With the equation <span>A₂+2B₂→2AB₂ we know that one mole of A₂ is necessary to react with two moles of B₂ and with that, we obtain two moles of the product. ( you can see this by just seeing what numbers are before the letters)
Since it's needed more quantity of B than A, B is the limited reagent.
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<span>Because it is capable of forming multiple covalent bonds. I hope 15 hours isn't too late to answer your question.
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Answer:
Explanation:
Hello,
In this case, given the acid, we can suppose a simple dissociation as:
Which occurs in aqueous phase, therefore, the law of mass action is written by:
![Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
That in terms of the change 
due to the reaction's extent we can write:
But we prefer to compute the Kb due to its exceptional weakness:
Next, the acid dissociation in the presence of the base we have:
![Kb=\frac{[OH^-][HA]}{[A^-]}=1x10^{6}=\frac{x*x}{0.1-x}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BOH%5E-%5D%5BHA%5D%7D%7B%5BA%5E-%5D%7D%3D1x10%5E%7B6%7D%3D%5Cfrac%7Bx%2Ax%7D%7B0.1-x%7D)
Whose solution is 
which equals the concentration of hydroxyl in the solution, thus we compute the pOH:
![pOH=-log([OH^-])=-log(0.0999)=1](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%280.0999%29%3D1)
Finally, since the maximum scale is 14, we can compute the pH by knowing the pOH:
Regards.
The difference between ergosterol and cholesterol is much more obvious in their interactions with POPC and DOPC. Whereas cholesterol induces a strong condensing effect that thickens both POPC and DOPC bilayers, ergosterol shows no condensing effect in POPC and DOPC at all.
Usually it is the CuSO4 that is the limiting reagent.
<span>if all of the color of the solution was gone, but there was still some zinc metal mixed in with the copper metal produced, then Zn is the excess reagent </span>
<span>f all of the color of the solution was not gone, but there was no zinc metal left in with the blue copper solution , then Zn is the limiting reagent Hope this helps.</span>
