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3 years ago

The image shows the formation of a fault-block mountain.

Chemistry

2 answers:

marissa [1.9K]3 years ago

7 0

1. Two parallel normal faults form.

4. The hanging wall on the left slides down relative to the footwall.

5. The hanging wall on the right slides down relative to the footwall.

lubasha [3.4K]3 years ago

4 0

Answer: A, D & E

Explanation: The image shows the formation of a fault-block mountain.

Which statements best explain the formation of these mountains? Check all that apply.

Answer: A, D & E Just took the edge test. 100%

(A) Two parallel normal faults form.

(D) The hanging wall on the left slides down relative to the footwall.

(E) The hanging wall on the right slides down relative to the footwall.

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Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine

yanalaym [24]

Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

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3 years ago

When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 27.6 g of carbon were burned in the presence of

In-s [12.5K]

Answer:

The answer to your question is: 101.2 g of CO2

Explanation:

C = 27.6 g

O₂ = 86.5 g remained 12.9 g

O₂ that reacted = 86.5 - 12.9 = 73.6 g

C + O₂ ⇒ CO₂ The equation is balanced

27.6 73.6 ?

MW 12 32 44

Rule of three

12 g of C------------------ 44 g CO2

27.6 g C ------------------ x

x = 27.6(44)/12 = 101.2 g of CO2

32 g of O2 --------------- 44 g of CO2

73.6 g of O2 ------------ x

x = 73.6(44)/32 = 101.2 g of CO2

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3 years ago

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Answer:6 atoms

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2 years ago

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Explanation:

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3 years ago

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Dafna11 [192]

Answer:

665 g

Explanation:

Let's consider the following thermochemical equation.

2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(l), ΔH°rxn= –5,314 kJ/mol

According to this equation, 5,314 kJ are released per 8 moles of CO₂. The moles produced when 1.00 × 10⁴ kJ are released are:

-1.00 × 10⁴ kJ × (8 mol CO₂/-5,314 kJ) = 15.1 mol CO₂

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3 years ago