how do you calculate the elctronegativity difference between elements?
482712
um...  subtract???  hahaha\n\nThe higher the differences in electronegativity, the more ionic the compound.  The higher electronegative atom/ion will want to "hog" all the electrons.  They are greedy like that.  They don't want to share (which would be a covalent bond).
749429
x = -7\n\nMultiply two negatives and you get a positive
691296
use that at equilibrium : n.p = ni^2\nand from charge neutrality: \n\nn + NA = p + ND
872957
Look at them carefully, find out the diff between each number and it should be obvious.\n\nEquations?
734519
(1) Take a container and measure the length of the container. Let us say the length of the container is L. \n\n(2) Take a beam of light (preferably a monochromatic source like laser beam) and shine it through the empty container. \n\n(3) Measure the intensity of the light at the end where the beam has gone through the empty container. Let this be Io. \n\n(4) Now fill the container with the liquid whose molar extinction coefficient is unknown. Now measure the light intensity as the beam travels the length L of the container. The measured intensity is I.\n\n(5) Beer-Lamberts law is I=Io exp(-a L) where all the quantities have been experimentally determined except a which is the extinction coefficient. \nRearrange the above formula and we get, a=(-1/L)*log(I/Io) where log is to the base e.
307451
Try\n\nhttp://www.purplemath.com/modules/negative.htm
1062254
Lol\nI actually did a test like this for my nephews science project last year.  Unless the formulas have changed the differences are completly negligent.  Their lasting time is about the same.
231126
The difference in Pauling's electronegativity\ndetermine percent ionic character\ncovalent and ionic are sort of relative positions on a continuum
201111
You need to measure the distance between the closely spaced lines (J values).  Trans and cis J values will be different
851182
Yes.  The easiest trick for determining polarity is to evaluate the electronegativity of the atoms.  Oxygen is more negative, while hydrogen is more positive.  If you write the partial charges on the atoms, you see that the charge balance is not equal.  If you had a molecule such as BCl4 (not real, just an example) the B is surrounded evenly by Cl atoms, so there is no polarirty
501994
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/posfe.html#c3\n\nThe link above may be of some help. Input the values you want or need.
930646
You probably want alpha, the common base current gain. In general, current gain is designated "beta," "alpha," or "gamma," for transistors connected in the common emitter, common base, or common collector configurations, respectively.  Beta = di(c)/di(b), the rate of change of collector current with respect to change in base current.  Alpha =di(c)/di(e), rate of change of collector current with respect to change in emitter current.  Gamma = di(e)/di(b), rate of change of emitter current with respect to change in base current.  Also, alpha = (beta)/(beta - 1), beta = (alpha)/(alpha - 1), and gamma = beta + 1.  For a detailed explanation, check out : http://www.tpub.com/neets/book7/25f.htm\n\nAnother possible answer: "amplification factor" for any amplifier circuit, regardless of configuration, or whether it uses tubes, BJTs, FETs, etc, refers to its voltage gain, dv(out)/dv(in), the change in output voltage with respect to a change in input voltage.
528390
minus should be negative and iy doesnt say if it is C or F
152635
You program the data in to a calculator (like a TI-83), graph, and model the data points using an exponential profile.  It will then give you the form of the equation that best fits your data points in the form of a*e^(b*x), which you can then graph over your data points.  In your case, "b" should be negative to indicate decay.
794082
assuming "perfect" polarization:\n\n50% through the first.   Then use trig to calculate the E field amplitude reduction caused by the 42deg  polariztion.
836917
WL = h/(mv)\n\nh is the planck constant\nm is the mass of the elecron\nLook for them in you book.  Plug the numbers.
1064705
I think knowledge of the terms electronegativity and bond energy address your question. \n\nElectronegativity refers to the difference by which two atoms tend to hold onto electrons. You should be able to find a periodic table that shows you the electronegative values of all the atoms. You can compute the electronegativity of a bond between two atoms by subtracting their values from each other (based on Linus Pauling's calculations). For example, Flourine (F) has an electronegativity value of 4 and Na = .9. The electronegative value of an F-Na bond is 3.1 (4 - .9). My old high school textbook (Heath) states that ionic bonds are those having an electronegative difference between 1.7 and 4; polar covalent bonds are between 0.2 and 1.7; covalent (non-polar) bonds are between 0 and 0.2. Thus an F-Na bond by this classification system is an ionic bond.\n\nBut I suppose to answer your question most directly, I would say the evidence that ionic bonds are strong concern something called bond energy. That is, the difference between how much energy is needed to make and break chemical bonds. Generally speaking, the higher the electronegative difference (which translates into more of an ionic bond--see calculation above), the more energy required to break such bonds. \n\nThe bottom line, is that I think this translates into something like the observed boiling points of various substances. Substances that have strong ionic bonds (again, determined by calculation above) I believe have a higher boiling point (because it takes more energy/heat) to break the bonds.