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I have two versions of this paper. Question 5 is different on the two versions. One version has Q11(d) as a 4 mark partial fractions problem, find A if \int_2^5\cfrac{8x - 1}{(2x - 1)(x + 1)}\ \!dx = \ln{A} whereas the other has 11(d) as a 2 mark integration of \ln{x} and a 2 mark 11(e) proof...

Looked at another way: \begin{align*} f'(x) &= g'(x) \qquad \text{proven in part (i)} \\ \text{Thus, on integrating:} \qquad \int f'(x)\ \!dx &= \int g'(x)\ \!dx \\ f(x) &= g(x) + C \qquad \text{for some constant $C$} \end{align*} Now, f(x) and g(x) have the same domain, x \in \big[-1,\...

Note that complexes of silver typically have two ligands, like: diamminesilver(I) cation = [Ag(NH3)2]+ dicyanoargentate(I) anion = [Ag(CN)2]- For most HSC examples, the equilibrium constants for their formation are large and so complexation is strongly favoured. For example, Q31 of the 2022...

The working for tywebb's factorisation: \begin{align*} 5z^4 - 11z^3 + 16z^2 - 11z + 5 &= z^2\left(5z^2 - 11z + 16 - \cfrac{11}{z} + \cfrac{5}{z^2}\right) \\ &= z^2\left[5\left(z^2 + \cfrac{1}{z^2}\right) - 11\left(z + \cfrac{1}{z}\right) + 16\right] \\ &= z^2\left[5\left(\left(z +...

Not *that* slow, but I do agree that tywebb's approach is preferable.

(5x2 + Ax + B) (x2 + Cx + D) \begin{align*} 5x^4 - 11x^3 + 16x^2 - 11x + 5 &= (5x^2 + Ax + B)(x^2 + Cx + D) \\ &= 5x^4 + (5C + A)x^3 + (5D + B + AC)x^2 + (BC + AD)x + BD \\ \\ \text{On equating coefficients:} \qquad A + 5C &= -11 \qquad \text{(1)} \\ 5D + B + AC &= 16 \qquad \text{(2)} \\ BC +...

Since all coefficients are real and rational, any complex roots will come in conjugate pairs. It must be factorisable as a product of two quadratics, and with a leading term of 5x4, you could start from (5x2 + Ax + B) (x2 + Cx + D) expand, and equate coefficients to get equations in A, B, C...

The crucial point about complexation is that these species are formed and held together by covalent bonds between the metal centre and each ligand. They are almost always coordinate covalent bonds, whereby the ligand donates a lone pair of electrons to form the covalent bond, rather than by one...

No, it would not be incorrect to say that, because it is true. The statements: HA is the conjugate acid of A- and A- is the conjugate base of HA are logically equivalent.

Sort of... The intensity before passing through the solution is defined to be I0. After passing through, the intensity is 42.6% of what it was before. That is, it has become I = 42.6% x I0 = 0.426I0. You then know the value of the ratio I0 / I = I0 / 0.426I0 = 1 / 0.426, on cancelling. This...

Roughly 42 ppm Cu2+ is a concentration of 42 mg / L divided by 63.55 g / mol to give [Cu2+] = 6.6089... x 10-4 mol / L Since copper(II) chloride is CuCl2, there are two chloride anions for every copper(II) cation, and so [Cl-] = 2 x 6.6089... x 10-4 = 1.32... x 10-3 mol / L And the answer is...

In part (ii), you showed that a^3 + b^3 + c^3 \ge 3abc so long as a, b, and c are real and positive. \text{Let } a = \sqrt[3]{\cfrac{x^3}{1+x^3}} \qquad \text{and} \qquad b = \sqrt[3]{\cfrac{y^3}{1+y^3}} \qquad \text{and} \qquad c = \sqrt[3]{\cfrac{1}{1+z^3}} where x, y, and z are suitably...

If pi/2 < y < pi, then it is an obtuse angle, and so the relationship to tan y can't rest on applying Pythagoras' Theorem from when y is acute. The method will work for both x > 0 and x < 0 cases, but not without justification / explanation. After all, the question is looking for a solution...

You haven't accounted for the case when x<0.

I think the policy is to mark your first answer, unless it is crossed out. So, it depends on which is written first. Also, be aware that contradictory material can lead to mark reductions.

Taking the unit vectors i and j as upwards and to the right, the forces acting are: a weight force of W = -mgj the tension in the string of T = +Fj the friction force of F(\cos{\alpha}\,\boldsymbol{i} + \sin{\alpha}\,\boldsymbol{j}) the normal force of N(-\sin{\alpha}\,\boldsymbol{i} +...

The solution is 12.5 ppm in Cu2+ ions but the solid formed is CuCO3. The question is also implying that all of the copper can be precipitated as CuCO3, which isn't true because there will be a solubility equilibrium, but almost all of it can be precipitated if enough carbonate is added without...

Most texts / teachers / tutors don't point out that the C-H bands can be informative. The bands at lower wavenumber for an aldehyde are the most obvious, but signals above the usual range can also help, reflecting H-C=C, H-C(aromatic, like a benzene ring), and H-C(alkyne). All are sharp and...

It's a ketone. Aldehydes have medium to strong bands (usually two of them) below the regular C-H band for the C-H of the terminal -CHO - somewhere like 2700 - 2800 cm-1. The IR spectrum for propanal can be seen at https: / /...

Extending on @nsw..wollongong's post: There are two isomeric pentanone's: 2-pentanone, CH3-CO-CH2-CH2-CH3, in which there are 5 x C environments and 4 x H environments, because the molecule is not symmetrical about the carbonyl group (-CO-), and 3-pentanone, CH3-CH2-CO-CH2-CH3, in which there...