Assuming it's a perfect gas, we have PV=nRT hence if T goes down, V goes down up. The volume will decrease.
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.
There are 7 digits from decimal to 1st digit, and it's coming from right, so exponent will be in negative 7
In short, Your Answer would be: Option C) <span>6.75 × 10-7
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Hope this helps!
<u>Answer:</u>
Law used: Combined Gas Law
<u>Explanation:</u>
We are given the following problem:
Carbon dioxide is in a steel tank at 20°C, 10 liters and 1 atm. What is the pressure on the gas when the tank is heated to 100°C?
To solve this, the most appropriate law that can be used it Combined Gas Law, which is the result of combining the Boyle's law, Charles' law, and Gay-Lussac's law together.
