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That is exactly why! It doesn't react, so it remains in the reaction vessel. The other products (NaCl, H2O) are neutral, so they don't affect the pH. Thus, the remaining HCl is the only chemical changing the pH from 7.

17. Alkanes: dispersion forces Alkanols: dispersion forces, polar hydroxyl group which allows for the formation of hydrogen bonds Alkanoic acids: dispersion forces, polar hydroxyl group and carboxyl group allowing for more hydrogen bond formation than alkanols Since hydrogen bonds are...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Ah right my bad. Thanks for the correction.

$18. The number of moles of each solution can be calculated using $c=\frac{n}{V}\to n=cV$. This formula shows us that there is 0.0016 moles of HCl and 0.0015 moles of NaOH. Since they react in a 1:1 ratio, the HCl is in excess by 0.0001 moles ($0.0016-0.0015$). Again using $c=\frac{n}{V}$, the...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $For $f(x)=ax+b$, where $a$ and $b$ are arbitrary constants, integrating $e^{f(x)}$ gives you $\frac{1}{f'(x)}e^{f(x)}+C$, whereas differentiating $e^{f(x)}$ gives you $f'(x)e^{f(x)}$. So the integral of $e^{kx}$ is...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread It isn't an indices law. You're factoring out 2^n like so: \begin{align*}2^{n+1}-2^{n}&=2^n\times2^1-2^n \\&=2^n(2-1) \\&=2^n\times 1 \\&=2^n\end{align*}

Re: Harder HSC Papers http://4unitmaths.com/hsc1981-1989.pdf. Go to the second last page.

Re: HSC 2015 3U Marathon b) For each, you pick a group of numbers where order does not matter. Once you have that group, there is only one way of arranging them in either ascending or descending order.

\begin{align*}\cos^2{x}+7\sin{x}\cos{x}&=3 \\\cos^2{x}+7\sin{x}\cos{x}&=3\left (\sin^2{x}+\cos^2{x} \right ) \\3\sin^2{x}+3\cos^2{x}-\cos^2{x}-7\sin{x}\cos{x}&=0 \\3\sin^2{x}-7\sin{x}\cos{x}+2\cos^2{x}&=0 \\3\sin^2{x}-6\sin{x}\cos{x}-\sin{x}\cos{x}+2\cos^2{x}&=0 \\3\sin{x} \left...

Of course! You can definitely do well in year 12. Just make sure you work hard and you'll be fine.

Re: HSC 2015 2U Marathon Here is a more cleaner version of what you wrote: \cos^{-1}{(\frac{2}{5})}=66.42. \, \cos{\theta}$ is positive in the 1st and 4th quadrants. $ \therefore \theta =66.42, (360-66.42) Use dollar signs ($) to enclose text which doesn't require mathematical characters...

From the syllabus: "The Preliminary course contains content that is considered assumed knowledge for the HSC course." I think it is mainly fundamental concepts/skills like mole calculations, intermolecular/intramolecular forces etc.

MRI is the hardest part of Medical Physics in my opinion. If you're having trouble I can try and help you out.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread \begin{align*}\ln{\frac{6}{e^3-e}}&=\ln\left [{\frac{1}{e} \times \frac{6}{e^2-1}} \right ] \\\textup{Since }\ln{ab}&=\ln{a}+\ln{b} \\&= \ln{\frac{6}{e^2-1}}+\ln{e^{-1}} \\&= \ln{\frac{6}{e^2-1}}-\ln{e} \\&=...

$Let the molecular mass of element $X=M_X$. The molecular mass of $XF_3$ is then $M_X+3\times19=M_X+57$. The number of moles of $XF_3$ in 2.5g is $\frac{2.5}{M_X+57}$ moles, since $n=\frac{m}{M}$. The concentration of 2.5g of $XF_3$ in 200mL is $\frac{\frac{2.5}{M_X+57}}{0.2}$M, since...

No worries :). But don't forget the restrictions of the locus. This is question 13 (c) (iii) from the 2014 HSC Extension 1 Paper, so if you can't figure it out have a read of Carrotsticks' solutions (page 10): https://www.dropbox.com/s/zt96vjplroj5xtm/HSC%202014%20Extension%201%20Solutions.pdf?dl=0

You're pretty close. The radius is a constant term (it doesn't involve y) so it isn't \sqrt{2ay}. What would you do if you were required to change x^2-4x+y^2=0 to the general equation of a circle?

A useful mnemonic is CAWCS: Carbonate + Acid -> Water + Carbon dioxide + Salt

\begin{align*} \tan{(A+B)}&=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}} \\ A+B+C&=180^{\circ} \textup{ (Angle sum of a triangle)} \\\tan{\left [(A+B)+C \right ]}&=\frac{\tan{(A+B)}+\tan{C}}{1-\tan{(A+B)}\tan{C}}...

2008 2U Question 10 (b)