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Everything posted by Dhamnekar Win,odd
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Indicators for Acids Titration
Dhamnekar Win,odd replied to Dhamnekar Win,odd's topic in Applied Chemistry
The following question is related to original question to some extent. Hence I am asking here.☺️ What is the explanation for the following acid-base behavior of Methyl Orange, an Indicator? I am working on the original question. Any Chemistry help will be accepted. -
How to choose indiacator here for these above mentioned Acids titation? Is the below-given chemical compound structure of 4- nitrophenol same as given in the above question?
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Answer to question (b) assumes Maxwell-Boltzmann statistics. But how to show that the probability of each distinguishable distribution is [math]n*(N)^r ?[/math] Why is Bose-Einstein statistics impossible here? Each distinguishable distribution have equal probability of [math]\frac{r!(n-1)!}{(n+r-1)!}[/math] under Bose-Einstein statistics.
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"AmmoniaConcentration" package in 'R'
Dhamnekar Win,odd replied to Dhamnekar Win,odd's topic in Homework Help
I tag this question as 'SOLVED' as I got correct answers to all my questions. ☺️ -
"AmmoniaConcentration" package in 'R'
Dhamnekar Win,odd replied to Dhamnekar Win,odd's topic in Homework Help
pKa = -Log(ka), pH = pKa + log (NH3/(NH3 + NH4+)). So, Ka= 4.33504889257e-10 Now, [math]\frac{1}{10^{pK_a- pH}}=\frac{NH_3}{NH_3 + NH_4^+}=3.44345173426e-3[/math] Why 1 is added to denominator NH3 + NH4+? I think Emerson et al. developed the equation pKa = 0.09018 + 2729.92/T on experimental basis. -
"AmmoniaConcentration" package in 'R'
Dhamnekar Win,odd replied to Dhamnekar Win,odd's topic in Homework Help
Rectified equation of f(fraction of NH3)= [math] \frac{1}{10^{(pKa-pH)} +1}[/math] -
Please read here and the answer the following questions if you know: How did the author derive the following equations? pKa = 0.09018 + 2727.92/T f = 1/(10(pKa − pH) + 1) How would you interpret the answers given by 'R' using this package? How is NH3_mgL computed in 'R' using this package? What is the denomination of total ammonia in the argument given by the author in the examples ? Is it mol/L or in grams?
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Maxima and Minima (vector calculus)
Dhamnekar Win,odd replied to Dhamnekar Win,odd's topic in Analysis and Calculus
> fdejong <- function(x,y){return(x^2 +y^2)* exp(-(x^2+y^2))} > x<- seq(-1,1 , length=2) > y <- x > z <- outer (x,y, fdejong) > require(lattice) Loading required package: lattice > wireframe(z, drape=T, col.regions=rainbow(100)) What is this? I don't get it.😕🤔😧 -
More details about the question: The following theorem gives sufficient conditions for a critical point to be a local maximum or minimum of a smooth function (i.e. a function whose partial derivatives of all orders exist and are continuous). What are the changes, we must make in this theorem, in case f(x,y, z) and f(w, x , y, z)? How can we use the aforesaid rectified theorem to answer the following question? Find three positive numbers x, y, z whose sum is 10 such that [math] x^2 \cdot y^2 \cdot z[/math] is a maximum. My attempt : The critical point is x=4, y=4, z =2 How to compute D in this case?
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Maxima and Minima (vector calculus)
Dhamnekar Win,odd replied to Dhamnekar Win,odd's topic in Analysis and Calculus
So, we will get the same first order correction for [math]1-\epsilon[/math] Isn't it? In which R package De Jong function is available ? -
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Rectified answer to (d): 2* (0.9)k - (0.81)k
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Rectified answer to (c) if k =5 :[math]\displaystyle\sum_{k=1}^{5}\binom{5}{k} (02-0.01)^2 (0.81)^{5-k}= 0.6513215599 [/math]= The probability that either 0 or 1 appears. So, neither 0 nor 1 appears is 1-0.6513215599 = 0.3486784401 = (0.9)5 × (0.9)5 So, we can express event (c) in terms of A and B as A*B or P(A)* P(B). QED.
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Answers to (a) and (b) are correct. [math](0.9)^5 \cdot (0.9)^5 = 0.3486784401 = (0.81)^5 [/math]whereas answer to (c) is (0.8)5 if k=5 So, there is minor difference of 0.1 in the probability in the answer to (c). I don't understand how did that arise?
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In hypergeometric distribution, n= population of n elements, r = sample size, n1 = elements recognized as having some defined criteria, n2 = n - n1 = remaining elements other than n1 . We seek the probability qk such that the sample size r contain exactly k recognized elements provided [math]k \geq 0, k \leq n_1 [/math] if n1 is smaller or [math] k \leq r[/math] if r is smaller. In such a case the probability [math]q_k =\frac{\binom{n_1}{k}\binom{n- n_1}{r - k}}{\binom{n}{r}} \tag {1}[/math] Now, I want to calculate n=8500, n1 =1000, r = 1000, k = 0 to 100. Inserting these values in (1), calculation of summation is very difficult. In such case, how can I use normal approximation to binomial distribution to find qk ? Do you have any clue or hint?
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