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chemhelper

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  1. A good starting place is define the equation. The unbalanced equation is ZnCl2 + Li --> LiCl + Zn Essentially the same problem as before except that the masses initially given need to be in terms of moles, and then moles will have to be converted back to mass in the end and then use the mass percent to find the total mass of the solution.
  2. Force = kQq/(r^2) Q=charge at point P and q=charge a distance r from P So the initial force would equal k(Q^2)/.25 The problem specifies that the force doubles as a result of its placement. Thus, the force in question becomes 2k(Q^2)/.25 = 8k(Q^2) Set this equal to the same force equation at the beginning 8k(Q^2) = kQq/(r^2) little q = 2Q by specification. as you can see, the Q^2 and k drops out and what is left is 8 = 2/(r^2) You'll get two answers for R. Try to figure out which side of the charge placing 2Q will increase the force on the charge as opposed to counte
  3. If you wish to use perl/cgi you can use one of the environmental variables http://www.cs.purdue.edu/homes/cs290w/perlLecs/PerlEnvVars.html This is a good site the explains how to use them
  4. An easier (and probably cheaper) method of altering shapes would be using magnetic fields instead. If you take a wire and place it in a magnetic field a force will act upon the wire, bending it in a direction corresponding to the direction of the magnetic field. What is your idea so that a more suitable way can be found
  5. Solubility constants are a lot like acid constants. For a compound AaBb..., Ksp = [A]^a*^b*... for each species, where the smaller case letters are the subscripts of the species. You are given the solubility of AgI as 8.3 x 10^-17. To begin with, you have a 0.04M solution of MgI2. In 1 liter, this would mean there is 0.040 mols MgI2 in the solution. This completely ionizes into 0.04mols Mg2+ and 0.080 moles of I-. Now, take Ksp = 8.3 x 10^-17 = [Ag] (the coefficients are one because subscripts of Ag and I are both 1) Make an ICE table AgI --><-- Ag+ + I-
  6. Presume that x,y and z are infinitesimally small. Essentially, you will get 0^0 + 0^0 + 0^0. Any quanity, regardless of what it is, raised to the zero power is equal to one by definition. Thus, you will have 1 + 1 + 1 > 2 when x,y,z > 0 This is a basic idea but I'm not certain if it is the actual proof ----- Have homework questions in math, physics or chemistry? Who Likes Homework -- http://www.wholikeshomework.com
  7. Please forgive me 3O2 + 2H2S -> 2H2O + 2SO2 So 146 would be correct --- Have homework questions in chemistry, physics or math? Who Likes Homework -- http://www.wholikeshomework.com
  8. Obviously, the bowling ball will have a larger mass than the ping pong ball. Presuming the two are moving at the same speed initially, two things could happen to make the momentum the same: 1. The bowling ball would have to slow down 2. The ping pong ball would have to speed up just remember, as was stated, p=m*v ---- Have homework questions in chemistry, physics or math? Who Likes Homework -- http://www.wholikeshomework.com
  9. Its not. A limiting reagent problem is when you are given all the reactant quantities. For example, if you were given 32 mol of O2 and 96.7 mol HS. However, it is asking how much oxygen is consumed. In other words, it means how much oxygen is needed to fully complete the reaction. If you write out the equation you get 2O2 + H2S -> 2H2O + SO2 This says that for every mole of hydrogen sulfide (coefficient=1), 2 moles of O2 are required to complete the reaction (coefficient=2). Thus, if you have 96.7 mol H2S, then you would need 96.7 H2S * (2 mol O2 / 1 mol H2S ) = 193.4 mol O2.
  10. To begin with, there are initially (0.2 L)(0.4M) = 0.08 mol NH3. (0.3 L)(0.1M) = 0.03 mol HCL is added, which completely ionizes. The 0.3 mol H+ neutralizes 0.03 mol NH3. Thus (0.08 - 0.03 mol) NH3 = 0.05 mol of NH3. The new concentration of NH3 is ( 0.05 mol ) / (0.2L + 0.3L) = 0.1M due to the volume of the HCL The reaction is now NH3(aq) -> NH4+ + OH- Kb = [NH4+][OH-]/[NH3] which yields 1.8 x 10^-5 = x^2 / (0.1-x) Solving for x=0.00133 mol/L and x=[OH-] -log [OH-] = pOH --> -log [0.00133] = 2.875 pOH + pH = 14 thus 14-2.875 = pH = 11.12 ---- Have
  11. (E)-2-butene-1-thiol and 3-methyl-1-butanethiol....aka skunk:-) http://www.humboldt.edu/~wfw2/chemofskunkspray.html ---- Have homework questions in chemistry, physics or math? Who Likes Homework -- http://www.wholikeshomework.com
  12. If you wanted to check your work: You have 0.2855 mol (8 g/28.02 g) N2 and 7.44 mol (15 g / 2.016 g) H2. The limiting reagent is the N2 0.29 mol N2 will require 0.2855*3 mole of H2 = 0.8565 mol H2 while using 7.44 mol H2 will require 7.44/3 mol N2 = 2.48 mol N2; thus you dont have enough N2 to use all of the hydrogen (H2 is in excess) Then using the equation eVon gave: Enthalpy = (0.2855 mol)*(46.2 kJ / mol ) = 13.19 kJ for the reaction. ------ Have homework questions in chemistry, math or physics? Who Likes Homework -- http://www.wholikeshomework.com
  13. I am confused by your question. The bond enthalpies of elements vary depending on what other element the atom bonds to. For example, the enthalpy (i.e. bond strength) of an O-H bond is going to be different than an O-C bond. Also, oxygen should have a stronger bond enthalpy than sulfur because it has a much smaller radius than sulfur. ---------- Have homework questions in chemistry, math or physics? Who Likes Homework -- http://www.wholikeshomework.com
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