Answer:
1. The dye that absorbs at 530 nm.
Explanation:
The dye will absorb light to promote the transition of an electron from the HOMO to the LUMO orbital.
The higher the gap, the higher the energy of transition. The energy can be calculated by E = hc/λ, in which h and c are constants and λ is the wavelength.
The equation shows that the higher the energy, the higher the gap and the lower the wavelength.
Therefore, the dye with absorption at 530 nm has the higher HOMO-LUMO gap.
Answer:
Endothermic reaction chemical equation
Reactnt A + Reactant B + Heat (energy) ⇒ Products
Exothermic reaction chemical equation
Reactnt A + Reactant B ⇒ Products + Heat (energy)
Explanation:
Endothermic Reaction
An endothermic reaction is a reaction that reaction that requires heat before it would take place resulting in the absorption of heat from the surrounding that can be sensed by the coolness of the reacting system
An example of an endothermic reaction is a chemical cold pack that becomes cold when the chemical and water inside it reacts
Exothermic Reaction
An exothermic reaction is one that rekeases energy to the surroundings when it takes place. This is as a result of the fact that the combined heat energy of the reactants is more than the chemical heat energy of the products. An example of an exothermic reaction is a burning candle
Answer:
I guess A, I am not sure...
Answer : The activation energy of the reaction is, 
Solution :
The relation between the rate constant the activation energy is,
![\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7BK_2%7D%7BK_1%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial rate constant = 
= final rate constant = 
= initial temperature = 
= final temperature = 
R = gas constant = 8.314 kJ/moleK
Ea = activation energy
Now put all the given values in the above formula, we get the activation energy.
![\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7B8.75%5Ctimes%2010%5E%7B-3%7DL%2Fmole%5Ctext%7B%20s%7D%7D%7B4.55%5Ctimes%2010%5E%7B-5%7DL%2Fmole%5Ctext%7B%20s%7D%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20%288.314kJ%2FmoleK%29%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7B468K%7D-%5Cfrac%7B1%7D%7B531K%7D%5D)
Therefore, the activation energy of the reaction is,
