Hot, molten rock deep below Earth’s surface is called lava mantle rock parent rock magma

What is the equilibrium equation for the dissociation of formic acid in water? hcooh (aq) + h2o (l) ⇌ h3o+ (aq) + hcoo- (aq)?

Pani-rosa [81]

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>

7 0

3 years ago

Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

$y_{O2} =4.3$ %

Explanation:

The ethanol combustion reaction is:

$C_{2}H_{5} OH+3O_{2}$ → $2CO_{2}+3H_{2}O$

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

$x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}$

Dividing the previous equation by x:

$1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}$

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

$3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )$

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

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

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

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

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

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

6 0

3 years ago

Scientists tracking the populations of certain freshwater fish have found an increase in the female to male ratio of several sma

vodomira [7]

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.

7 0

2 years ago

Read 2 more answers

Label the liquids from least to most dense, Vegetable oil, water, Dish soap, Corn syrup, Honey

masya89 [10]

Answer:

1. Vegetable oil

2. Water

3. Dish soap

4. Corn syrup

5. Honey

8 0

3 years ago

How would you prepare 1.00 L of a 0.50 M solution of each of the following?

Lunna [17]

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:

$n_1=n_2$

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

$M_1V_1=M_2V_2$

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

$V_1=\frac{M_2V_2}{M_1} \\\\V_1=\frac{1.00L*0.5M}{12M}\\ \\V_1=0.0417L=41.7mL$

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.

Best regards.

6 0

3 years ago