithmetic operation either, though it looks like one) is both futile and meaningless. I thought that the limit of f(x)=x^(0.5) or (x+(x+4)^0.5)^0.5 is infinity as x approaches infinity. However, the graph of y= x^0.5 appears to have an horizontal asymptote (it does however, appear to be indefinite) Why is there such an asymptote? Is the square root of infinity some other indefinite constant which is not infinity? Or is the square root still infinity, though it is part of a different, yet still infinite set? Is the numerical value of the square root of infinity, infinity if we try to assign a value to it? Or is the square root of infinity impossible to define in terms of infinity (since both are indefinite), and thus could only be expressed as itself, the square root of infinity? (surd infinity is surd infinity) Incidentally, does a limit exist if the function approaches infinity as x approaches infinity? Since there is no actual boundry, a limit, apparently does not exist (to me). Furthermore, the limit appears to be one-sided, since it seems impossible to approach infinity from the right (or negative infinity from the left, for that matter)
  10. Thanks for the advice. Would a layer of non-conductive material beneath the armour such as padding or a doublet help to further reduce the current? It would be analogous to a person wearing a skinsuit in water. By the way, if I recall correctly, a Faraday cage is any enclosure constructed using electrical-conductive material(s) while an electrical shunt is a device which allows electrical current to pass through the circuit via another point when the voltage is high enough. (acting somewhat like an antifuse) I believe a Faraday cage can only block radio waves (or any other electromagnetic radiation) if it is thick enough(for radio waves, most are though).
  11. Just curious.. but is a person wearing full plate more susceptible to electricty, or will it act as a Faraday cage, counterintuitively decreasing the amount of electrical current flowing through his body? Furthermore, will other metallic armour work the same way to a lesser extent? (I heard rumours that chainmail is used this way) As a side note, please relocate this thread if it is in the wrong category. Thank you.
  12. Just curious... since most names are transliterated into Japanese by using katakana, how are East-Asian names, particularly Chinese translated into Japanese? Are Chinese names transliterated by using katakana, using the hanzi as kanji directly (if it exists in both and modified first), any combination of the above or another method?
  13. TWJian

    Human reaction time

    I read somewhere on the Internet (a police research I think) that drivers on average requires 750 ms to break as a response to a certain emergency. Quote: Mean reaction time (motor response to a simple visual stimulus e.g. a light) is between 300 - 500ms. So, how did I get a mean reaction time of 170 ms? Anticipation perhaps? Or maybe the programming is bugged? By the way, thank you guys for responding quickly. As a side note, does anybody know how quick can a neuron emit a single impulse?
  14. Anybody knows how much time we take to respond to a simple stimulus? (on average). I try googling and found a lot of conflicting reports. Did some experiments by playing reaction time games. Example: http://www.newgrounds.com/portal/view/282683 Did manage to get a mean reaction time of around 170 ms.. (repeated 20 times) If this post violates forum rules (since newgrounds does contains unsuitable material), please delete this thread.
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