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The more latex-like invocation would be \(...\) for inline mode and \[...\] for display mode. This does not work as backslashes and brackets have a special meaning on stack exchange and oft...
Write and balance the half-reactions (reductions) for the following redox couples and determine $\Delta G^{\circ}$ and $E^{\circ}$ for each half-reaction from $G^{\circ}_f$ data.
$\ce{Pb3O4}$ / $\ce{Pb^{2+}}$
$\ce{NO2-}$ / $\ce{NH4+}$
$\ce{HSeO3−}$ / $\ce{SeO^{2−}_4}$
I know h...
that's true, but shouldn't matter, it's in the title tags
<title>organic chemistry - Why do unsaturated hydrocarbons burn with a sooty flame while saturated hydrocarbons burn with a clear flame when burnt in the air? - Chemistry Stack Exchange</title>
Yes, interestingly enough, "organic chemistry - why-do-unsaturated-hydrocarbons-burn-with-a-sooty-flame-while-saturated-hydrocar" has length 100, I wonder if it's a coincidence
lol... one never knows... these things are a sensitive issue
although I am less afraid of google facebook and co tracking my every move than I am afraid of the (any) government requesting a backdoor to my encrypted conversations.
@Martin-マーチン they are, given how new virtual movements start on twitter so quickly with simple retweets; that's why I stay away from these social networks
@Martin-マーチン well the gov won't have anything to backdoor request if google, fb, co won't track you in the first place
that's not the only way countries waste their money though; nuclear weapons are a big threat, every country has them these days, "trendy stuff" I should say :P imagine if one goes off by mistake...
Stuff like that makes me very sad. It's not like that there are multiple answers explaining that it is not that simple; but the way it is written is so absolute and without any evidence. (@pentavalentcarbon couldn't find the down-vote button?!)
Write and balance the half-reactions (reductions) for the following redox couples and determine $\Delta G^{\circ}$ and $E^{\circ}$ for each half-reaction from $G^{\circ}_f$ data.
$\ce{Pb3O4}$ / $\ce{Pb^{2+}}$
$\ce{NO2-}$ / $\ce{NH4+}$
$\ce{HSeO3−}$ / $\ce{SeO^{2−}_4}$
I know h...
The Periodic Table
The Periodic Table