When performing a gram stain, it is necessary to remove crystal violet stain from any gram negative bacteria on the slide. Microbiologists use a decolorizing solution to accomplish this. A common recipe is equal parts 95% ethanol and acetone. How many mL of water would you need to add if you have 99% ethanol and want to end up with 500 mL of decolorizer

10.11 mL

Explanation:

Given that :

Total volume of decolorizer = 500 mL

A common recipe is equal parts 95% ethanol and acetone.

So;

The volume of 95% ethanol in the decolorizer = 500mL/2 = 250 mL

Let represent V(ml) for the 99% of ethanol needed to make 95% of the 250 mL of ethanol;

Therefore:

V(ml) × 0.99 = 250 × 0.95

V(ml) × 0.99 = 237.5

V(ml)  = 237.5/0.99

V(ml) = 239.89 mL

Hence; the amount of water to be added to 95% of ethanol = ( 250 -  239.89 )mL = 10.11 mL

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.