What did Aristotle believe?

Determine el PH y el % de disociación de una solución de ácido débil, sabiendo que se disuelven 20 gramos del ácido (masa molar=

IceJOKER [234]

Answer:

pH = 4.27. Porcentaje de disociación: 0.03%

Explanation:

El pH de un ácido débil, HX, se obtiene haciendo uso de su equilibrio:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

Donde la constante de equilibrio, Ka, es

Ka = 1.65x10⁻⁸ = [H⁺] [X⁻] / [HX]

Como los iones H⁺ y X⁻ vienen del mismo equilibrio podemos decir:

[H⁺] = [X⁻]

[HX] es:

20g * (1mol/55g) = 0.3636moles / 2.100L = 0.1732M

Reemplazando es Ka:

1.65x10⁻⁸ = [H⁺] [H⁺] / [0.1732M]

2.858x10⁻⁹ = [H⁺]²

5.35x10⁻⁵M = [H⁺]

pH = -log[H⁺]

El porcentaje de disociacion es [X⁻] / [HX] inicial * 100

Reemplazando

5.35x10⁻⁵M / 0.1732M * 100

5 0

2 years ago

Static charges can be applied to neutral objects by friction, induction or conduction. What do all of these methods utilize to c

MatroZZZ [7]

Electrons are valence and free moving so they take place in charge transfer

5 0

2 years ago

Write Lewis structures for the following molecules: (a) ICl, (b) PH3, (c) P4 (each P is bonded to three other P atoms), (d) H2S,

spin [16.1K]

Answer : The Lewis-dot structure for the following molecules are shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

Now we have to determine the Lewis-dot structure for the following molecules.

(a) The given molecule is, $ICl$

As we know that iodine and chlorine have '7' valence electrons.

Therefore, the total number of valence electrons in $ICl$ = 7 + 7 = 14

According to Lewis-dot structure, there are 2 number of bonding electrons and 12 number of non-bonding electrons.

(b) The given molecule is, $PH_3$

As we know that phosphorous has '5' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in $PH_3$ = 5 + 3(1) = 8

According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.

(c) The given molecule is, $P_4$

As we know that phosphorous has '5' valence electrons.

Therefore, the total number of valence electrons in $P_4$ = 4(5) = 20

According to Lewis-dot structure, there are 6 number of bonding electrons and 14 number of non-bonding electrons.

(d) The given molecule is, $H_2S$

As we know that sulfur has '6' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in $H_2S$ = 6 + 2(1) = 8

According to Lewis-dot structure, there are 4 number of bonding electrons and 4 number of non-bonding electrons.

(e) The given molecule is, $N_2H_4$

As we know that nitrogen has '5' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in $N_2H_4$ = 2(5) + 4(1) = 14

According to Lewis-dot structure, there are 10 number of bonding electrons and 4 number of non-bonding electrons.

(f) The given molecule is, $HClO_3$

As we know that chlorine has '7' valence electrons, oxygen has '6' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in $HClO_3$ = 1 + 7 + 3(6) = 26

According to Lewis-dot structure, there are 12 number of bonding electrons and 14 number of non-bonding electrons.

(g) The given molecule is, $COBr_2$

As we know that bromine has '7' valence electrons, oxygen has '6' valence electrons and carbon has '4' valence electrons.

Therefore, the total number of valence electrons in $COBr_2$ = 4 + 6 + 2(7) = 24

According to Lewis-dot structure, there are 8 number of bonding electrons and 16 number of non-bonding electrons.

3 0

2 years ago

Which of the following contains the most atoms?

aksik [14]

Explanation:

9.0122 g be Maybe, sorry i don't think so

8 0

2 years ago

Read 2 more answers

Using the standard reduction potentials, Pb 2+(aq) + 2e– => Pb(s), E° = –0.13 V Fe 2+(aq) + 2e– => Fe(s), E° = –0.44 V Zn

mojhsa [17]

Answer:

Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$

Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$

Explanation:

Standard reduction potential denotes ability to consume electrons from another species.

Hence, higher the standard reduction potential, higher will be the ability to oxidize another species.

Metal with $E_{red}^{0}$ value lower than 1.51 V will donate electron to $Mn^{3+}$ and thus reduces $Mn^{3+}$ to $Mn^{2+}$ .

So, Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$ .

Metal with $E_{red}^{0}$ value lower than -0.40 V will donate electron to $Cr^{3+}$ and thus reduces $Cr^{3+}$ to $Cr$ .

So, Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$ .

6 0

2 years ago