# Formatting Sandbox III: please test stuff here

Spoiler warning: Be aware that this page contains a lot of MathJax, so it will probably need quite a while to load completely.

Sandbox II has become as clogged as Sandbox I was (and is), especially for users with 10,000 reputation or above. So, here is a new new post...

Before you delete a post here, please reduce it to one line without MathJax.

Old formatting sandboxes:

• Excellent! Many thanks for this new Sandbox! – Ed V Jun 2 '20 at 15:39
• Why do we have to reduce it to one line without MathJax? – Micelle Jun 19 '20 at 14:53
• @Micelle deleted posts are viewable by users with >10,000 reputation, so it's just a courtesy thing. Long mathjax posts make the page load slowly. – orthocresol Jun 19 '20 at 14:55
• I totally forgot to thank you for making a new one. :D – Martin - マーチン Jul 9 '20 at 16:21

### Somebody complained that the reaction I asked about doesn't exist. Why is this a problem?

Chemistry is an experimental science first and foremost, and this is especially true of synthetic chemistry, whether organic or inorganic.

What this means is that: we don't come up with theories from first principles, then use them to predict reactions. [We're getting better at doing this using quantum mechanics, but it's still very early days.] Instead, we find out that a reaction happens, and then we work backwards to come up with a model that explains it.

The ultimate source of "truth" in chemistry is not defined by our theories, but rather by our experimental observations. The theories only exist because they can explain experimental evidence.

[Incidentally, that's why there are so many exceptions to the theories. Many of them have a limited range of validity, in that they can only explain a certain subset of the experimental observations we have. A simple example is the octet rule. It works for quite a lot of organic molecules, but can completely fall apart in other contexts.]

So, asking "why does this reaction occur?" is only sensible if that reaction has actually occurred!

If nobody has done it before in real life, then we have no way of knowing whether it would actually occur. And secondly, if it doesn't actually occur and we come up with a theory to explain it, then there is no guarantee that that theory would be correct.

Comparing two back-calculated x values and their confidence intervals

1. Start with a data set for OLS processing. Here is a decent one, from the International Standards Organization (ISO):

or use your own data sets for OLS processing.

1. Perform OLS (ordinary least squares) and compute $$t_{slope}$$. This is shown in the figure below. Also note the critical t value, $$t_{critical}$$, for the desired confidence and correct degrees of freedom.

1. Compute the "g" convenience term, as shown in the figure below:

1. Perform the t test of the slope. The calibration curve is viable, i.e., may be used to compute valid back-calculated x values, if and only if g < 1. See the slide below:

1. Now the fun part. As per the next slide, compute the back-calculated x value, i.e., $$x_u$$. Also compute the confidence interval limits $$x_1$$ and $$x_2$$.

In general, the confidence interval is asymmetric about $$x_u$$. If g is negligible, then the simplified expressions (expressions 25 and 26) give a symmetric confidence interval about $$x_u$$. The literature reference is at the bottom of the figure and similar references exist. Personally, I use expressions 23 and 24 unless g < 0.001. Some sources say g < 0.05 is acceptable, but I ran Monte Carlo simulations that show that is too sloppy for me.

1. So now you compare the two confidence intervals and decide what to do: I have no guidance to give on that. Best of success!
• Unfortunately, the formula doesn't account for bounty system. However, the general trend "less Q&As — more rep" is indeed a sign of high-quality posts. – andselisk Dec 15 '20 at 10:16
• @andselisk Yeah, it is very crude and broad brush. Just a late night thought before I called it a day. – Ed V Dec 15 '20 at 13:35
• You might be interested in: data.stackexchange.com/chemistry/revision/1349860/1660907/… This simply takes the average score of all posts of a user. If the results don't show up hit "Run Query" near the bottom. A lot of the users there are no longer active... I find that upvotes were more generously given out in the early days of the site. (Your score is 4.12.) [PS I don't know anything about databases, I just copied this query off somebody else!] – orthocresol Dec 19 '20 at 16:38
• @orthocresol Wow, many thanks for the link! Very cool and helpful: I definitely see I have lots of room for improvement! And it helps to know that upvotes were more generously given in the early days, but that is very likely as it should be. So I will delete my crude late night idea. – Ed V Dec 19 '20 at 17:00
• No problem! I personally wouldn't consider it room for improvement, actually: I'm sure you noticed it already, but upvotes are pretty random and often the Q's/A's that get lots of upvotes are ones that are more accessible to the lay person (partly due to the Hot Network Question feature of SE, where you can "jump" to popular questions on other sites and vote). I've previously dumped some of my thoughts on the matter here. – orthocresol Dec 19 '20 at 17:08

$$h=\frac{I_{\text{const}}\cdot R_{\text{ref}}(1+\alpha\Delta T)}{A_{\text{filament}}(T-T_{\text{flow}})}$$

$\require{begingroup}\begingroup$ $\def\pi{\neq 3.14}$

$$\require{begingroup}\begingroup$$ $$\def\pi{\neq 3.14}$$

The ratio of a circle's circumference to its diameter is $\pi$.

The ratio of a circle's circumference to its diameter is $$\pi$$.

$\endgroup$

$$\endgroup$$

• $\pi here doesn't leak through:$\pi\$ – orthocresol Jan 30 at 15:15