Answer:
[Acetic acid] = 0.07 M
[Acetate] = 0.13 M
Explanation:
pH of buffer = 5
pKa of acetic acid = 4.76
![pH=p_{Ka} + log\frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3Dp_%7BKa%7D%20%2B%20log%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now using Henderson-Hasselbalch equation
![5=4.76 + log\frac{[Acetate]}{[Acetic\;acid]}](https://tex.z-dn.net/?f=5%3D4.76%20%2B%20log%5Cfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3Bacid%5D%7D)
![log\frac{[Acetate]}{[Acetic\;acid]} = 0.24](https://tex.z-dn.net/?f=log%5Cfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3Bacid%5D%7D%20%3D%200.24)
![\frac{[Acetate]}{[Acetic\;acid]} = 1.74](https://tex.z-dn.net/?f=%5Cfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3Bacid%5D%7D%20%3D%201.74)
....... (1)
It is given that,
[Acetate] + [Acetic acid] = 0.2 M ....... (2)
Now solving both the above equations
[Acetate] = 1.74[Acetic acid]
Substitute the concentration of acetate ion in equation (2)
1.74[Acetic acid] + [Acetic acid] = 0.2 M
[Acetic acid] = 0.2/2.74 = 0.07 M
[Acetate] = 0.2 - 0.07 = 0.13 M
When something burns it usually means something is being damaged or being changed such as colour, smell, or shape.
The moles of water that are produced from 373 moles of ammonia present in any chemical reaction is 746 moles.
<h3>What is the stoichiometry?</h3>
Stoichiometry of the reaction gives idea about the amount of entities present in any chemical reaction before and after the reaction.
Given chemical reaction is:
3Al + 3NH₄ClO₄ → Al₂O₃ + AlCl₃ + 3NO + 6H₂O
From the stoichiometry of the reaction it is clear that:
3 moles of Al = produces 6 moles of H₂O
373 moles of Al = produces 6/3×373 = 746 moles of H₂O
Hence required moles of water is 746 moles.
To know more about stoichiometry, visit the below link:
brainly.com/question/21931988
In region C the enzyme is saturated with substrate.
The fee of reaction whilst the enzyme is saturated with the substrate is the maximum charge of the response, Vmax. the relationship between the price of response and awareness of substrate relies upon the affinity of the enzyme for its substrate.
whilst all lively sites are occupied, the enzyme is saturated. At this saturation point, adding extra substrate makes no difference to the response rate.
Learn more about enzyme saturated here: brainly.com/question/13280099
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Answer:
A. DH° = –36 kJ
Explanation:
It is possible to obtain DH° of a reaction by the sum of DH° of half reactions. The DH° of the reaction:
B₂H₆(g) → 2B(s) + 3H₂(g)
Could be obtained from:
<em>(1) </em>2B(s) + 1.5O₂(g) → B₂O₃(s) DH° = –1273kJ
<em>(2) </em>B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(g) DH° = –2035kJ
<em>(3) </em>H₂(g) + 0.5O₂(g) → H₂O(g) DH° = –242kJ
The sum of (2) - (1) gives:
B₂H₆(g) + 1.5O₂(g) → 2B(s) + 3H₂O(g) DH° = -2035kJ - (-1273kJ) = -762kJ
Now, this reaction - 3×(3):
B₂H₆(g) → 2B(s) + 3H₂(g) DH° = -762kJ - (3×-242kJ) = -36kJ
Thus, right answer is:
<em>A. DH° = –36 kJ</em>
