bohr model of potassium?
149440
http://www.chemicalelements.com/elements/k.html\n\nhttp://www-spof.gsfc.nasa.gov/stargaze/Q6.htm
495461
Atomic radius (or size) increases as you go down and to the left on the periodic table.  Therefore, out of these, the largest would be K.
788044
i can assure you that  the answer is \n\nA. the Bohr model of the atom, with discrete orbits. \n\ni have been analyzing this work  for 3 years.\n\n\np.s the answer is so funny C."C. the Einstein model of the atom."
613346
The previous responses are right but ko is not a potential. It's the (average) momentum for a particle in free space. With psi defined by:\n\npsi(x) = A exp(ikx - x^2/(2sigma))\n==>\npsi*(x)psi(x) = |A|^2 exp(-x^2/sigma)\n\nYou need the integrals:\n\n&#92;int_(-oo)^(+oo) exp(-x^2/sigma) dx = sqrt(pi*sigma)\n&#92;int_(-oo)^(+oo) exp(-x^2/sigma) x^2 dx = (sqrt(pi)/2) sigma^(3/2)\n\n<x^2> is the second expression divided by the first:\n\n<x^2> = 1/(2 sigma)\n\nSince H = p^2/2m, \n\nH psi(x)\n= -(1/2m)(d^2/dx^2) psi(x)\n= -(1/2m) [(ik-x/sigma)^2 - 1/sigma] psi(x)\n= (1/2m) [k^2-x^2/sigma^2 + 2ikx/sigma + 1/sigma] psi(x)\n\n==>\n\n<H>\n=(1/2m)[k^2-<x^2>/sigma^2 + 2ik<x>/sigma + 1/sigma]\n=(1/2m)[(k^2+1/sigma-(1/2sigma)]\n=(1/2m)[(k^2+1/(2sigma)]
491001
Homework, huh?\n\nWell, it's because K+ contains more electrons than the Na+. Which means that K+ has 3 layers of electrons (filled in the 1s2 2s2 2p6 3s2 3p6) while while Na+ has only 2 layers of Electrons (1s2 2s2 2p6)
147821
- -----------O\n------------ . .-----------  _\n  K+ ( O: Mn  :O )\n-------------. .\n------------ O\n\nomit the long dotted lines
737924
You can figure it out.\n\nIf a neutral atom of zinc has an atomic number of 30, and a mass number of 65, then it has 30 protons, 35 neutrons, and 30 electrons.\n\nProtons and neutrons are found in the nucleus.\nSo, draw a circle. Write "30p+" and "35 n" inside it.\nThis represents the nucleus.\n\nIn Bohr's model, electrons orbit the nucleus at specific energy levels. The first level holds up to 2 electrons.\nThe second holds up to 8.\nThe third holds up to 8.\nThe fourth holds up to 18.\nWe use dots to represent electrons.\n\nDraw a ring around the nucleus.\nAdd two dots to the ring, usually at noon and 6:00.\n\nDraw another ring around the first one.  Draw eight dots: two at noon, two at 3:00, two at 6:00, and two at 9:00.\n\nContinue adding rings and electrons.\nYou should end up with twelve dots in the outermost ring.\n\n(Which doesn't correspond with the  Lewis diagram for zinc, but that's  another story....)
907330
Potassium (chemical symbol K) has 19 each of protons and electrons, because it has an atomic number of 19.  Potassium has more than one naturally occuring isotope.  Potassium can have an atomic mass of 39, 40, or 41.  However, over 90% of naturally occuring Potassium is K-39, with 20 neutrons.  K-41, with 22 neutrons, is also stable.  K-40 has 21 neutrons, and does occur in nature, but with an abundance of only 0.012%.  In addition, it is not truly stable, although its half-life is over a billion years.
573821
I think this question is incomplete because the Ka of HF hasn't been provided.  The amount of H+ that is formed from HF can only be calculated from its Ka value.
553682
When a Potassium atom becomes a Potassium ion, it looses a single electron to become K+.\nThe one electron it looses is located in its outermost, valance, electron shell (the S orbital).\nWhen this electron is taken away, the S orbital is empty and the radius of the resulting ion is now smaller than the previous atom.
887049
K2Cr2O7\nPostassium ion: K+\nDichromate ion: Cr2O72(2-)\nso to there has to be 2 potassium ions for each dichromate ions. You can also see http://en.wikipedia.org/wiki/Potassium_dichromate\nif you want more info.
659878
They are:\n\n1. Principal Quantum Number\n2. Azimuthal Quantum Number\n3. Magnetic Quantum Number\n4. Spin Quantum Number\n\nThat makes four.\n\nThe Bohr model was a one-dimensional model that used one quantum number to describe the distribution of electrons in the atom. The only information that was important was the size of the orbit, which was described by the n quantum number. Schrodinger's model allowed the electron to occupy three-dimensional space. It therefore required three coordinates, or three quantum numbers, to describe the orbitals in which electrons can be found.\n\nThe three coordinates that come from Schrodinger's wave equations are the principal (n), angular (l), and magnetic (m) quantum numbers. These quantum numbers describe the size, shape, and orientation in space of the orbitals on an atom.\n\nThe principal quantum number (n) describes the size of the orbital. Orbitals for which n = 2 are larger than those for which n = 1, for example. Because they have opposite electrical charges, electrons are attracted to the nucleus of the atom. Energy must therefore be absorbed to excite an electron from an orbital in which the electron is close to the nucleus (n = 1) into an orbital in which it is further from the nucleus (n = 2). The principal quantum number therefore indirectly describes the energy of an orbital.\n\nThe angular quantum number (l) describes the shape of the orbital. Orbitals have shapes that are best described as spherical (l = 0), polar (l = 1), or cloverleaf (l = 2). They can even take on more complex shapes as the value of the angular quantum number becomes larger.\n\nThere is only one way in which a sphere (l = 0) can be oriented in space. Orbitals that have polar (l = 1) or cloverleaf (l = 2) shapes, however, can point in different directions. We therefore need a third quantum number, known as the magnetic quantum number (m), to describe the orientation in space of a particular orbital. (It is called the magnetic quantum number because the effect of different orientations of orbitals was first observed in the presence of a magnetic field.)\n\n• The three quantum numbers (n, l, and m) that describe an orbital are integers: 0, 1, 2, 3, and so on.\n\n• The principal quantum number (n) cannot be zero. The allowed values of n are therefore 1, 2, 3, 4, and so on.\n\n• The angular quantum number (l) can be any integer between 0 and n - 1. If n = 3, for example, l can be either 0, 1, or 2.\n\n• The magnetic quantum number (m) can be any integer between -l and +l. If l = 2, m can be either -2, -1, 0, +1, or +2.\n\n\nYou should probably read the sources...
176528
http://www.chem1.com/CQ/ionbunk.html\n\nskim this article.
706351
KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
711310
boron has five protons and electrons
286410
yes, it will explode.  the reaction\n\n2K + 2H2O -> 2KOH + H2\n\nis extremely exothermic and can get very hot.  hot enough, in fact, to ignite the hydrogen, which in an oxygen atmosphere (like air) will explode.  for this reason potassium metal is always stored under oil or kerosene that will not mix with water.\n\nit is essentially the same reaction as the well-known sodium metal in water; lithium, sodium and potassium are in the same group of the periodic table and therefore have similar chemistry.\n\nhere are some pictures of potassium and water reacting, about halfway down the page:\nhttp://www-chem.ucdavis.edu/chemclub/Photos/pday.htm\n\nnote that was probably a small amount of potassium; a larger amount would probably cause a pretty spectacular kaboom.
671814
Firstly mash up a pile of bananas\n\nWeight out a set amoult of banana mash.\n\nAdd a x amoult of solvent to the banana mash that will\nextact the potassium from the mash.\n\nNow you have a chose of techinquies\nbut the best one I would go for is\nflame spectroscopy for potassium.\n\nFrom the solvent (taking note of the amoult\nyou got) Use in the spectroscopy machine\nthis should give you a ppm reading.\nFrom this work out the weight of potassium\nin the solvent.\n\nFrom that work out the weight of potassium\nin xxx bananas.