Answer:
It lights on fire
Explanation:
The friction sparks the match causing it to go on fire.
Answer:
80mL of 1.00M NaOH
Explanation:
Using H-H equation, we can determine oH of a buffer as acetate buffer. First, we need to determine amount of acetate ion and acetic acid at pH 3.50 and 5.07. Then, with the reaction of NaOH with acetic acid we can find the amount of 1.00M NaOH that must be added:
At pH 3.50:
pH = pka + log [C₂H₃O₂⁻] / [HC₂H₃O₂]
3.50 = 4.74 + log [C₂H₃O₂⁻] / [HC₂H₃O₂]
0.057544 = [C₂H₃O₂⁻] / [HC₂H₃O₂] (1)
Using and replacing in (1):
[HC₂H₃O₂] + [C₂H₃O₂⁻] = 0.250 M
[HC₂H₃O₂] + 0.057544[HC₂H₃O₂] = 0.250 M
1.057544 [HC₂H₃O₂] = 0.250M
[HC₂H₃O₂] = 0.2364M * 0.500L = 0.1182 moles of acetic acid at first pH
At pH 5.07:
pH = pka + log [C₂H₃O₂⁻] / [HC₂H₃O₂]
5.07 = 4.74 + log [C₂H₃O₂⁻] / [HC₂H₃O₂]
2.13796= [C₂H₃O₂⁻] / [HC₂H₃O₂] (1)
Using and replacing in (1):
[HC₂H₃O₂] + 2.13796[HC₂H₃O₂] = 0.250 M
3.13796 [HC₂H₃O₂] = 0.250M
[HC₂H₃O₂] = 0.07967M * 0.500L = 0.0398 moles of acetic acid at first pH
Now, NaOH reacts with HC₂H₃O₂ as follows:
NaOH + HC₂H₃O₂ → NaC₂H₃O₂ + H₂O
As moles of acetic acid decreases from 0,1198 moles - 0,0398 moles = 0,08 moles of acetic acid are consumed = 0,08 moles of NaOH
0,08 mol NaOH * (1L / 1mol) = 0,08L of 1.00M NaOH =
B) 13 nM
C) 8.2 pM
D) 6.9 mM
Answer:
Binding affinity measures the strength of the interaction between a molecule to its ligand; it is expressed in terms of the equilibrium dissociation constant; and the higher value of this constant, the more weaker the binding between the molecule and the ligand is. On the other hand, small constans means that the interaction is tight. So "C" binds most tightly to the enzyme and "D" binds least tightly.
Answer:
The value of Keq is 4e-9. See the solution below
Explanation:
We need to balanced rhe equation and use the formula of the Keq
The quantity of acetic acid that is needed to prepare the 500 mL buffer is 9.0075 grams.
Given the following data:
First of all, we would write the equilibrium chemical reaction for acetate-acetic acid as follows:
Next, we would calculate HA by applying Henderson-Hasselbalch equation:
Where:
Substituting the given parameters into the formula, we have;
For the concentration of both acids, we have:
For acetate ion:
At a volume of 0.5 liters, we have:
By stoichiometry:
Total moles = = 0.15 moles.
Mass = 9.0075 grams.
Read more on moles here: brainly.com/question/3173452
Answer:
You will need 9,0 g of acetic acid
Explanation:
The equilibrium acetate-acetic acid is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,76
Using Henderson-Hasselbalch you will obtain:
pH = pka + log₁₀
Where HA is acetic acid and A⁻ is acetate ion
4,90 = 4,76 + log₁₀
1,38 = (1)
As acetate concentration is 0,300M:
0,300M = [HA] + [A⁻] (2)
Replacing (2) in (1):
[HA] = 0,126 M
And:
[A⁻] = 0,174 M
As you need to produce 500 mL:
0,5 L × 0,126 M = 0,063 moles of acetic acid
0,5 L × 0,174 M = 0,087 moles of acetate
To produce moles of acetate from acetic acid:
CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O
Thus, moles of acetate are equivalents to moles of NaOH and all acetates comes from acetic acid, thus:
0,087 moles of acetate + 0,063 moles of acetic acid ≡ 0,15 moles of acetic acid × = 9,0 g of acetic acid
I hope it helps!
Answer: Option (c) is the correct answer.
Explanation:
Entropy is defined as the degree of randomness that is present within the particles of a substance.
As is ionic in nature. Hence, when it is added to water then it will readily dissociate into ammonium ions () and nitrate ions ().
Therefore, it means that ions of ammonium nitrate will be free to move from one place to another. Hence, there will occur an increase in entropy.
Thus, we can conclude that ammonium nitrate () dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol because the overall entropy of the system increases upon dissolution of this strong electrolyte.
The formula for osmotic pressure is:
where is osmotic pressure, is van't Hoff's factor, molarity, is Ideal gas constant, and T is Temperature.
= 132 atm
The van't Hoff's factor for glucose, = 1
Substituting the values in the above equation we get,
So, the molarity of the solution is .