All categories

3 years ago

Determine the [h3o+] of a 0.210 m solution of formic acid.

1 answer:

3 0

When the Pka for formic acid = 3.77
and Pka = -㏒ Ka
3.77 = -㏒ Ka
∴Ka = 1.7x10^-4

when Ka = [H+][HCOO-}/[HCOOH]

when we have Ka = 1.7x10^-4 &[HCOOH] = 0.21 m
so by substitution: by using ICE table value
1.7x10^-4 = X*X / (0.21-X)
(1.7x10^-4)*(0.21-X) = X^2 by solving this equation for X

∴X = 0.0059
∴[H+] = 0.0059
∴PH= -㏒ [H+]
= -㏒ 0.0059
= 2.23

You might be interested in

The period on the periodic table(1-7) tells you the number of ________ in an atom.

Ber [7]

Answer:

Explanation:

as it depends on electronic configuration

3 0

3 years ago

Read 2 more answers

Why don't the praticles in a solid move past one another

Ne4ueva [31]

Answer:

Particles in a solid object are super tight and close to eachother. They do not move past eachother because it is a solid, not a liquid or gas.

Explanation:

3 0

3 years ago

Read 2 more answers

What is the total number of valence electrons un xenon(xe)

Ber [7]

Answer:

Explanation:

Hope this helps you :)

6 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

The first-order decay of radon has a half-life of 3.823 days. How many grams of radon remain after 7.22 days if the sample initi

Nesterboy [21]

substitute: t1/2=ln(2)k→k=ln(2)t1/2 Into the appropriate equation: [A]t=[A]0∗e−kt [A]t=[A]0∗e−ln(2)t1/2t [A]t=(250.0 g)∗e−ln(2)3.823 days(7.22 days)=67.52 g

6 0

3 years ago