Anon E. Moose アナンイムース

I know that the Taylor Series formula is:

f(x) = Σ[n=0,∞] fⁿ(x₀) * (x - x₀)ⁿ / n!

However, what if I want to take the Taylor Series as x tends towards infinity? ∴ x₀ = ∞, and this formula becomes nonsensical. Is there a specific formula to perform this operation?

Example:

I have the function:

f(x) = 1/√[1 - 1/(4x²)]

When I ask Wolfram for the Taylor Series, it gives me:

1 + 1/(8x²) + 3/(128x^4) + ...

[Source: http://www.wolframalpha.com/input/?i=taylor+series... ]

How did they get this answer?

I know this seems like a stupid question, but why do we separate digits with commas every 3 digits? Is there some sort of reason for this? is it easier to count over 3 digits, rather than 5 or 10? I realize it's partly because of our naming system (thousand, million, billion, trillion), but 3 seems like a very arbitrary number to me.

100 Quintillion

1 * 10^20

100,000,000,000,000,000,000

1,00000,00000,00000,00000

1 Quadripenten (new number naming system?)

1,0000000000,0000000000

1 Billidecen

I realize you would need new number systems, so I guess

10 = ten

100 = duen

1000 = trien

10000 = quadren

100000 = penten

------------------------

1000000 = sexten

10000000 = septen

100000000 = octen

1000000000 = nonen

10000000000 = decen

I figure if our number system is base ten, why not separate our digits in groups of 10? or maybe 5? something that's easier to simplify exponents with, rather than by 3 (doesn't divide well with 10).

I suppose the choice of 3 digits might be due to human memory; maybe someone decided that groups of 3 digits were the easiest to remember.

Another reason might be that it's faster to say

Example:

1,243,463

one million two hundred fourty three thousand four hundred sixty three

12,43463

twelve millipenten four quadren three trien four duen sixty three

Although the base-10^5 uses less characters, I suppose it should depend more on the amount of syllables. Both of them end up having 17 syllables as well.

Etymology:

http://www.etymonline.com/index.php?term=thousand

http://dictionary.reference.com/browse/thousand

I'm just curious as to what everyone's opinions on this are. I doubt any change would ever happen, just like converting from the US system of weights and measures to the metric system, or from pi to tau.

I don't know how to solve this. Could someone please help? It's not an assignment or anything, I just haven't taken diffeq or anything, and I'm not sure how to solve something like this. Be it Laplace Transforms or whatever, just please tell me how to solve this.

x"(t) + k/x² = 0

Where x'(0) = v0

And x(0) = x0

Part A -

Calculate the pH of a solution prepared by dissolving 0.750 mol of NH3 and 0.250 mol of NH4Cl in water sufficient to yield 1.00 L of solution. The Kb of ammonia is 1.8 x 10^(-5).

pOH = pKb + log[(conj. acid) / (base)]

pOH = -log(1.8*10^(-5)) + log[0.750 / 0.250]

pOH = 5.22

pH = 14 - pOH = 14 - 5.22 = 9.78

Answer 1: pH = 9.74

Part B -

Calculate the pH of the above solution after addition of 0.25 mol of HCl, assuming there is no change in volume.

Answer 2: 9.56

This part I don't understand how to do. Could someone show or explain it to me? I tried assuming that HCl completely dissociates; I tried writing it with an initial concentration for each: [NH4] = 0.250, [NH3] = 0.750, [H+] = 0.250. But this didn't give me any sort of definite answer, and I was unsure how to proceed. I have a test on Monday, so it would be great if someone could help me on this today. Thanks in advance.

Calculate [OH-] and pH of a 0.10 M NaCN solution. HCN (Ka = 4.9*10^(-10))

Do I write the equation as follows:

NaCN(aq) → Na+(aq) + CN-(aq)

Assuming that NaCN is an ionic compound (composed of a monatomic cation and a polyatomic anion) it will completely dissociate in water, therefore [NaCN] = [CN-]

Then, since HCN (Prussic Acid) is a weak acid, write the equilibrium equation between the acid and its conjugate base:

..... H2O(l) + CN-(aq) → HCN(aq) + OH-(aq)

Pure solids and liquids are ignored in equilibrium reactions, so water is ignored.

I ........ .......... 0.1 . . | . . . . 0 . . . . . 0

C ...... ............ -x . . | . . . .+x . . . . +x

E ...... .......... 0.1 . . | . . . . x . . . . . x

Since Ka was given, change it to Kb by using:

Ka * Kb = Kw

4.9*10^(-10) * Kb = 10^(-14)

Kb = 2.0*10^(-5)

Then set up the base dissociation constant equation (since it was written with a CN- (base) on the left and HCN (acid) on the right). It is assumed that x is significantly smaller than [CN-], just to make the calculations easier (avoiding quadratics).

Kb = [HCN] [OH-] / [CN-]

2.0*10^(-5) = x² / 0.1

x² = 2.0*10^(-6)

x = [OH-] = 1.4*10^(-3) M

pOH = -log[OH-] = -log(1.4*10^(-3)) = 2.8

pH = 14 - pOH = 14 - 2.8 = 11.2

Is this solution correct? I tried doing it myself, but I'm pretty unsure of my work, so I would appreciate if someone could verify it. I have a test on Monday, but it would be nice if I could get immediate feedback on how I'm doing.

The acid-dissociation constants of sulfurous acid (H2SO3) are Ka1 = 1.7*10^(-2) and Ka2 = 6.4*10^(-8) at 25.0 °C. Calculate the pH of a 0.163 M aqueous solution of sulfurous acid.

I tried following the example in my book, but my answer didn't match any of the choices.

Ka1 = [H+][HSO3-]/[H2SO3] = x²/(0.163 - x) = 1.7*10^(-2)

→ x ≈ 0.0448221 M

Ka2 = [H+][SO3^2-]/[HSO3-] = (0.0448221 + y)(y)/(0.163 - 0.0448221) = 6.4*10^(-8)

→ y ≈ 1.68742*10^(-7) M

[H+] = x + y = 0.0448221 + 1.68742*10^(-7) = 0.0448222687 M

pH = -log[H+] = -log(0.0448222687) = 1.35

However, my choices are:

A) 1.30

B) 1.86

C) 4.53

Could someone please clarify?

Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.85 cm2 and insulation thickness of 0.0400 mm.

I tried using the formula, C = ε₀ ∙ A / d:

Giving me C = (8.854e-12)(1.85e-4)/(0.04e-3) = 4.09e-11 * (1e+12) = 40.9 pF.

However the software tells me this answer is wrong. The answer is somewhere in the range of 10~100pF, but I don't want to randomly guess at it. Could someone explain to me how to solve this problem?

Cr(OH)3(s) + 3HNO3(aq) --> Cr(NO3)3(aq) + 3H2O(l)

This is the correct balanced chemical equation.

I have tried entering both:

3OH-(aq) + 3H+(aq) --> 3H2O(l)

3OH-(s) + 3H+(aq) --> 3H2O(l)

But neither are marked correct. Could someone please tell me what is wrong with my answer?