Answer:
Na₂CO₃•H₂O
Explanation:
After it is heated, the remaining mass is the mass of sodium carbonate.
30.2 g Na₂CO₃
Mass is conserved, so the difference is the mass of the water:
35.4 g − 30.2 g = 5.2 g H₂O
Convert masses to moles:
30.2 g Na₂CO₃ × (1 mol Na₂CO₃ / 106 g Na₂CO₃) = 0.285 mol Na₂CO₃
5.2 g H₂O × (1 mol H₂O / 18.0 g H₂O) = 0.289 mol H₂O
Normalize by dividing by the smallest:
0.285 / 0.285 = 1.00 mol Na₂CO₃
0.289 / 0.285 = 1.01 mol H₂O
The ratio is approximately 1:1. So the formula of the hydrate is Na₂CO₃•H₂O.
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:
Explanation:
H ₂ S O ₄ + 2 N a O H ⟶ 2 H ₂ O + N a ₂ S O ₄
29.09 mL of 0.639 M N a O H is mixed with 213.8 mL of H ₂ S O ₄
Let the concentration of H ₂ S O ₄ be S₂ .
In terms of normal or equivalent solution is will be 2 N solution
From the formula S₁ V₁ = S₂ V₂
= 29.09 x .639 = 213.8 x S₂
S₂ = .087 N solution
In terms of molar solution it will be .087 / 2 M
= .0435 M
Answer:
D- the grass population and hawk population would decrease. The snake population would increase..