To estimate the number of ladybugs in the entire garden, Ajoy needs the dimension of the garden.
<h3>What is a dimension?</h3>
Dimensions in mathematics are the measure of the size or distance of an object or region or space in one direction.
Dimensional measurement is of fundamental importance for interchangeability and global trade.
Hence, to estimate the number of ladybugs in the entire garden, Ajoy needs the dimension of the garden.
Learn more about dimensions here:
brainly.com/question/13314350
#SPJ1
Answer:
the HOMO-LUMO energy difference in ethylene is greater than that of cis,trans−1,3−cyclooctadiene
Explanation:
The λmax is the wavelength of maximum absorption. We could use it to calculate the HOMO-LUMO energy difference as follows:
For ethylene
E= hc/λ= 6.63×10^-34×3×10^8/170×10^-9= 1.17×10^-18J
For cis,trans−1,3−cyclooctadiene
E= hc/λ=6.63×10^-34×3×10^8/230×10^-9=8.6×10^-19J
Therefore, the HOMO-LUMO energy difference in ethylene is greater than that of cis,trans−1,3−cyclooctadiene
False, gravity is a force, matter has mass and volume.
Answer:
A compound is a unique substance that forms when two or more elements combine chemically. Compounds form as a result of chemical reactions. The elements in compounds are held together by chemical bonds. A chemical bond is a force of attraction between atoms or ions that share or transfer valence electrons
Answer: The value of 
for the half-cell reaction is 0.222 V.
Explanation:
Equation for solubility equilibrium is as follows.
Its solubility product will be as follows.
![K_{sp} = [Ag^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)
Cell reaction for this equation is as follows.
Reduction half-reaction: 
, 
Oxidation half-reaction: 
, 
= ?
Cell reaction: 
So, for this cell reaction the number of moles of electrons transferred are n = 1.
Solubility product,
= 
Therefore, according to the Nernst equation
![E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%5Cfrac%7B0.0592%20V%7D%7Bn%7D%20log%20%5Cfrac%7B%5BAgCl%5D%7D%7B%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D%7D)
At equilibrium, 
= 0.00 V
Putting the given values into the above formula as follows.
![0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}](https://tex.z-dn.net/?f=0.00%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%5Cfrac%7B0.0592%20V%7D%7B1%7D%20log%20%5Cfrac%7B1%7D%7B%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D%7D)
= 
= 0.577 V
Hence, we will calculate the standard cell potential as follows.
= 0.222 V
Thus, we can conclude that value of for the half-cell reaction is 0.222 V.
