They are all things you can do to elements on the periodic table?
Answer:
0.075
Explanation:
First obtain the mean of the measurement;
Mean = 10.15 + 9.95 + 9.99 + 10.02/4 = 10.03
Then obtain d^2= (mean-score)^2 for each score;
(10.15-10.03)^2 = 0.0144
(9.95-10.03)^2 = 0.0064
(9.99-10.03)^2 = 0.0016
(10.02-10.03)^2= 0.0001
∑d^2= 0.0144 + 0.0064 + 0.0016 + 0.0001
∑d^2= 0.0225
Variance = ∑d^2/N = 0.0225/4 = 0.005625
Standard deviation= √0.005625
Standard deviation= 0.075
First, recognize that this is an elimination reaction in which hydroxide must leave and a double bond must form in its place. It is likely an E2 reaction. Here is an efficient mechanism:
1) Pre-reaction: Protonate the -OH to make it a good leaving group, water. H2SO4 or any strong H+ donor works. The water is positively charged but still connected to the compound.
2) E2: Use a sterically hindered base, such as tert-butoxide (tButO-) to abstract the hydrogen from the secondary carbon. [You want a sterically hindered base because a strong, non-sterically hindered base could also abstract a hydrogen from one of the two methyl groups on the tertiary carbon, and that leads to unwanted products, which is not efficient]. As the proton of hydrogen is abstracted, water leaves at the same time, creating an intermediate tertiary carbocation, and the 2 electrons in the C-H bond immediately are used to make a double bond towards the partial positive charge.
In the products we see the major product and water, as expected. Even though you have an intermediate, remember that an E2 mechanism technically happens in one step after -OH protonation.
This question may only be ansewered by frequent mattrrs
The dilution formula can be used to find the volume needed
c1v1 = c2v2
Where c1 is concentration and v1 is volume of the concentrated solution
And c2 is concentration and v2 is volume of the diluted solution to be prepared
c1 - 0.33 M
c2 - 0.025 M
v2 - 25 mL
Substituting these values in the equation
0.33 M x v1 = 0.025 M x 25 mL
v1 = 1.89 mL
Therefore 1.89 mL of the 0.33 M solution needs to be diluted up to 25 mL to make a 0.025 M solution