Answer : The fraction of carbonic acid present in the blood is 5.95%
Explanation :
The mixture consists of carbonic acid ( H₂CO₃) and bicarbonate ion ( HCO₃⁻). This represents a mixture of weak acid and its conjugate which is a buffer.
The pH of a buffer is calculated using Henderson equation which is given below.
![pH = pKa + log \frac{[Base]}{[Acid]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%20%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D)
We have been given,
pH = 7.5
pKa of carbonic acid = 6.3
Let us plug in the values in Henderson equation to find the ratio Base/Acid.
![7.5 = 6.3 + log \frac{[base]}{[acid]}](https://tex.z-dn.net/?f=7.5%20%3D%206.3%20%2B%20log%20%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
![1.2 = log \frac{[base]}{[acid]}](https://tex.z-dn.net/?f=1.2%20%3D%20log%20%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
![\frac{[Base]}{[Acid]} = 10^{1.2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%20%3D%2010%5E%7B1.2%7D)
![\frac{[Base]}{[Acid]} = 15.8](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%20%3D%2015.8)
![[Base] = 15.8 \times [Acid]](https://tex.z-dn.net/?f=%5BBase%5D%20%3D%2015.8%20%5Ctimes%20%5BAcid%5D)
The total of mole fraction of acid and base is 1. Therefore we have,
![[Acid] + [Base] = 1](https://tex.z-dn.net/?f=%5BAcid%5D%20%2B%20%5BBase%5D%20%3D%201)
But Base = 15.8 x [Acid]. Let us plug in this value in above equation.
![[Acid] + 15.8 \times [Acid] = 1](https://tex.z-dn.net/?f=%5BAcid%5D%20%2B%2015.8%20%5Ctimes%20%5BAcid%5D%20%3D%201)
![16.8 [Acid] = 1](https://tex.z-dn.net/?f=16.8%20%5BAcid%5D%20%3D%201)
![[Acid] = \frac{1}{16.8}](https://tex.z-dn.net/?f=%5BAcid%5D%20%3D%20%5Cfrac%7B1%7D%7B16.8%7D)
![[Acid] = 0.0595](https://tex.z-dn.net/?f=%5BAcid%5D%20%3D%200.0595)
[Acid] = 0.0595 x 100 = 5.95 %
The fraction of carbonic acid present in the blood is 5.95%
<h3>Answer:</h3>
Limiting reactant is Lithium
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of Lithium as 1.50 g
- Mass of nitrogen is 1.50 g
We are required to determine the rate limiting reagent.
- First, we write the balanced equation for the reaction
6Li(s) + N₂(g) → 2Li₃N
From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Second, we determine moles of Lithium and nitrogen given.
Moles = Mass ÷ Molar mass
Moles of Lithium
Molar mass of Li = 6.941 g/mol
Moles of Li = 1.50 g ÷ 6.941 g/mol
= 0.216 moles
Moles of nitrogen gas
Molar mass of Nitrogen gas is 28.0 g/mol
Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol
= 0.054 moles
- According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
- On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.
Thus, Lithium is the limiting reagent while nitrogen is in excess.
Answer:
cornea, pupil, lens, vitreous humor
Answer:
you could get a job if you are of age if not ask your parents if you could do chores for allowance
Answer:
Sodium chloride is formed from a single replacement reaction.
Explanation:
