Mass of solute ( m1 ) = 50.0 g
mass of solvent ( m2 ) = 150.0 g
Therefore:
m/m = ( m1 / m1 + m2 )
m/m = ( 50.0 / 50.0 + 150.0 )
m/m = ( 50.0 / 200 )
m/m = 0.25
Answer:
The length of the fence is 22.5ft and the width of the fence is 7.5ft.
Explanation:
To solve this problem we will turn this question into an equation.
The equation for the perimeter of a rectangle is 2l + 2w = perimeter.
In the question we were given the perimeter so we can go ahead and plug this in.
2l + 2w = 60
In the question we are also told that the fence is 3 times as long as it is wide. This means that 1l = 3w. We can now subsitute l for 3w so that we only have one variable.
2(3w) + 2w = 60
Now we can simply use algebra to solve for w.
6w + 2w = 60
8w = 60
w = 7.5
Now that we know the value of w we can find the value of l since we know 1l = 3w.
l = 3w
l = 3(7.5)
l = 22.5
The length of the fence is 22.5ft and the width of the fence is 7.5ft.
Answer:
[IBr] = 0.049 M.
Explanation:
Hello there!
In this case, according to the balanced chemical reaction:
It is possible to set up the following equilibrium expression:
![K=\frac{[IBr]^2}{[I_2][Br_2]} =0.0110](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BIBr%5D%5E2%7D%7B%5BI_2%5D%5BBr_2%5D%7D%20%3D0.0110)
Whereas the the initial concentrations of both iodine and bromine are 0.50 M; and in terms of 
(reaction extent) would be:
Which can be solved for to obtain two possible results:
Whereas the correct result is 0.0245 M since negative results does not make any sense. Thus, the concentration of the product turns out:
![[IBr]=2x=2*0.0249M=0.049M](https://tex.z-dn.net/?f=%5BIBr%5D%3D2x%3D2%2A0.0249M%3D0.049M)
Regards!
Spectrophotometric cell or a cuvette is made of quartz for UV spectrophotometers. These cuvettes are used as sample holders for the spectrophotometric determination of the analytes. The material that makes up the cuvette and the condition of the cuvette is to be taken care of in order to avoid erroneous absorbance readings. The sample holder or the cuvette must be removed from the spectrophotometer in between two successive readings. This is to ensure that the light sensing detector of the instrument is not affected.
