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3 years ago

Cryosphere definition earth science

2 answers:

5 0

The cryosphere is those portions of Earth's surface where water is in solid form, including sea ice, lake ice, river ice, snow cover, glaciers, ice caps, ice sheets, and frozen ground.

Olin [163]3 years ago

4 0

The cryosphere is those portions of Earth's surface where water is in solid form, including sea ice, lake ice, river ice, snow cover, glaciers, ice caps, ice sheets, and frozen ground. Thus, there is a wide overlap with the hydrosphere

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Which type of wires (copper, aluminum, or string) are ferromagnetic metals and why?

nalin [4]

Answer: Copper isn't ferromagnetic,

Aluminum isn't ferromagnetic,

String has a ferromagnetic property

Explanation: first of all it is important to understand that all materials have diamagnetic effect.

From this we can explain two distinct property which is Paramagnetism and Ferromagnetism.

1.What do we understand by a paramagnetic material?

Ans- this describes a material that doesn't retain it's magnetic property even when the magnetic field has been removed, an example is Aluminum.

2. Ferromagnetic materials describes those type of materials that even after the removal of magnetic field retains it's magnetism. A good example of this is Iron, nickel etc.

By definition ferromagnetism is a basic property ( which depends on temperature, crystal Structure, chemical composition, etc.) That gives a material that attraction to magnet's and can form permanent magnet.

So from the examples listed in the question,

1. copper doesn't follow as a ferromagnetic material because it requires so much magnetic field to operate and is so weakly magnetized.

2. We already established aluminium as a paramagnetic material because of its weak ability to retain magnetism in the absence of magnetic field.

3. String because of its component which is (iron+carbon) has the ability to form ferrous metals.

4 0

3 years ago

1. Analysis of an unknown substance formerly used in rocket fuel reveals a composition of 93.28% nitrogen and 6.72% hydrogen by

Nitella [24]

Answer:

The formula of the compound is:

N2H2

Explanation:

Data obtained from the question:

Nitrogen (N) = 93.28%

Hydrogen (H) = 6.72%

Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:

N = 93.28%

H = 6.72%

Divide by their molar mass

N = 93.28 /14 = 6.663

H = 6.72 /1 = 6.7

Divide by the smallest

N = 6.663 / 6.663 = 1

H = 6.72 /6.663 = 1

Therefore, the empirical formula is NH.

Now, we can obtain the formula of the compound as follow:

The formula of a compound is simply a multiple of the empirical formula.

[NH]n = 30.04

[14 + 1]n = 30.04

15n = 30.04

Divide both side by 15

n = 30.04/15

n = 2

Therefore, the formula of the compound is:

[NH]n => [NH]2 => N2H2

6 0

3 years ago

Mary discovered that the cell membrane of a bacteria act like a fence because it keeps some things in and other things out

Diano4ka-milaya [45]

Answer:

The Answer is B. Analogy

7 0

3 years ago

Read 2 more answers

Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica

Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = $d_{ideal} = \dfrac{P\times M}{RT}$

here:

P = pressure of the O₂ gas = 310 bar

= $310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}$

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂ gas = 32 g/mol

∴

$d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}$

$d_{ideal}$ = 287.37 g/L

To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol² &

b = 0.0319 L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

$P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}$

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

$310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}$

$310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}$

$310V^3 -44.389V^2+1.382V-0.044=0$

After solving;

V = 0.1152 L

∴

$d_{Van \ der \ Waal} = \dfrac{32}{0.1152}$

$d_{Van \ der \ Waal}$ = 277.77 g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

5 0

3 years ago

Which of the following is an isotope of scandium?<br> 40 X 40 X 4X 2X

Liula [17]

688x

Explanation- Your welcome

6 0

3 years ago