Calculating Reaction Speed Changes: Second and First Order Kinetics at 298K

[Questions999]

The reaction A+B ->Products is of second order related to A ,and of first order related to B.The reaction occurs on 298 K.When A=0.75 mol L^-1 and B=0,8 mol l^-1 the speed of the consumption of A and B is 0.00347 mol l^-1 min ^-1.What will be the speed of consumption of A and B after these changes happen: in 100 ml of the mixture we pour 200 ml of clean water?And in 200 ml of the reaction we add 300 ml of clean water?

So the equation of the speed must look like : V=k*[A]^2*...how do I relate this to the changes//?

 

[epenguin]

The reaction A+B ->Products is of second order related to A ,and of first order related to B.The reaction occurs on 298 K.When A=0.75 mol L^-1 and B=0,8 mol l^-1 the speed of the consumption of A and B is 0.00347 mol l^-1 min ^-1.What will be the speed of consumption of A and B after these changes happen: in 100 ml of the mixture we pour 200 ml of clean water?And in 200 ml of the reaction we add 300 ml of clean water?

So the equation of the speed must look like : V=k*[A]^2*...how do I relate this to the changes//?

Your equation is OK. Use it. Frankly if there is a problem with this question it is that they are actually giving you more information than you need to solve it.