Answer:
It favors the forward reaction.
Explanation:
According to Le Chatelier's Principle, when a system at equilibrium suffers a perturbation, the system will react in order to counteract the effect of such perturbation.
If more reactant is added, the system will try to decrease its concentration. It will do so by favoring the forward reaction, decreasing the concentration of the reactant and increasing the concentration of the products, in order to re-establish the equilibrium.
<u>Answer:</u> The percent composition by mass of hydrogen in given compound is 6.33 %
<u>Explanation:</u>
We are given:
A chemical compound having chemical formula of 
It is made up by the combination of 1 nitrogen atom, 5 hydrogen atoms, 1 carbon atom and 3 oxygen atoms
To calculate the percentage composition by mass of hydrogen in the compound, we use the equation:
Mass of compound = ![[(1\times 14)+(5\times 1)+(1\times 12)+(3\times 16)]=79g/mol](https://tex.z-dn.net/?f=%5B%281%5Ctimes%2014%29%2B%285%5Ctimes%201%29%2B%281%5Ctimes%2012%29%2B%283%5Ctimes%2016%29%5D%3D79g%2Fmol)
Mass of hydrogen = 
Putting values in above equation, we get:
Hence, the percent composition by mass of hydrogen in given compound is 6.33 %
If you have a book, read it!! I promise you, it tells you this answer!
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
H3PO4 has molecular weight of approximately 98 grams per
mole. 4.50 M is equal to 4.50 mole per 1000 mL solution of H3PO4. 255 mL times
4.50 mol /1000 mL times 98 g/mol is equal to 112.455 grams. Note that I
automatically equate 1 Liter to 1000 mL since the given volume is in mL for
easier computation.
