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3 years ago

A small rock lies on a slope near a tropical sea. What are three ways that the rock could be eroded?

Chemistry

2 answers:

vlabodo [156]3 years ago

7 0

Answer: - The heat of the sun could break down the rock.

- Tall, powerful waves could carry the rock into the ocean.

- The rock could react to seawater, break down, and be carried away by the sea.

Erosion is a natural phenomena in which the top surface of the soil or the rock is displaced or carried away to different location. It disturbs the fertility of the soil. The physical agents responsible for erosion are water, wind, heat and others.

The three ways the small rock that lies on a slope near a tropical sea could be eroded are:

The heat of the sun could break down the rock: The sunlight incident on the rock can weak the external structure of the rock and the loosen particles are displaced by either wind and water.

Tall, powerful waves could carry the rock into the ocean: The waves can displaced the rocks away from actual place causing erosion.

The rock could react to seawater, break down, and be carried away by the sea: The sea water can bring chemical change in the rock structure and will enhance it's weathering.

Mashutka [201]3 years ago

5 0

Erosion is the removal of small particles from something, it’s not to be confused with weathering, which breaks a large thing into smaller things. By this definition of erosion, the first one and the last two are the answers. The second and third option describe weathering, but not erosion.

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A. The Russian chemist Dmitri Mendeleev

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Answer:

O B. Convert the 10 g of NaCl to moles of NaCl.

Explanation:

The formula for finding the molality is m=moles of solute/kg of solvent. The solute for this question is NaCl and the solvent is water.

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3 years ago

Use Hess's Law to calculate the enthalpy change for the reaction

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Answer:

ΔH = 125.94kJ

Explanation:

It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:

1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ

2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ

-1/2 (1):

WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ

3/2 (2):

3H2(g) + 3/2O2(g) → 3H2O(g) ΔH = -716.76kJ

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3 years ago

Which type of reaction is the Haber process: N2(g) + 3 H2(g) → 2 NH3(g) + heat? *

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Explanation:

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3 years ago

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0

3 years ago