Answer:
Pro exercise con suffication
Explanation:
...
Answer:
B:the second answer
Explanation:
Because the all have leaves
Answer:
Animals do not photosynthesize.
Explanation:
Plants use chloroplasts to capture sunlight to convert to energy in order to photosynthesize. Animals do not do this, therefore it is not needed.
Answer:
0.702 /s
Explanation:
Rate constant at 
Rate constant at 


Activation energy, 
Use the following equation to calculate
Use the following equation to calculate
Therefore,
![\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{1}{298 \mathrm{~K}}-\frac{1}{350 \mathrm{~K}}\right]](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%28%5Cfrac%7BK_%7B2%7D%7D%7B3.46%20%5Ctimes%2010%5E%7B-2%7D%20%5Cmathrm%7B~s%7D%5E%7B-1%7D%7D%5Cright%29%20%26%3D%5Cfrac%7B50.2%20%5Ctimes%2010%5E%7B3%7D%20%5Cmathrm%7B~J%7D%20%2F%20%5Cmathrm%7Bmol%7D%7D%7B8.314%20%5Cmathrm%7BJK%7D%5E%7B-1%7D%20%5Cmathrm%7B~mole%7D%5E%7B-1%7D%7D%5Cleft%5B%5Cfrac%7B1%7D%7B298%20%5Cmathrm%7B~K%7D%7D-%5Cfrac%7B1%7D%7B350%20%5Cmathrm%7B~K%7D%7D%5Cright%5D)
![\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{52 \mathrm{~K}}{298 \mathrm{~K} \times 350 \mathrm{~K}}\right]](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%28%5Cfrac%7BK_%7B2%7D%7D%7B3.46%20%5Ctimes%2010%5E%7B-2%7D%20%5Cmathrm%7B~s%7D%5E%7B-1%7D%7D%5Cright%29%20%26%3D%5Cfrac%7B50.2%20%5Ctimes%2010%5E%7B3%7D%20%5Cmathrm%7B~J%7D%20%2F%20%5Cmathrm%7Bmol%7D%7D%7B8.314%20%5Cmathrm%7BJK%7D%5E%7B-1%7D%20%5Cmathrm%7B~mole%7D%5E%7B-1%7D%7D%5Cleft%5B%5Cfrac%7B52%20%5Cmathrm%7B~K%7D%7D%7B298%20%5Cmathrm%7B~K%7D%20%5Ctimes%20350%20%5Cmathrm%7B~K%7D%7D%5Cright%5D)




hence, the rate constant at
is 0.702