Chrome File Edit View History Bookmarks People Window Help 4) a 100% G Wed 9:36

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Chrome File Edit View History Bookmarks People Window Help 4) a 100% G Wed 9:36 PM Apple opposes order to ha y Take Test: Benzoin Oxid at Definitions of oxidation an x CNE fi -https://courses.cscc.edu/webapps/assessment/take/launch.jsp?course assessment id 4458491&course; id 108901 1&content; id 7412182 1&step-null; a E ABP QUESTION 1 0.5 points Save Answer Indicate whether the following reaction is an oxidation or reduction reaction. 1. NaBH4 Oxidation Reduction QUESTION 2 0.5 points Save Answer Indicate whether the following reaction is an oxidation or reduction reaction. HNO3 Reduction Oxidation QUESTION 3 2 points Save Answer To prepare benzil you are heating a solution of benzoin with concentrated nitric acid. F treat/dispose of the left over liquid from this step? ollowing this, you will add water, cool the mixture, and collect the solid. How will you It must be neutralized with sodium bicarbonate and can then be washed down the drain. It can be directly poured into the organic waste container 17

Explanation / Answer

You have multiple questions in a same post and due to guidelines I cannot answer all of them. So, I will answer some of the questions, and the rest, post them on a different question thread so you can be answered faster.

Question 1.

It's a reduction because the NaBH4 is a common reagent used to reduce carbonyl group to alcohols. You can view by the fact that the oxigen atom in the carbonyl has a neutral charge of 0, while the OH if you substract the H atom, it will have a 1- charge. PAssing from 0 to -1, means that is a reduction (gaining electrones).

Question 2.

It's an oxidation. In this case, it occurs the opposite of question 1. We are passing from 1- to 0 (lose electrons) in that way OH can be converted to carbonyl group.

Question 6.

In this case, we need the molecular weight of benzoin (reported 212.24 g/mol)

now, calculate the moles of benzoin in this mass of 0.3 g

moles = 0.3 / 212.24 = 1.41x10-3 moles

Now assuming a 1:1 relation, we can assume the following expression too:

M1V1 = M2V2  

and moles = M*V so:

15.8V1 = 1.41x10-3

V1 = 1.41x10-3 / 15.8

V1 = 8.92x10-5 L

Expressed in mL it would be:

8.92x10-5 * 1000 = 0.089 mL

Question 4 and 5, post them in a different question thread. Hope this helps

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