(B) is clearly the answer. A condensation polymer can be two monomers long (dimer).
EDIT: Yes, okay if you want to be incredibly nit-picky, you could say (B) is an oligomer (a molecule consisting of only a few repeating subunits - see http://goldbook.iupac.org/O04286.html). IMO it boils down to...
That equilibrium question for industrial chem is an interesting one. Typically, only a change in temperature will affect the value of K.
The initial and final values of 10 and 11 respectively (2 sf) seem pretty disparate to me.
re: HSC Chemistry Marathon Archive
Throwback Thursday - this was the paper I sat.
Think about the net ionic equations for each type of reaction:
\flushleft{\ce{H^+ + CO_3^{2-} \rightarrow H_2O + CO_2}} \\ {\ce{H^+ + HCO_3^- \rightarrow H_2O + CO_2}}
In both cases, we have a 1:1...
Re: HSC 2015 4U Marathon - Advanced Level
Suppose we set up a system of lines to accommodate this. This would imply at least 2 intersections, each of which are the meeting places for at least 2 lines (one intersection would have lines 1 and 2, and the second, lines 3 and 4). Our sequence would...
Re: HSC 2015 4U Marathon - Advanced Level
I would like to propose my own solution to the lines-in-a-plane problem. It's a simple two-case interpretation, which I hope it's comprehensive.
Looking at our definitions, we know that parallel lines never meet in a plane. Any one non-parallel line...
Why is it that https://student.unsw.edu.au/grade states the threshold for a HD WAM is 90?
EDIT: Never mind, this only applies "where grades have been entered without an associated mark". Had a bit of a shock there
Re: HSC 2015 4U Marathon
Perhaps a more systematic variation of your response would mention two things:
(1) the quadratic x_{n-1}-x_n is monotonically increasing above zero for x > 3, meaning that successive values of x starting from 4 get larger and larger. That is, we continue to move...
re: HSC Chemistry Marathon Archive
"Stick to one side"
Just to clear up what seems to be a vaguely articulated debate about the health effects of radioisotopes - first of all, it's the ionizing ability and not the penetrating power that's the real concern here. Radioisotopes knock electrons...
re: HSC Chemistry Marathon Archive
Work out the oxidation number/state of each element in redox pairs concerning water, for each reaction. Note that oxygen retains an OS of 2- unless it's in a peroxide, whilst hydrogen retains an OS of 1+ unless it's in a metal hydride. Additionally, pure...
I agree. 4 unit methods are quite often acknowledged in marker's comments. The only caveat being that many get too ambitious and misuse them, leading to remarks like "candidates tried to use Extension 2 methods where Extension 1 methods would have sufficed, resulting in convoluted scripts that...
Alternatively, we find R(x) first, using the remainder theorem: "the remainder of the division of a polynomial f(x) by some linear polynomial bx-a is equal to f(a/b)".
P(2/3) = 5/3, so that's our remainder.
We take the remainder away from what we're dividing to find what's exactly divisible...
A continuation of Drogonski's response - dividing the right-hand side by 3x-2, we get
2x + \frac{x+1}{3x-2} (*)
Looking at the fraction, we can express it as
\frac{1}{3}\cdot\frac{3(x+1)}{3x-2}
Now the second factor here can be manipulated into
\frac{3x-2}{3x-2} +\frac{5}{3x-2}...
Re: HSC 2015 3U Marathon
Frank's one is a bit tedious (unless, of course, there is a shorter way). Accounting for the saw-tooth shape of y = arcsin(sinx), such that each term of the sum oscillates in
y\in\left[-\frac{\pi}{2}, \frac{\pi}{2} \right]
we posit the following:
y = \begin{cases}...
Re: HSC 2015 3U Marathon
Consider selecting n members for an exec. team from a pool of 2n candidates, where men and women are equally numbered.
On the left-hand side, we use the symmetric property of combinations to change the summation to:
\sum_{k=0}^n\binom{n}{k}\binom{n}{n-k}
Now...
Re: HSC 2015 4U Marathon - Advanced Level
Alternatively, observing the standard limit:
\lim_{\frac{1}{x}\to0}\frac{\tan ^{-1}\left( \frac{1}{x} \right)}{\left( \frac{1}{x} \right)}
we can say that as x tends to infinity this summation tends to the harmonic series. By looking at the upper...
Re: HSC 2015 3U Marathon
This could work by induction on n. For the base case n = 1, we are forced to have k = 0:
\sum_{m=k}^0 \binom{m}{0} =\binom{1}{1}
which is true since both sides equate to 1.
Assuming this is true for 0\leq n\leq s-1, where 0\leq k<n we make the hypothesis...