Answer:
NaOH
Explanation:
The sodium hydroxide is strongest base because it fully dissociate into ions.
It dissociate to gives the sodium and hydroxyl ions and solution became more basic.
NaOH → Na⁺(aq) + OH⁻(aq)
Its pH is 12. It is also known as caustic soda.
Other given Options:
The pH of blood is vary from 7.35 - 7.45. it is slightly alkaline.
The pH of HCl is 3.
The pH of saliva is 6.2 - 7.6.
We know that neutral pH is 7 the pH less than 7 is acidic so HCl is acidic because its pH is 3.
The pH is greater then 7 is basic so saliva and blood are basic but less than the NaOH.
No because it is a gas so (in space) so, it would most likely stay in one place. Your welcome!
Answer:
Layer 2 and layer 9 is the same relative age since it is the same type of rock and has the same fossils.
Hope this helped!
Answer:
V₂ = 107.84 L
Explanation:
Given data:
Initial volume = 100 L
Initial pressure = 80 KPa (80/101 =0.79 atm)
Initial temperature = 200 K
Final temperature =273 K
Final volume = ?
Final pressure = 1 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁T₂ /T₁P₂
V₂ = 0.79 atm × 100 L × 273 K / 200 K × 1 atm
V₂ =21567 atm.L.K /200 K.atm
V₂ = 107.84 L
Answer:
The molar solubility of lead bromide at 298K is 0.010 mol/L.
Explanation:
In order to solve this problem, we need to use the Nernst Equaiton:
![E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}](https://tex.z-dn.net/?f=E%20%3D%20E%5E%7Bo%7D%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D)
E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.
At equilibrium, E = 0, therefore:
![E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\](https://tex.z-dn.net/?f=E%5E%7Bo%7D%20%20%3D%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%20%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D%20%5C%5C%5C%5Clog%20%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D%20%3D%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%20%5C%5C%5C%5Clog%5Bred%5D%20%3D%20%20log%5Box%5D%20-%20%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%5C%5C%5C%5C%5Bred%5D%20%3D%2010%5E%7B%20log%5Box%5D%20-%20%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%7D%20%5C%5C%5C%5C%5Bred%5D%20%3D%2010%5E%7B%20log0.733%20-%20%20%5Cfrac%7B2x5.45x10%5E%7B-2%7D%20%20%7D%7B0.0591%7D%7D%5C%5C%5C%5C)
[red] = 0.010 M
The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.
