A 4.00 (±0.01) mL Class A transfer pipet is used to transfer 4.00 mL of a 0.302 (±0.004) M Cu2+ stock solution to a 100.00 (±0.08) Class A volumetric flask. Deionized water is used to bring the flask to volume. Calculate the concentration of the diluted soluion in the volumetric flask and report its absolute uncertainty.

concentration of diluted solution = 0.0125 ( ± 0.0002)M

Uncertainty = ± 0.0002  

Explanation:

Given that

Initial volume of Cu2+ = 4.00 (±0.01) mL

Initial molarity 0f Cu2+ = 0.302 (±0.004) M

transferred to  100.00 (±0.08) Class A volumetric flask

first we get amount of water added

100.00 (±0.08) - 4.00 (±0.01)  = 96 ± (0.09)

Now according to  law of dilution

The concentration of Cu2+ after adding water

M1V1 = M2V2

we substitute

0.302 (±0.004) * 4.00 (±0.01) = x * 96 ± (0.09)

Now the multiplication of two digits with uncertainty is

(0.004/0.302) * 100 =  1.32% ;    (0.01/4.00) * 100 = 0.25%

= [0.302 ( ± 1.32% )] * [ 4.00 ± (0.25%)]

= 1.208 ±(1.57%)

1.57/100 * 1.208 = 0.0189

so

= (1.208 ± 0.0189)

now substitute in our previous equation

1.208 ± (0.0189) = x * 96 ± (0.09)

x = 1.208 ± (0.0189) / 96 ± (0.09)

{ 0.09/96 * 100 = 0.094% }

so x = 1.208 ± (1.57%) / 96 ± (0.094% )  

x = 0.0125 ± ( 1.664)

now( 1.664/100 * 0.0125)

= ± 0.000208

Hence

concentration of diluted solution = 0.0125 ( ± 0.0002)M

Uncertainty = ± 0.0002  

most sediment washes or falls into the river as a result of mass movement and runoff. other sediment eroded from the bottom or sides of the river.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g