Answer:
B. 1-Butene rightarrow (1) BH3: THF (2)H202, OH-
Explanation:
In the hydroboration of alkenes, an alkene is hydrated to form an alcohol with anti-Markovnikov orientation.
the reagent BH₃:THF is the way that borane is used in organic reactions. The BH₃ adds to the double bond of an alkene to form an alkyl borane. Peroxide hydrogen in basic medium oxidizes the alkyl borane to form an alcohol. Indeed, hydroboration-oxidation converts alkenes to alcohols by adding water through the double bond, with anti-Markovnikov orientation.
Potassium is placed where it is based on its properties and it's reactivity. It's also placed there based on it's atomic number.
Answer:
Solution A is a Weak Alkali, Solution B is a strong Acid.
Explanation:
At pH 10, the colour is blue, therefore it's a weak alkali.
At pH 1, the colour is red, therefore it's a strong Acid.
When the products of a reaction are hotter than the reactants, an exothermic reaction is happening. An exothermic reaction is a reaction that releases energy to the surroundings. The energy released should be more than what is absorbed in order to maintain the reaction.
<h3>Al + O2 -> Al2O3</h3>
Balance it:
<h3>2Al + 3O2 -> 2Al2O3</h3><h3 />
So you need 2 Al and 3 O2 to make 2 Al2O3 (aluminum oxide).
I'm going to assume you have all the O2 you need.
Since 2 mols of Al is needed to make 2 mols of the product, it's a 1:1 ratio. You get as much aluminum oxide for as much aluminum you burn.
So 12.5 mols if there is not a lack of the O2.
