how much energy must be released by 50.0 g of steam to decrease its temperature from 125.0 degrees Celsius to 100.0 degrees Cels
1 answer:
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The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.
<h3>Explanation</h3>What's the initial color of the solution?
is a salt soluble in water.
dissociates into ions completely when dissolved.
.
The first test tube used to contain .
is a weak base that dissociates partially in water.
.
There's also an equilibrium between and
ions.
.
ions from
will shift the equilibrium between
and
to the right and reduce the amount of
in the solution.
The indicator equilibrium will shift to the right to produce more ions along with the colored indicator ions. The solution will show a pink color.
What's the color of the solution after adding NH₄Cl?
Adding will add to the concentration of
ions in the solution. Some of the
ions will combine with
ions to produce
.
The equilibrium between and
ions will shift to the left to produce more of both ions.
The indicator equilibrium will shift to the left as the concentration of increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.
Answer:
Reagents: 1) 2)
,
Mechanism: Hydroboration
Explanation:
In this case, we have a <u>hydration of alkene</u>s reaction. But, in this example, we have an <u>anti-Markovnikov reaction</u>. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: <u>"hydroboration"</u>.
The <u>first step</u> of this reaction is the addition of borane () to the double bond. Then in the <u>second step</u>, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the <u>third step</u>, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond "
". In <u>step number 4</u> we have the migration of the C-B bond to oxygen. Then in <u>step number 5</u>, we have the attack of
on the
to produce an alkoxide. Finally, the water molecule produce in step 2 will <u>protonate</u> the molecule to produce the alcohol.
See figure 1
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