Answer : The correct option is, (D) 100 times the original content.
Explanation :
As we are given the pH of the solution change. Now we have to calculate the ratio of the hydronium ion concentration at pH = 5 and pH = 3
As we know that,
![pH=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH_3O%5E%2B%5D)
The hydronium ion concentration at pH = 5.
![5=-\log [H_3O^+]](https://tex.z-dn.net/?f=5%3D-%5Clog%20%5BH_3O%5E%2B%5D)
![[H_3O^+]=1\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1%5Ctimes%2010%5E%7B-5%7DM)
..............(1)
The hydronium ion concentration at pH = 3.
![3=-\log [H_3O^+]](https://tex.z-dn.net/?f=3%3D-%5Clog%20%5BH_3O%5E%2B%5D)
![[H_3O^+]=1\times 10^{-3}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1%5Ctimes%2010%5E%7B-3%7DM)
................(2)
By dividing the equation 1 and 2 we get the ratio of the hydronium ion concentration.
![\frac{[H_3O^+]_{original}}{[H_3O^+]_{final}}=\frac{1\times 10^{-5}}{1\times 10^{-3}}=\frac{1}{100}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH_3O%5E%2B%5D_%7Boriginal%7D%7D%7B%5BH_3O%5E%2B%5D_%7Bfinal%7D%7D%3D%5Cfrac%7B1%5Ctimes%2010%5E%7B-5%7D%7D%7B1%5Ctimes%2010%5E%7B-3%7D%7D%3D%5Cfrac%7B1%7D%7B100%7D)
![100\times [H_3O^+]_{original}=[H_3O^+]_{final}](https://tex.z-dn.net/?f=100%5Ctimes%20%5BH_3O%5E%2B%5D_%7Boriginal%7D%3D%5BH_3O%5E%2B%5D_%7Bfinal%7D)
From this we conclude that when the pH of a solution changes from a pH of 5 to a pH of 3, the hydronium ion concentration is 100 times the original content.
Hence, the correct option is, (D) 100 times the original content.
Cellular respiration is the process by which the chemicalenergy of "food" molecules is released and partially captured in the form of ATP. Carbohydrates, fats, and proteins can all be used as fuels in cellular respiration, but glucose is most commonly used as an example to examine the reactions and pathways involved.
Answer:
A: dedrite
B: cell body
c: axon terminal
d: axon
e: nucleus
Explanation:
I took it and got it right
Answer:
P = 0.0009417 atm
Or,
P = 9.417 × 10⁻⁴ atm
Or,
P = 0.0954157 kPa
Or,
P = 0.715677 mmHg (Torr)
Explanation:
Data Given:
Moles = n = 3.2 mol
Temperature = T = 312 K
Pressure = P = ?
Volume = V = 87 m³ = 87000 L
Formula Used:
Let's assume that the gas is acting as an Ideal gas, the according to Ideal Gas Equation,
P V = n R T
where; R = Universal Gas Constant = 0.082057 atm.L.mol⁻¹.K⁻¹
Solving Equation for P,
P = n R T / V
Putting Values,
P = (3.2 mol × 0.082057 atm.L.mol⁻¹.K⁻¹ × 312 K) ÷ 87000 L
P = 0.0009417 atm
Or,
P = 9.417 × 10⁻⁴ atm
Or,
P = 0.0954157 kPa
Or,
P = 0.715677 mmHg (Torr)
