Answer:
Green, Blue, Indigo, Violet
Explanation:
According to Einstein's equation of photoelectric effect, the kinetic energy of emitted photoelectron is the difference between the energy of the incident photon and the work function of the metal. The work function of the metal referee to the minimum energy that must be supplied in order to eject an electron from a metal surface. The energy of the incident photon must exceed the work function of the metal.
When we look at the electromagnetic spectrum, only the selected colours have frequency above the threshold frequency as shown by the image attached below.
In buffer solution there is an equilibrium between the acid HA and its conjugate base A⁻: HA(aq) ⇌ H⁺(aq) + A⁻(aq).
When acid (H⁺ ions) is added to the buffer solution, the equilibrium is shifted to the left, because conjugate base (A⁻) reacts with hydrogen cations from added acid, according to Le Chatelier's principle: H⁺(aq) + A⁻(aq) ⇄ HA(aq). So, the conjugate base (A⁻) consumes some hydrogen cations and pH is not decreasing (less H⁺ ions, higher pH of solution).
A buffer can be defined as a substance that prevents the pH of a solution from changing by either releasing or absorbing H⁺ in a solution.
Buffer is a solution that can resist pH change upon the addition of an acidic or basic components and it is able to neutralize small amounts of added acid or base, pH of the solution is relatively stable
Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after 
is 0.00345 mol/L.
Explanation:
Rate Law: ![k[HI]^2 ](https://tex.z-dn.net/?f=k%5BHI%5D%5E2%0A)
Rate constant of the reaction = k = 
Order of the reaction = 2
Initial rate of reaction = 
Initial concentration of HI =![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
![1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-7%7D%20mol%2FL%20s%3D%286.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%29%5BHI%5D%5E2)
![[A_o]=5 mol/L](https://tex.z-dn.net/?f=%5BA_o%5D%3D5%20mol%2FL)
Final concentration of HI after t = [A]
t =
Integrated rate law for second order kinetics is given by:
![\frac{1}{[A]}=kt+\frac{1}{[A_o]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3Dkt%2B%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
![\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D6.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%5Ctimes%204.53%5Ctimes%2010%5E%7B10%7D%20s%2B%5Cfrac%7B1%7D%7B%5B5%20mol%2FL%5D%7D)
![[A]=0.00345 mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D0.00345%20mol%2FL)
The concentration of HI after is 0.00345 mol/L.
Answer:
n = 7.86 mol
Explanation:
This question can be solved using the ideal gas law of PV = nRT.
Temperature must be in K, so we will convert 22.5C to 295 K ( Kelvin = C + 273).
R is the ideal gas constant of 0.0821.
(2.24atm)(85.0L) = n(0.0821)(295K)
Isolate n to get:
n = (2.24atm)(85.0L)/(0.0821)(295K)
n = 7.86 mol
