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3 years ago

What is the land part of the earth called?

Chemistry

2 answers:

Vladimir79 [104]3 years ago

5 0

Answer:

Lithosphere - Earth

The lithosphere contains all of the cold, hard solid land of the planet's crust (surface), the semi-solid land underneath the crust, and the liquid land near the center of the planet.

Explanation:

Molodets [167]3 years ago

4 0

The Land Hemisphere is the part of the Earth that encompasses the largest portion of land on the planet. Despite being called the Land Hemisphere, the area of water in the hemisphere is greater than that of the area of land present here. The Land Hemisphere has its center at 47°13′N 1°32′W which is located near the French city of Nantes. The Land Hemisphere includes most of Asia and South America and the whole of Africa, Europe, and North America. Seven-eighths of the land area of the Earth is part of this hemisphere

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Hydrogen can be extracted from natural gas according to the following equilibrium.

antiseptic1488 [7]

Answer:

2.16x10⁻²

Explanation:

First, let's find out the molar concentrations of the reactants. The molar mass of CH4 is 16 g/mol, and of CO2 is 44 g/mol. The number of moles is the mass divided by the molar mass:

nCH4 = 24.0/16 = 1.5 moles

nCO2 = 88.0/44 = 2 moles

The concentration is the number of moles divded by the volume, thus:

[CH4] = 1.5/1 = 1.50 M

[CO2] = 2/1 = 2.00 M

For the equilibrium reaction, let's do an equilibrium chart:

CH4(g) + CO2(g) ⇄ 2CO(g) + 2H2(g)

1.50 2.00 0 0 Initial

-x -x +2x +2x Reacts (stoichiometry is 1:1:2:2)

1.50-x 2.00-x 2x 2x Equilibrium

As sateted in problem, [CH4] = 2.70*[CO]

1.50 - x = 2.70*2x

1.50 - x = 5.4x

6.4x = 1.50

x = 0.2344

Thus, at equilibrium:

[CH4] = 1.50 - 0.2344 = 1.2656 M

[CO2] = 2.00 - 0.2344 = 1.7656 M

[CO] = 2*0.2344 = 0.4688 M

[H2] = 2*0.2344 = 0.4688 M

The equilibrium constant is the multiplication of the concentrations of the products elevated by their coefficients, divided by the multiplication of the concentration of the reactants elevated by their coefficients.

K = ([CO]²*[H2]²)/([CH4]*[CO2])

K = (0.4688²*0.4688²)/(1.2656*1.7656)

K = 0.0483/2.2345

K = 2.16x10⁻²

7 0

3 years ago

A sample of an unknown gas takes 371 s to diffuse through a porous plug at a given temperature.

blsea [12.9K]

Answer:

125.84 g/mol is the molar mass of the unknown gas.

Explanation:

Let the volume of the gases effusing out be V.

Effusion rate of the unknown gas = $R=\frac{V}{371 s}$

Effusion rate of the nitrogen gas = $r=\frac{V}{175 s}$

Molar mass of unknown gas = m

Mass of nitrogen gas = 28 g/mol

Graham's law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$