User talk:Lamoureaux32

where can I find the 2 dimensional structures of compounds like C8H8O? Or other compounds like it benzylideneaniline? Does anyone know anything about 3 dimensional chemical software? I want to make a simulator using compounds in 3 dimensions...How can I do this? If I want to draw the compound to scale how could this be done. AutoCad, OrCad, some kind of CAD...I want the compounds to be see in 3d with all the orbital rotations of all of the Atoms inside a simulator so I could predict the actual transformations of the reactants <-> Products. How can I do this?

Avogadro's number
I am trying to find out how Avogadro's number is calculated. First they say it is a carbon 6 with 6 protons + 6 electrons and 6 neutrons. So I try to add the number of moles of a 12 gram sample of C6... My Physic's book says Electron Mass	9.10939x10-31 grams Proton Mass	1.672623x10-27 grams Neutron Mass	1.674929x10-27 grams

a carbon atom is 12.011 grams per mole so a 12 gram carbon sample is 12 grams x (1 mole/12.011g) x (1/Avogadros number) Avogadro's number is 1 mole = 6.022136736x1023 particals

is Avogadros number the sum of 6 protons + 6 electrons + 6 neutron divided by the mass of the sample 12 grams of Carbon. So the number should be 12 grams/mole * 1 partical/(sum(6p + 6e + 6n) grams) = 5.97288973x1026 particals the closest I've ever been is (1 mole/144.132 gram) / (1 partical/(6xMassP + 6xMassE + 6xMassN)) = 1.726683919x1023

Did avogadro use a Buckminsterfullerine sample of C60 with no H. Or was he really using a saturated sample of Benzene or some kind of cyclohexane? where am I going wrong. How is it that the number of particals can also be used as the number of molecules? Is the book referring to the partical as the molecule?

I am taking into consideration that this is an atom not a compound or a substance for example it is simply the molcular weight of a carbon atom not the molecular weight of Benzene C6H6 which is cabon 6 x 12.01 = 72.066 grams per mole hydrogen 6 x 1.0079 = 6.0474 grams per mole equaling 78.1134 grams per mole

if the calculation is going to be exact. I need the mass of proton + neutron + electron count the periodic tables common atomic structure (meaning no isotopes) yielding grams. Then take the sample 12 grams of carbon divide by the molecular weight of the atom then divid by the common weight of the proton + neutron + electron and that should be avogadro's number for the atom inside the periodic table. Unfortunately I am totally incorrect....

wait I got it its... 12 g C• (1 mol C)/(12.011 g C)•6•((9.109389x〖10〗^(-28))/(Electron Mass)+(1.672623x〖10〗^(-24))/(Proton mass)+(1.674929x〖10〗^(-24))/(Neutron Mass))=2.0072378x〖10〗^(-23)

adhesion and cohesion
So there must be some other way of understand chemistry. I am considering the idea that Adhesion-The physical attraction or joining of two substances, especially the macroscopically observable attraction of dissimilar substances. Cohesion-The intermolecular attraction by which the elements of a body are held together.

so cement is a mixture of sand, water and cement mix

cement mix is made up of four main compounds: tricalcium silicate (3CaO · SiO2), dicalcium silicate (2CaO · SiO2), tricalcium aluminate (3CaO · Al2O3), and a tetra-calcium aluminoferrite (4CaO · Al2O3Fe2O3). In an abbreviated notation differing from the normal atomic symbols, these compounds are designated as C3S, C2S, C3A, and C4AF, where C stands for calcium oxide (lime), S for silica, A for alumina, and F for iron oxide. Small amounts of uncombined lime and magnesia

sand the chemical formula for sand is SiO2, so it is a compound, made from Silicon and Oxygen,

Water H20

when I was working in construction I used a cement mixer for the first time. First we added the Sand, then the water and later the cement mix. the mixer sturred the reactants together and formed a cement compound. Is this an example of Adhesive or is it Cohesive forces acting on the mixutre?

higher than 1st order differential equations
4) y^''+2y^'+2y=10sin⁡4x First lets find the complementary function of the equation by solving for the rational roots P(m)=m²+2m+2=0 We must use the quadratic equation x=(-b±√(b²-4ac))/2a x=(-2±√(2²-4(1)(2)))/2↔x=-1±i Because there is a complex number involved in the root of the equation we must use this formula to solve for the complementary function yc, where b = -1 and c = 1 e^mx=e^((b+ic))=e^bx cos⁡(cx)+e^bx  sin⁡〖(cx)〗 The complementary function is e^(-x)  cos⁡〖(x)+e^(-x)  sin⁡〖(x)〗 〗 using a Wronskian Matrix to check to see if the set is a linear dependent/independent set, W(e^(-x)  cos⁡〖(x)+e^(-x)  sin⁡〖(x)〗 〗 )=[■(e^(-x) cos⁡(x)&e^(-x)  sin⁡〖(x)〗@-e^(-x)  cos⁡〖(x)-e^(-x)  sin⁡〖(x)〗 〗&-e^(-x)  sin⁡〖(x)+e^(-x)  cos⁡〖(x)〗 〗 )]= e^(-x)  cos⁡(x) [-e^(-x)  sin⁡〖(x)+e^(-x)  cos⁡(x) 〗 ]-e^(-x)  sin⁡〖(x)〗 [-e^(-x)  cos⁡〖(x)-e^(-x)  sin⁡〖(x)〗 〗 ]= (e^(-x)  cos⁡(x) )^2+(e^(-x)  sin⁡〖(x)〗 )^2=e^(-x)≠0 Therefore the set is a linearly independent set of values… y_c=c_1 e^(-x) cos⁡(x)+c_2 e^(-x)  sin⁡〖(x)〗 y_c^'=〖-c〗_1 e^(-x) cos⁡(x)-c_1 e^(-x)  sin⁡〖(x)〗-c_2 e^(-x)  sin⁡〖(x)〗+c_2 e^(-x)  cos⁡〖(x)〗 y_c^''=c_1 e^(-x) cos⁡(x)+c_1 e^(-x)  sin⁡〖(x)+〗 c_1 e^(-x)  sin⁡〖(x)〗-c_1 e^(-x)  cos⁡〖(x)〗+ c_2 e^(-x) sin⁡〖(x)〗-c_2 e^(-x)  cos⁡〖(x)-c_2 e^(-x)  cos⁡〖(x)-c_2 e^(-x)  sin⁡〖(x)〗 〗 〗 y_c^''=2c_1 e^(-x) sin⁡〖(x)-〖2c〗_2 e^(-x)  cos⁡(x) 〗 When we plug the complementary set into the ODE we get 2c_1 e^(-x) sin⁡〖(x)-〖2c〗_2 e^(-x)  cos⁡(x) 〗+2(〖-c〗_1 e^(-x)  cos⁡(x)-c_1 e^(-x)  sin⁡(x)-c_2 e^(-x)  sin⁡(x)+c_2 e^(-x)  cos⁡(x) )+ 2(c_1 e^(-x) cos⁡(x)+c_2 e^(-x)  sin⁡(x) )=0 Now we solve for the particular integral by finding the undetermined coefficient set of 10 sin⁡4x. The UC Set formula is {〖 sin〗⁡〖(bx+c),cos⁡〖(bx+c)〗}〗 y_p=Asin (4x)+Bcos(4x) y_p^'=4Acos (4x)-4Bsin(4x) y_p^''=-16Asin (4x)-16Bcos(4x) -16Asin (4x)-16Bcos(4x)+2(4Acos (4x)-4Bsin(4x))+2(Asin (4x)+Bcos(4x)) (-16A-8B+2A) sin⁡〖(4x)+(-16B+8A+2B)  cos⁡〖(4x)=10 sin⁡4x 〗 〗 In order to get the equation particular integral correct we must get A and B (-14A-8B)=10 (-14B+8A)=0 -(10+8B)/14=A↔-14B+8(-(10+8B)/14)=-196/14 B-80/14-64B/14↔ -14/260•-260/14 B=80/14•(-14/260)=-80/260=-4/13 -14A-8•(-4/13)→A=+7/13 The particular integral is y_p=(7/13) sin⁡4x-(4/13)  cos⁡4x We check to see if the integral is correct…only if A = -7/13 y_p=-(7/13) sin⁡4x-(4/13)  cos⁡4x y_p^'=-(28/13) cos⁡4x+(16/13)  sin⁡4x y_p^''=+(112/13) sin⁡4x+(64/13)  cos⁡4x

-(112/13) sin⁡4x+(64/13)  cos⁡4x+2((28/13)  cos⁡4x+(16/13)  sin⁡4x )+2((7/13)  sin⁡4x-(4/13)  cos⁡4x ) or (+112/13+32/13-14/13) sin⁡4x+(64/13-56/13-8/13)  cos⁡〖4x=10 sin⁡〖(4x)〗 〗 Therefore the general solution for this ODE is y_p+y_c=(7/13) sin⁡4x-(4/13)  cos⁡4x+c_1 e^(-x)  cos⁡(x)+c_2 e^(-x)  sin⁡〖(x)〗