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Re: UNSW chit chat thread 2016 Define "good".

Re: MATH1231/1241/1251 SOS Thread Woops, I did a mistake in dimension and thought that Ax is still 3x5 matrix

Re: MATH1231/1241/1251 SOS Thread Is there missing detail like Ax = 0 or Ax is an element of ___

Re: UNSW chit chat thread 2016 It's a message

Re: Discrete Maths Sem 2 2016 Another method: if gcd(a, b) = d, then by Bézout's identity, there exists integers x and y such that ax + by = d. Dividing through by d yields (a/d)x + (b/d)y = 1 because d cannot be 0. Then reverse Bézout's identity to conclude that gcd(a/d, b/d) = 1.

Yes

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$Let $I = \int_0^1 x^m (1-x)^n dx $ and keep integrating by parts to reduce $x^m$ down to $x^0$,$ \\ \begin{align*} I &= \left[ -\frac{x^n(1-x)^{n+1}}{n+1} \right]_0^1 + \frac{m}{n+1} \int_0^1 x^{m-1}(1-x)^{n+1} dx \\&= \frac{m}{n+\color{red}{1}}\int_0^1...

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread (1+x)^{n-1} = \binom{n-1}{0} + \binom{n-1}{1}x + \dots + \binom{n-1}{n-2} x^{n-2} + \binom{n-1}{n-1}x^{n-1} \\ $Put $x = 1, \\\\ 2^{n-1} = \binom{n-1}{0} + \underbrace{\binom{n-1}{1} + \dots + \binom{n-1}{n-2}}_{\text{isolate...

Sorry, I don't know this one. I forgot nearly all of this but try going through each step of the experiment and see if you can spot any inconsistencies that arise

Sure

Yeh, don't do that. Pb(NO3)2 and PbSO4 were in 3:3 ratio. Doing that puts it in 3:9 ratio

yeh

Sorry, I misunderstood what you were saying. Pb(NO3)2 and PbSO4 coincidently have a mole ratio of 3:3, so 1:1. This does not mean you should always use the moles of a reactant (which supposedly has same mole ratio to the desired product) to find the moles of a product because it may be in...

Did you mean 0.085?

That's finding the number of moles of Al2(SO4)3 and you cannot use this to find the moles of PbSO4 because it's the excess. You must use the limiting reagent to find the moles of PbSO4

Why is the moles of PbSO4 = 0.0325? The rest is fine, well the idea not the numbers.

Yep :) Nice work! wait, don't times the molar mass of lead by 6

Yeh, the limiting reagent is Pb(NO3)2. Now what do you do find the mass of PbSO4?

Please do share your opinion on why you think tutoring is not worth it. I am very interested.