Identify the two main materials that can be found in the inner and outer core layers of the earth.

Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr

VikaD [51]

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy

F = Faraday constant = 96500 C

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

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

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

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

5 0

3 years ago

Which of these is NOT a lipid?

Minchanka [31]

Answer: RNA

Explanation:

6 0

2 years ago

A solution of ammonia and water contains 2.10×1025 water molecules and 7.10×1024 ammonia molecules. How many total hydrogen atom

Natali5045456 [20]

There are 6.33 × 10²⁵ hydrogen atoms in this solution in total.

<h3>Explanation</h3>

There are two hydrogen atoms in each water $\text{H}_2\text{O}$ molecule.
There are three hydrogen atoms in each ammonia $\text{NH}_3$ molecule.

2.10 × 10²⁵ water molecules and 7.10 × 10²⁴ ammonia molecules will contain

$2 \times 2.10 \times 10^{25} + 3 \times 7.10 \times 10^{24} = 6.33 \times 10^{25}$ hydrogen atoms in total.

8 0

3 years ago

How many moles of water are produced if 5.43 mol PbO2 are consumed

Mariana [72]

It has to be understood that 2 moles of oxygen are there in each mole of PbO2. Then it has to be calculated for 2 moles of oxygen.

Amount of oxygen = 2 * 5.43 moles
= 10.86 moles
Now it is also a fact that each mole of H2O contains 1 mole of oxygen. Then it can be easily concluded that 10.86 moles of water will be produced. I hope the procedure is clear enough for you to understand.

5 0

3 years ago

Read 2 more answers

Can you please answer the question shown below?

Yuki888 [10]

Answer:

Nothing is shown below!!!

6 0

3 years ago

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