Solids maintain their shape, whereas fluids do not because <span>the molecules in solids maintain a regular pattern and only vibrate, or move very slowly. The correct option among all the options that are given in the question is the last option or option "d". I hope the answer has come to your help.</span>
Empirical formula is the simplest way the molecular formula can be wrote so here 7 goes into all of these so it would be CH2O
I think it's 2:1 or 2:1:4. I mostly think it's 2:1 though. (:
Answer:
19.79%
Explanation:
mass % = (mass solute / total mass) * 100
total mass = 129.54 + 525 = 654.54
solute = C6H12O6
(129.54/654.54) = .1979
.1979 * 100 = 19.79%
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⁻ .