To answer the questions,
(1) Activation energy is the amount of energy that is needed for the reaction to proceed, converting the reactant to products. The answer is letter B.
(2) The rate of chemical reaction normally increases as the reactant concentration is increased. The answer is letter C.
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Answer :
The correct answer for primary component of phosphate buffer at pH = 7.4 is H₂PO₄⁻ and HPO₄²⁻ .
<u>Buffer solution :</u>
It is a solution of mixture of weak acid and its conjugate base OR weak base and its conjugate acid . It resist any change in solution when small amount of strong acid or base is added .
<u>Capacity of a good buffer : </u>
A good buffer is identified when pH = pKa .
From Hasselbalch - Henderson equation which is as follows :
![pH = pka + log \frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=pH%20%3D%20pka%20%2B%20log%20%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If [A⁻] = [HA] ,
pH = pka + log 1
pH = pKa
This determines that if concentration of weak acid and its conjugate base are changed in small quantity , the capacity of buffer to maintain a constant pH is greatest at pka . If the amount of [A⁻] or [HA] is changed in large amount , the log value deviates more than +/- 1M and hence pH .
Hence Buffer has best capacity at pH = pka .
<u>Phosphate Buffer : </u>
Phosphate may have three types of acid-base pairs at different pka ( shown in image ).
Since the question is asking the pH = 7.4
At pH = 7.4 , the best phosphate buffer will have pka near to 7.4 .
If image is checked the acid - base pair " H₂PO₄⁻ and HPO₄²⁻ has pka 7.2 which is near to pH = 7.4 .
Hence we can say , the primary chemical component of phosphate buffer at pH = 7.4 is H₂PO₄⁻ and HPO₄²⁻ .
Answer:
17.6 moles of oxygen gas.
Explanation:
STP Variables:
P=1 atm
R=0.082
T=273 K
Use the PV=nRT, then plug in.
n=PV/RT
n= (1 atm)(395 L)/(0.082) (273 K)
Simplify.
n=17.6 moles of O2
Hoped this helped.
The 2 L of sucrose stock solution would contain similar
concentration with the 100 mL aliquot. Therefore the concentration of aliquot
is still 2 M.
The molar mass of sucrose is 342.3 g / mol. Therefore the
mass in a 100 mL (0.1 L) aliquot is:
mass = (2 mol / L) * 0.1 L * (342.3 g / mol)
<span>mass = 68.46 g</span>