- The answer is shorter wavelength and equal speed.
That is, compared to ultraviolet light, an electromagnetic wave that has a higher frequency will also have shorter wavelength and equal speed.
This can be seen by the reaction given below:

h= Planck's constant
c=speed of the light
=frquency
=wavelength
So, higher is the frequency, lesser is the volume while speed remains constant as c is speed of light.
Answer:
Solution A that will form a precipitate with Ksp = 2.3 x 10−4
Explanation:
Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
3S S
Where S = Solubility(mole/lit) and Ksp = Solubility product
⇒ Ksp = (3S)³ x (S)
⇒ 27S⁴ = 2.3x10−4
⇒ S = 0.05 mol/lit
Concentration of Li₃PO₄ precipitate = 0.05
<u>Solution A </u>
0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole
0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole
3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄
(Mole/Stoichiometry)

= 0.05 = 0.24
Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.
So concentration of Li₃PO₄ is equal to 0.05.
Because elemental mercury is a liquid at a room temperature
Answer:
emf generated by cell is 2.32 V
Explanation:
Oxidation: 
Reduction: 
---------------------------------------------------------------------------------
Overall: 
Nernst equation for this cell reaction at
-
![E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE_%7Bcell%7D%5E%7B0%7D-%5Cfrac%7B0.059%7D%7Bn%7Dlog%7B%5BAl%5E%7B3%2B%7D%5D%5E%7B2%7D%5BI%5E%7B-%7D%5D%5E%7B6%7D%7D)
where n is number of electrons exchanged during cell reaction,
is standard cell emf ,
is cell emf ,
is concentration of
and
is concentration of 
Plug in all the given values in the above equation -
![E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D2.20-%5Cfrac%7B0.059%7D%7B6%7Dlog%5B%284.5%5Ctimes%2010%5E%7B-3%7D%29%5E%7B2%7D%5Ctimes%20%280.15%29%5E%7B6%7D%5DV)
So, 
Answer:
is the oxidizing agent
Explanation:
An oxidizing agent is an element in a reaction that accepts the electrons of another element. It is typically hydrogen, oxide, or any halogen. In this case, it is oxygen. The answer is 02.