Formic acid when in water would dissociate into ions just like any acids. It would dissociate into the hydrogen ion and the formate ion. The equilibrium dissociation equation would be written as:
<span>HCOOH (aq) + H2O (l) ⇌ H+ (aq) + HCOO- (aq)
Formic acid is a weak acid which means that when in aqueous solution it does not completely dissociate into its corresponding ions. Only a certain amount that would be dissociated so in the solution there will be HCOOH, HCOO- and H+ molecules. It is also known as Methanoic acid and an important substance for the synthesis of a number of substances. It is naturally occurring in ants.</span>
Answer:
%
Explanation:
The ethanol combustion reaction is:
→
If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

Dividing the previous equation by x:

We would need 3.30 oxygen moles per ethanol mole.
Then we apply the composition relation between O2 and N2 in the feed air:

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

Calculate the number of moles of CO2 and water considering the same:


The total number of moles at the reactor output would be:

So, the oxygen mole fraction would be:
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I believe that the answer to the question provided above is that <span>they formed two very different hypotheses since it is what they saw according to what they knew. Also hypothesis is just an intelligent guess.</span>
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.
Answer:
Take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water.
Explanation:
Hello,
In this case, for the dilution process from concentrated 12-M hydrochloric acid to 1.00 L of the diluted 0.50M hydrochloric acid, the volume of concentrated HCl you must take is computed by considering that the moles remain constant for all dilution processes as shown below:

Which can also be written in terms of concentrations and volumes:

Thus, solving for the initial volume or aliquot that must be taken from the 12-M HCl, we obtain:

It means that you must take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water for such preparation.
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