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3 years ago

15

Planet A is an inner planet with no moon and hardly any atmosphere. Planet B is an inner planet with no moon but with a dense at

mosphere. Which pair of planets is being described?

1 answer:

Nady [450]3 years ago

7 0

Planet A would be Mercury, Planet B would be Venus.

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Consider the Gibbs energies at 25 ∘C.

zysi [14]

Answer:

A)ΔGrxn∘=55.7kJ/mol

B)Ksp=1.75×10^−10

C)ΔGrxn∘=70.0kJ/mol

D)Ksp=5.45×10^−13

Explanation:

ΔGrxn∘=ΔGf,products∘−ΔGf,reactants∘

To calculate for the

Ksp

of the dissolution reaction can be claculated

ΔGrxn∘=−RTlnKsp

where R is the proportionality constant equal to 8.3145 J/molK.

A)

ΔGrxn∘=[ΔGf,Ag(aq)+∘+ΔGf,Cl(aq)−∘]−ΔGf,

AgCl(s)⇌Ag(aq)++Cl(aq)−

ΔG∘rxn=[77.1kJ/mol+(−131.2kJ/mol)]−(−109.8kJ/mol)

ΔGrxn∘=55.7kJ/mol

b) Calculate the solubility-product constant of AgCI.

ΔGrxn∘=−RTlnKsp

55.7kJ/mol=−(8.3145×10−3J/molK)(298.15K)InKsp

Ksp=1.75×10^−10

c) Calculate

To calculate ΔG°rxn

for the dissolution of AgBr(s).

ΔGrxn∘=[ΔGf,Ag(aq)+∘+ΔGf,Br(aq)−∘]−ΔGf,

ΔGrxn∘=[77.1kJ/mol+(−104.0kJ/mol)]−(−96.90kj/mol

ΔGrxn∘=70.0kJ/mol

d)To Calculate the solubility-product constant of AgBr.

ΔGrxn∘=−RTlnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K)lnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K

Ksp=5.45×10^−13

8 0

3 years ago

Classify the chemical reaction shown here:

lesantik [10]

Answer:

Single-Replacement Reaction

General Formulas and Concepts:

<u>Chemistry - Reactions</u>

Synthesis Reactions: A + B → AB
Decomposition Reactions: AB → A + B
Single-Replacement Reactions: A + BC → AB + C
Double-Replacement Reactions: AB + CD → AD + BC

Explanation:

<u>Step 1: Define RxN</u>

Mg + H₂SO₄ → MgSO₄ + H₂

<u>Step 2: Identify RxN</u>

Mg + H₂SO₄ → MgSO₄ + H₂

Single-Replacement Reaction A + BC → AB + C

A = Mg
BC = H₂SO₄
AB = MgSO₄
C = H₂

7 0

3 years ago

Determine the percent ionization of a 0.230 m solution of benzoic acid.

Fudgin [204]

We need the dissociation constant of benzoic acid which is 6.3x10^'5. Then using the dissociation formula, ka = x2 / (Mo - x) where Mo is the initial concentration. x is determined then. percent ionization is computed as (x/Mo)* 100%. This is then the final answer.

5 0

4 years ago

Consider the reaction: A <=> B. Under standard conditions at equiliubrium, the concentrations of the compounds are [A] = 1

Katen [24]

Answer:

$Keq'>1\\\Delta G'$

Explanation:

Hello,

In this case, for the given reaction, the equilibrium constant turns out:

$Keq=\frac{[B]}{[A]}=\frac{0.5M}{1.5M} =1/3$

Nonetheless, we are asked for the reverse equilibrium constant that is:

$Keq'=\frac{1}{Keq}=3$

Which is greater than one.

In such a way, the Gibbs free energy turns out:

$\Delta G'=-RTln(Keq')\\$

Now, since the reverse equilibrium constant is greater than zero its natural logarithm is positive, therefore with the initial minus, the Gibbs free energy is less than zero, that is, negative.

7 0

3 years ago

Read 2 more answers

How many sig figs are in 1401.00

matrenka [14]

Answer:

6 We know it is closer to 1401.00 than 1400.99 or 1401.01 therefor the 6 sig figs

Explanation:

7 0

3 years ago