Answer: The formula mass (formula weight) of a molecule is the sum of the atomic weights of the atoms in its empirical formula. The molecular mass (molecular weight) of a molecule is its average mass as calculated by adding together the atomic weights of the atoms in the molecular formula.
Hope this helps.... Stay safe and have a great weekend!!!!! :D
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%
I believe the answer is C
Answer:
active because he built up antibodies due to exposure to the flu
Explanation:
Answer:
Kc = 50.5
Explanation:
We determine the reaction:
H₂ + I₂ ⇄ 2HI
Initially we have 0.001 molesof H₂
and 0.002 moles of I₂
If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.
H₂ + I₂ ⇄ 2HI
In: 0.001 0.002 -
R: x x 2x
Eq: 0.001-x 0.002-x 0.00187
x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted
So in the equilibrium we have:
0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵ moles of H₂
0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂
Expression for Kc is = (HI)² / (H₂) . (I₂)
0.00187 ² / 6.5×10⁻⁵ . 1.065×10⁻³ = 50.5
