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sleep is for the weak

no they are 100% correct :) didn't you read?

a neutralisation reaction will occur between the NaOH and HCl, so just look at how much of each you have, which is in excess, and what the pH will be due to this excess after neutralisation has completed (I think)

screamo/skram : * noisy sins of the insect *, the birds are spies they report to the trees, saetia, amalthea, 1000 travels of jaharawlal, analena, +other stuffz i don't listen to this stuff as much any more but it was really cool musix

moss icon (80s), indian summer, native nod, don martin three, bob tilton, days in december yeah jazz, mineral, ethel meserve etc. type stuff is good too

well you draw a line of best fit with your experimental data points, so it may or may not pass through the origin

sum of n square terms starting at 1, is n(n+1)(2n+1)/6

this is what i get too

HCO3- + H+ <=> H2CO3 by shifting the equilibrium to the right (ie more carbonic acid) we reduce the concentration of H+ .'. increase the pH. Note that the bicarbonate is acting as a base. H2CO3 would usually decrease a solutions pH when it is added, as it donates hydrogen ions, but in this case...

molar mass of Ba(OH)2 = 137.3 + 2(16.00 + 1.008) = 171.3g so there are 0.75/171.3 = 4.38 * 10-3 moles of Ba(OH)2 in the 2.0L So [OH-] = 2*(4.38 * 10-3)/2.0 as there are 2 OH- for each Ba(OH)2 [OH-] = 4.38 * 10-3 using [OH-][H+] = 10-14 for water [H+] = 10-14/ [OH-] = 2.28 * 10-12 pH =...

if z = Rcis@ then z3 = R3cis[3@] for the condition to hold, what ranges of values can 3@ have? so what range does @ have?

do you mean x² + y² + x? so z = 0, -1

ostentatious: the further away two masses are, the higher their GPE is. For convenience, we say that at the maximum distance, ie. as it approaches infinity, the absolute GPE is zero. So at any other point it will be less than this. The amount below zero it will be is equal to the amount of work...

basically like you said apply halving the interval, which shows the root is between 0.5 and 0.75, so it lies closer to 0.5 i think that should be clear enough

that's not what 6ii is asking ^ you just draw the circle w/ radius 1, centre (2,0) -> this is the graph of z+2, where |z|=1 then you find the lines through the origin which are tangent to the circle to find the min/max arg

it's easier to equate coefficients as timothy + drongoski pointed out for mine, with the roots, you get things in the form n +- rt(n^2 - 1), so if a,b,c = cos[1,5,7 pi/9] you can then group things like (a + rt(a2 - 1) + a - rt(a2-1))(b + rt(b2-1) + b - rt(b2-1)) [3 like this] + (a + rt(a2))(a...

you could solve 10) iii) by solving each quadratic, then using the sum of the roots 2 at a time = 0 for the original equation once you group the pairs together it simplifies down to 4cos[pi/9]cos[5pi/9] + 4cos[5pi/9]cos[7pi/9] + 4 cos[7pi/9]cos[pi/9] + 3 = 0 which gives the desired result

assume true for n=k 3^{3k} + 2^{k+2} = 5M show true for n = k+1 3^{3(k+1)} + 2^{(k+1)+2)}\\ = 27\times3^{3k} + 2\times2^{k+1}\\ = 25\times3^{3k} + 2\times(3^{3k} + 2^{k+2})\\ = 5\times5\times3^{3k} + 2\times5M\\ = 5(5\times3^{3k} + 2M) maybe you were only adding one to the 3 power rather than...

1/(1+tn) <= 1/(1+tn+1) for 0 <= t <= 1 but the equality only exists when t = 0, 1 as the integral also has all other values in the range, it can't be equal

i don't think they can be equal