Answer:1. ![Rate=k[CHCl_3]^1[Cl_2]^\frac{1}{2}](https://tex.z-dn.net/?f=Rate%3Dk%5BCHCl_3%5D%5E1%5BCl_2%5D%5E%5Cfrac%7B1%7D%7B2%7D)
2. The rate constant (k) for the reaction is 
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
![rate=k[CHCl_3]^x[Cl_2]^y](https://tex.z-dn.net/?f=rate%3Dk%5BCHCl_3%5D%5Ex%5BCl_2%5D%5Ey)
k= rate constant
x = order with respect to 
y = order with respect to 
n = x+y= Total order
1. a) From trial 1:
(1)
From trial 2:
(2)
Dividing 2 by 1 :![\frac{0.0069}{0.035}=\frac{k[0.020]^x[0.010]^y}{k[0.010]^x[0.010]^y}](https://tex.z-dn.net/?f=%5Cfrac%7B0.0069%7D%7B0.035%7D%3D%5Cfrac%7Bk%5B0.020%5D%5Ex%5B0.010%5D%5Ey%7D%7Bk%5B0.010%5D%5Ex%5B0.010%5D%5Ey%7D)
therefore x=1.
b) From trial 2:
(3)
From trial 3:
(4)
Dividing 4 by 3:
therefore 
![rate=k[CHCl_3]^1[Cl_2]^\frac{1}{2}](https://tex.z-dn.net/?f=rate%3Dk%5BCHCl_3%5D%5E1%5BCl_2%5D%5E%5Cfrac%7B1%7D%7B2%7D)
2. to find rate constant using trial 1:

Answer:
0.0192 mL per min.
Explanation:
IV rate = 36 mg per 30 min.
IV concentration = 125 mg per 2.0 mL
36 mg per 30 min. IV rate = 36/30 = 1.2 mg per min
If 125 mg methylprednisolone is present in 2.0 mL of the IV nag, how many mL would contain 1.2 mg?
= 2x1.2/125
= 0.0192 mL
<em>Therefore, the flow rate of the IV must be </em><em>0.0192 mL per min</em><em>. in order to be able to deliver 36 mg per 30 min. </em>