I think it’s the cell membrane
True? I really don’t know but ones someone
Since chlorine is one of the 7 diatomic elements we know that chlorine appears as Cl₂ gas naturally. That means that the molar mass of a chlorine gas is 70.9g/mol. That being said, first you need to find the number of moles of chlorine gas that are present in a 35.5g sample. To do this divide 35.5g by the molar mass of chlorine gas (70.9g/mol) to get 0.501mol of chlorine. Then you have to multiply 0.501mol by 6.02×10²³ to get the number of chlorine gas molecules. Therefore 3.01×10²³ molecules of chlorine gas are present in a 35.5g sample.
I hope that helps. Let me know in the comments if anything is unclear.
4 mol / 205g H2O = 4/.205 = 19.5 mol/kg boiling point = 100 + 19.5 • 0.51 = 109 ºC
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.