Because of the lack of reactives in a laboratory, you wish to prepare a solution of agno3 0.1 n, given two solutions, the first 0.0825n and the second 0.7280n. what volume of each of them you'll have to mix to obtain 1lt of the solution you want?

Given:
c1=0.0825 M
c2=0.7280 M
c3=0.1 M
V3 =1 dm3
Setting up the equation:
c1V1+c2V2 = c3V3
now you work a little magic:
V1+V2=V3
Solving for V1
V1= V3 - V2
substitute V1 in first equation:
c1(V3-V2) + c2V2=c3V3

Solving: 

c1V3-c1V2 + c2V2=c3V3

Rearranging the values:

c1V3-c3V3=c1V2-c2V2

V3(c1-c3)=V2(c1-c2)

So,
V2 = V3(c1-c3) / c1-c2
Substituting known values:
V2= 1 dm3 * (0.0825-0.1) / 0.0825 - 0.7280
and you get:
V2= 0.027 dm3
Hence,
V1 = V3-V2 V1
     = 0.973 dm2

Ithink that the use of creativity in the sciences entails that one produce something out of nothing or from another thing. with this, i think that the he wrote down his results as four parts of the atomic theory is the manifestation of the  creativity of john dalton.

Following are the  advantages of the parametric equations: 1) they can be used to graph curves that are not functions, like the unit circle.2) the parameter can be used to represent the quantity that is  useful to  provide us with better information about the graph.3)  it is useful to convert between rectangular equations and parametric equations.with specific reference to chemistry,  parametric equations are employ for solving hydrogen and hydrogen-like ions, quantum mechanically. herein, cartesian coordinates are converted in spherical polar coordinates. the solution obtained, when plotted on spherical polar graph paper, provides visualization of atomic orbital such as s,p,d and f.