Recent content by Styx

I never knew chemistry could taste so good! Thanks gradeaswimr, makes sense to me.

I am doing this question as well but I am not sure whether or not I am correct... What I did was: HA = 0.10 - 0.001 = 0.099 H+ = 0.001 [H+] = [A-] Therefore A- = 0.001 k = [H+][A-] / [HA] k = (0.001)(0.001) / (0.099) = 10^-5

The formation of products is strongly favoured in this acid-base system: HX + B^- \rightleftharpoons HB + X^- a) Identify the bases competing for protons b) Which base is stronger? c) Which is the weaker acid, HX or HB? d) Does the K for this system have a large or small value? e) How is...

Saved again, thanks Cristo

Here is an example in my book... Determine the coefficient of x^6 in the expansion of (2-3x^2)(1+2x)^8 (2-3x^2)(1+2x)^8 = 2(1+2x)^8 - 3x^2(1+2x)^8 The general term in the expansion of (1+2x)^8 is C(8,k)(2x)^k Therefore, the term containing x^6 in the expansion of (2-3x^2)(1+2x)^8 is...

thanks cepheid

Use the binomial theorem to expand each of the following. Simpify your answers (1-y^2)^5 Let a = 1 and b = -(y^2) Then using binomial theorem, you have: (a+b)^5 = C(5,0)a^5 + C(5,1)a^4 b + C(5,2)a^3 b^2 + C(5,3)a^2 b^3 + C(5,4)a b^4 + C(5,5)b^5 Substitute a = 1 and b = -(y^2) (1-y^2)^5 =...

Ok, so ((-y)^2)^3 = (-y)^6

but that would make the answer -625, not 625

This is probably a very easy question but it is messing me up. (-y^2)^2 = (-y)^4 but (-y^2)^3 = (-y^6) why does one exponent need to be inside the brackets and the other outside? Example: (-5^2)^2 = 625 = (-5)^4 (-5^2)^3 = -15625 = (-5^6)

Thanks Dick. It must hurt to help sometimes...

Ok, so I was right before. n!/(n-r)!/n = (n-1)!/(n-r)! but (n-r)! is equal to ((n-1)-(r-1))! therefore, n!/(n-r)!/n = (n-1)!/((n-1)-(r-1))! (n-1)!/((n-1)-(r-1)) = P(n-1, r-1) therefore, n!/(n-r)!/n = P(n-1, r-1) You have no idea how much trouble my algebra gets me into. Here I am doing...

I understand everything except how r/n becomes r-1 n!/n = (n-1)! (n-r)!/n = ((n-1)-(r/n))!

Ok, how about this: P(n,r)/n = n!/(n-r)!/n = (n-1)!/(n-r)! = P(n-1, r-1) EDIT: Actually, wouldn't (n-1)!/(n-r)! be P(n-1, r)? 6!/3! = 120, 120/6 = 20 5!/3! = 20 while 5!/2! = 60

I know that the statement is true but I don't think I can use a specific example as that would be conjecture and not a proof. P(n,r)= n!/(n-r)! = n x (n-1) x (n-2) x ... x (n - r + 1)