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2 years ago

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Bond energies do not account for the energy associated with the formation of aqueous solutions. Explain what energy is not accou

nted for AND how this energy contributes to the difference in calculated ∆Hrxn values using ∆Hf versus bond energy data ( -524 kJ/mol bond energy and -162.59 kJ/mol ∆Hf)

1 answer:

Tems11 [23]2 years ago

5 0

Answer:

Why do we all not know the answer to this on the practical

Explanation:

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A gas occupies 2.0 m3 at 100.0k and exerts a pressure of 100.0kPa. What volume will the gas occupy if the temperature is increas

Svetradugi [14.3K]

According to ideal gas equation, we know for 1 mole of gas: PV=RT
where P = pressure, T = temperature, R = gas constant, V= volume
If '1' and '2' indicates initial and final experimental conditions, we have
$\frac{P1V1}{P2V2} = \frac{T1}{T2}$

Given that: V1 = 100.0 kPa, T1 = 100.0 K, V1 = 2.0 m3, T2 = 400 K, P2 = 200.0 kPa

∴ on rearranging above eq., we get V2 = $\frac{P1V1T2}{T1} = \frac{100 X 2 X 400}{200X100}$
∴ V2 = 4 m3

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2 years ago

What is the mass of 1.71 ✕ 1023 molecules of h2so4?

Anuta_ua [19.1K]

The equation for calculating a mass is as follows:

m=n×M

Molar mass (M) we can determine from Ar that can read in a periodical table, and a number of moles we can calculate from the available date for N:

n(H2SO4)=N/NA

n(H2SO4)= 1.7×10²³ / 6 × 10²³

n(H2SO4)= 0.3 mole

Now we can calculate a mass of H2SO4:

m(H2SO4) = n×M = 0.3 × 98 = 27.8 g

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3 years ago

A voltaic cell consists of a Zn>Zn2+ half-cell and a Ni>Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn

nlexa [21]

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M is, 0.52 V

(c) The concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

$E^0_{[Ni^{2+}/Ni]}=-0.23V$

$E^0_{[Zn^{2+}/Zn]}=-0.76V$

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $Zn\rightarrow Zn^{2+}+2e^-$ $E^0_{[Zn^{2+}/Zn]}=-0.76V$

Reaction at cathode (reduction) : $Ni^{2+}+2e^-\rightarrow Ni$ $E^0_{[Ni^{2+}/Ni]}=-0.23V$

The balanced cell reaction will be,

$Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)$

First we have to calculate the standard electrode potential of the cell.

$E^o=E^o_{cathode}-E^o_{anode}$

$E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}$

$E^o=(-0.23V)-(-0.76V)=0.53V$

(a) Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}$

$E_{cell}=0.49V$

(b) Now we have to calculate the cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M.

New concentration of $Ni^{2+}$ = 1.50 - x = 0.500

x = 1 M

New concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}$

$E_{cell}=0.52V$

(c) Now we have to calculate the concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}$

Now put all the given values in the above equation, we get:

$0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}$

$x=1.49M$

The concentration of $Ni^{2+}$ = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1.49 = 1.59 M

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3 years ago

Which branch of the chemical industry usually requires advanced degrees? Marketing product development R & D corporate manag

fredd [130]

Answer: R & D is the correct option.

Explanation:

Research and development department of a chemical industry requires individuals holding advanced degrees because :

This will help the chemical industry to come up with new developments and advancements in preparation of the chemical products like :green methods to prepare certain chemicals which usually involves handling of toxic substances.
Formulation and development of the new chemical substitutes
Improvement of chemical product at microscopic level.

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2 years ago

What is the formula of iron(iii) oxide?

SCORPION-xisa [38]

Answer:

Fe2O3

Explanation:

8 0

2 years ago

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