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.
The answer is: the mass of oxygen is 16.95 grams.
The overall balanced photosynthesis reaction:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.
m(C₆H₁₂O₆) = 15.90 g; mass of glucose.
n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 15.9 g ÷ 180.18 g/mol.
n(C₆H₁₂O₆) = 0.088 mol; amount of glucose.
From chemical reaction: n(C₆H₁₂O₆) : n(O₂) = 1 : 6.
n(O₂) = 6 · 0.088 mol.
n(O₂) = 0.53 mol; amount of oxygen.
m(O₂) = 0.53 mol · 32.00 g/mol.
m(O₂) = 16.95 g; mass of oxygen.
Answer:
I have no clue what the question is
Answer:
They are all alkali earth metals.
Explanation:
Their valence shell each has 2 electrons. Also, they are all shiny, silvery-white, somewhat reactive metals at standard temperature and pressure. They form alkaline solutions, hydroxides, when reacting with water and their oxides are found in the earth’s crust.
