Because on both of them you are typing/writing something.
The choices can be found elsewhere and as follows:
<span>A.The enzyme pectinase speeds up the breakdown of pectin in fruits, producing more juice.. .
B.The enzyme pectinase enhances the taste of fruit juices, making them more popular.. .
C.The enzyme pectinase speeds up the breakdown of toxins in fruits, producing more juice.
The correct answer would be A. </span>The enzyme pectinase speeds up the breakdown of pectin in fruits, producing more juice. With this, the manufacturer company will get more volume of product from same amount of resource.
Answer: B) Particles can be filtered from a suspension.
Explanation: Colloids are solutions with particle size intermediate between true solutions and suspensions. They exhibit tyndall effect that is scattering of light.
Suspensions have large sized particles which settle when left undisturbed for sometime and thus can be filtered off easily.
The particle size in colloids is less and hence they do not settle under the effect of gravity.
A solution can be homogeneous in which the composition is uniform or heterogeneous in which is the composition is not uniform.
The overall reaction is given by:
The fast step reaction is given as:
The slow step reaction is given as:
(slow step 
)
Now, the expression for the rate of reaction of fast reaction is:
![r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]](https://tex.z-dn.net/?f=r_%7B1%7D%3Dk_%7B1%7D%5BNO%5D%5BBr_%7B2%7D%5D-k_%7B-1%7D%5BNOBr_%7B2%7D%5D)
The expression for the rate of reaction of slow reaction is:
![r_{2}=k_{2}[NOBr_{2}] [NO]](https://tex.z-dn.net/?f=r_%7B2%7D%3Dk_%7B2%7D%5BNOBr_%7B2%7D%5D%20%5BNO%5D)
Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as ![[NOBr_{2}]](https://tex.z-dn.net/?f=%5BNOBr_%7B2%7D%5D)
takes place in this reaction.
The expression of rate of formation is:
= ![k_{2}[NOBr_{2}][NO]](https://tex.z-dn.net/?f=k_%7B2%7D%5BNOBr_%7B2%7D%5D%5BNO%5D)
(1)
Now, consider that the fast step is always is in equilibrium. Therefore, 
![k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]](https://tex.z-dn.net/?f=k_%7B1%7D%5BNO%5D%5BBr_%7B2%7D%5D%3D%20k_%7B-1%7D%5BNOBr_%7B2%7D%5D)
![[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]](https://tex.z-dn.net/?f=%5BNOBr_%7B2%7D%5D%20%3D%20%5Cfrac%7Bk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5BBr_%7B2%7D%5D)
Substitute the value of in equation (1), we get:
![\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28NOBr%29%7D%7Bdt%7D%3Dk_%7B2%7D%5BNOBr_%7B2%7D%5D%5BNO%5D)
=![k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]](https://tex.z-dn.net/?f=k_%7B2%7D%20%5Cfrac%7Bk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5BBr_%7B2%7D%5D%5BNO%5D)
= ![\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]](https://tex.z-dn.net/?f=%5Cfrac%7Bk_%7B1%7Dk_%7B2%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5E%7B2%7D%5BBr_%7B2%7D%5D)
Thus, rate law of formation of 
in terms of reactants is given by .
