Chemistry, 04.01.2020 15:31, llamasking

Alaboratory experiment requires 4.8 l of a 2.5 m solution of sulfuric acid (h2so4), but the only available h2so4 is a 6.0 m stock solution. how could you prepare the solution needed for the lab experiment? show all the work used to find your answer.

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answered: azaz1819

Take 2.0 L of the H2SO4 stock solution of 6.0 M and dilute it to 4.8 L

Explanation:

For The exercise we are going to need the next equation:

(Final concentration)(final volume) = (initial concentration)(initial volume) or CfVf=CiVi

The final solution is the one required to the experiment, the one with a concentration of 2.5 M and a volume of 4.8 L, and the initial solution is the one available of concentration 6.0 M. So, we need to know the volume of the initial solution we need to take to prepare the require solution. For the equation above we have:

$Vi=\frac{(Cf)(Vf)}{Ci}$

Replacing the values:

$Vi=\frac{(2.5M)(4.8L)}{6.0M}=2.0L$

Then, to prepare the solution needed for the experiment we must take 2.0 L of the H2SO4 stock solution of 6.0 M and dilute it to 4.8 L of solution, that is to say, add 2.8 L of water, so to get a 2.5 M H2SO4 solution.

answered: Guest

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( $n_2$ ) .

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.

answered: Guest

c- cold periods in earth's history during which glaciers covered much of the surface.

explanation:

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Alaboratory experiment requires 4.8 l of a 2.5 m solution of sulfuric acid (h2so4), but the only ava...