V₁ = 400.0 mL
T₁ = 22.0ºC + 273 = 295 K
V₂ = ?
T₂ = 30.0ºC + 273 = 303 K
V₁ / T₁ = V₂ / T₂
400.0 / 295 = V₂ / 303
295 x V₂ = 400.0 x 303
295 V₂ = 121200
V₂ = 121200 / 295
V₂ = 410.84 mL
hope this helps!
For a first order reaction, the half life is inversely proportional to the rate constant.
The formula is
half life = ln(2)/k = 0.693/k
where k is the rate constant
t = 5.50 minutes
k = ln(2)/5.50 = 0.126 min^-1
Your rate constant is 0.126 min^-1.
Explanation:
the table is not given plz send the table
When a liquid goes from a liquid state to a gaseous state it is called evaporation
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.
