homogeneous mixture question?
597810
To be a homogenous a mixture must be the same throughout.  If you are only having 29 protons in the nucleus, then it is only one element.  Element number 29 is copper.  Try some nice copper wire.  Common and chemical names are both "copper".
24810
Cu + 2H2SO4 --> 2H2O + SO2 + CuSO4\nDouble the H2SO4 and the H2O.
811304
let thetemperature of the mix be 't'*Cfind the heat lost by coffee\nfind the heat gained by ice in three stages\na)from -2.2*C to 0*C\nb)for melting and becoming water at 0*C\nc)for getting to the equilibrium temperature of the mixture\nequate and solve
395302
mixture is combining substances, solution is dissolving a substance in another
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
349201
Write out the sample space:\nHHH, HHT, HTH, THH, HTT, THT, TTH, TTT.\n\nThere are 8 different possibilities - 2^3.\n\nTwo of the outcomes: HHH and TTT are all the same -\nProbability is 2/8 = 1/4 = .25.
435235
I assume you mean mixtures like solutions where the components of each separate phase are distributed homogeneously (i.e. metal alloys, liquids, polymer blends) and the solution is at equilibrium.  \n\nPut simply, the main reason ternary (or binary or quaternary..) soltuions can remain separate has to do with the thermodynamics of the solution.  Under a given set of equilibrium conditions (temperature, pressure, etc.) different phases can form if they correspond to a minimum in what is called the free energy.  The free energy represents the balance between what are called enthalpic interactions (molecules/atoms interacting with each other or within themselves) and entropic contributions (the amount of disorder within the system).\n\nNote:  I've limitied my answer to a VERY rudimentary explanation.  There are other factors that determine whether phase separation actually occurs (i.e. kinetics).  I urge you to visit any of the many websites that can give a better explanation than I have given here.  For example, try to find links through www.acs.org.
542164
0.2x+0.5(300-x) = 123\n0.2x+150-0.5x = 123\n-0.3x +150 = 123\n-0.3x = -27\nx= 90\nyou will need 90 gallons of 20% saline solutions and 210 gallons of 50% saline solution to obtain the mixture.\nHope this helps!!!
877862
From one Big Daddy to another:\n\n.75(x) + .9(40 - x) = .78(40)\n.75x + 36 -.9x  = 31.2\n-.15x = -4.8\nx = 32\n\nThus, they should use 32L of the 75% mixture, and 8L of the 90%.
677509
(30 x 4 x 1)/(1 x 3 x 2) = 20 ml\n\n(30ml)(4.0M)(OH=+1)/\n(\n(1ml)(3.0M)(SO4=-2)\n)
1186593
Mix with a 0% solution in the relation 33:2\n\nSo use 2 units of your solution (like 2 liters) and 33 units of the 0% solution (like 33 liters).\n\nThat's all.
877652
These are GAS LAW problems, so you need to use PV=nRT.  However, for these problems, nRT will be constant so you only need to use P(pressure) and V(volume).  Therefore, for both of these questions, use the equation:\nP1V1 + P2V2 = PtVt   where P1 is the pressure in the first vessel, P2 is the pressure in the second vessel, and Pt is the total pressure of the connected system.\n\nProblem #1\nP1V1 + P2V2 = PtVt  \n(7.0atm)(15.0L) + (P2)(8.4L) = (10.0atm)(23.3L)  \nP2 = 15.4 atm\n\nProblem#2\n(7.2atm)(15.0L) + (4.1atm)(8.3L) = (Pt)(23.3L)  \nPt = 6.1 atm
365444
0.500L x 0.250moles/L x 58.44g/mol = 7.31g\n(a)
917667
Part A:\nThe answer is 12 L H2.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(3 mol H2 / 1 mol N2)(22.4 L H2 / 1 mol H2)\nso (4*1*3*22.4) / (22.4*1*1) = 12 L H2\n**note that in in ideal gas there are 22.4 liters/mol**\n\nPart B:\nThe answer is 8 L NH3.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(2 mol NH2 / 1 mol N2)(22.4 L NH3 / 1 mol NH3)\nso (4*1*2*22.4) / (22.4*1*1) = 8 L NH3\n\nHope this helps!
458226
compound is a mixture of 2 differ net elements. homogeneous mixture of 2 or more substances that keep their identifies. therefore they are similar in that they are mixture of two or more diffident substances, but different in that the homogeneous mixture has the substances that keep their identifies. Homogeneous mixtures are also called solutions because u cannot identify their ingredients like in a compound.
1090587
the homogeneous equation is (dy/dt) +2y = 0\nor equivalently, dy/dt = -2y\n=> (1/y)(dy/dt) = -2\nintegrate both sides wrt to t:\nintegral (1/y) dy = integral (-2) dt\n=> ln y = -2t + k   , k is a constant\n=> y = exp{-2t + k} = Ce^(-2t)  , C is another constant where C= e^k\n\n\nto solve the nonhomogeneous equation, try a particular solution of the form y = a*sin t + b*cos t\nthen, (dy/dt) = a*cos t - b*sin t\n\nsubstitute into the orginal equation to get\na*cos t - b*sin t + 2*(a*sin t + b* cos t) = - cos t\n=> (a + 2b) cos t + (2a - b) sin t = - cos t\nequating coefficients:\na+2b = -1 and 2a-b = 0\nsolve for a and b:\na = -1/5, b = -2/5\n\nhence, a particular solution to the nonhomogeneous equation is y= -1/5 sin t - 2/5 cos t\n\nso, the complete general solution to the nonhomogeneous equation is\ny = C e^(-2t) - 1/5 sin t - 2/5 cos t\nwhere C has to be determined from initial conditions
752805
The first thing you need to do is figure out which of the three liquids is going to be the one that limits the maximum quantity of the mixture that can be made.  Since the ratios are 5:2:1 you need 5 parts x, 2 parts y, and 1 part z.  You have 25 gallons of x and, if that's 5 'parts' then 1 'part' must be 5 gallons.  Now you need 2 'parts' (or 10 gallons) of y, and you do have that much in stock.  Lastly 1 'part' (5 gallons) of z is required and you have that much in stock as well.  So you'll mix 25 gallons (5 'parts') of x, 10 gallons (2 'parts') of y, and 5 gallons (1 'part') of z for a total of 25+10+5 = 40 gallons of the mixture.\n\nHope that helps.\n\n\nDoug