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3 years ago

Which of the following affects cell potential?

Chemistry

1 answer:

Slav-nsk [51]3 years ago

3 0

Answer:

the standard cell potential value

Explanation:

For every cell, we can calculate its standard electrode potential from the table of standard electrode potentials listed in many textbooks.

However, from Nernst's equation;

Ecell= E°cell - 0.0592/n log Q

Hence the standard cell potential (E°cell) affects the value of the calculated cell potential Ecell from Nernst's equation as stated above.

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7. The youngest part of the ocean floor is found ____.

aev [14]

Answer:

The answer is Near The Ocean Ridges.

5 0

2 years ago

Calculate the standard potential for the following galvanic cell: Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s) which has the overall bala

mylen [45]

Answer:

1.06 V

Explanation:

The standard reduction potentials are:

Ag^+/Ag E° = 0.7996 V

Ni^2+/Ni E° = -0.257 V

The half-cell and cell reactions for Ni | Ni^2+ || Ag^+ | Ag are

Ni → Ni^2+ + 2e- E° = 0.257 V

Ni + 2Ag^+ → Ni^2+ + 2Ag E° = 1.0566 V

To three significant figures, the standard potential for the cell is 1.06 V .

8 0

3 years ago

The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

hammer [34]

Answer:

A) 1059 J/mol

B) 17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

$C_p=C_v+R$

We can determine $C_v$ from above if we make $C_v$ the subject of the formula as:

$C_v$ $=C_p-R$

$C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314$

$C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4$

A).

The formula for calculating change in internal energy is given as:

$dU=C_vdT$

If we integrate above data into the equation; it implies that:

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 1059J/mol$

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 17,920 J/mol$

3 0

3 years ago

The diagram below is an artist’s impression of a single atom of element Be. The neutrons are shown with stripes, the protons are

Ahat [919]

Answer is: beryllium-10.

The diagram shows that atom has 4 protons, 5 neutrons and 2 valence electrons.

Atomic number is the number of protons, which is characteristic of a chemical element, beryllium (Be) is an element with atomic number 4.

Two valence electrons means that atom is from 2. group of periodic table, only beryllium is from that group; sodium (1. group), boron (13. group) and carbon (14. group).

Beryllium-10 has 6 neutrons, so it is isotope (different number of neutrons or mass number).

3 0

3 years ago

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What is the name of the "invisible force" that holds you to Earth's surface

Alex

Answer:

Gravity

Explanation:

7 0

2 years ago

Read 2 more answers