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2 years ago

What is the meaning of cryosphere

Chemistry

2 answers:

nikklg [1K]2 years ago

8 0

Answer:

The part of the earth’s surface where, at temperatures below 32°F (0°C), the water is frozen solid. Even a small pond while frozen over during the winter is part of the cryosphere, but cryosphere refers more often to large regions covered in snow and ice for much or all of the year, as the ice sheets of Antarctica and Greenland, glaciers, icebergs, and expanses of permafrost, all of which are especially sensitive to global shifts in climate:

Explanation:

irinina [24]2 years ago

3 0

Answer:

The cryosphere is the frozen water part of the Earth system.

One part of the cryosphere is ice that is found in water.

Explanation:

examples: mountain glaciers, continental ice sheets,seasonal snow and ice cover on land.

PLASE GIVE BRAINLIEST AND 5 STARS.

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Diamond has a density of 3.52 g/ml. what is the volume in cubic centimeters of a diamond with a mass of 15.1 g? 4.3 cm3 4.29 cm3

Kamila [148]

Density of 3.52g/ml means for each milliliter of space there are 3.52 grams of diamond.

To set the stage:
$\frac{3.52}{1} = \frac{15.1}{x}$
Set the denominators equal:
$\frac{3.52x}{x} = \frac{15.1}{x}$
Remove the denominators:
$3.52x = 15.1$
Isolate x by dividing both sides by 3.52:
$x = 4.289772 {cm}^{3}$

6 0

3 years ago

Read 2 more answers

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

2 years ago

How many liters of 4.0 M NaOH solution will react with 0.60 liters 3.0 M H2SO4?

slava [35]

Answer:

A. 0.90 L.

Explanation:

NaOH solution will react with H₂SO₄ according to the balanced reaction:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.

1.0 mole of H₂SO₄ reacts with 2.0 moles of NaOH.

For NaOH to react completely with H₂SO₄, the no. of millimoles should be equal.

∴ (MV) NaOH = (xMV) H₂SO₄.

x for H₂SO₄ = 2, due to having to reproducible H⁺ ions.

∴ V of NaOH = (xMV) H₂SO₄/ M of NaOH = 2(0.6 L)(3.0 M)/(4.0 M) = 0.90 L.

4 0

2 years ago

the vapor pressure of a naqueous solution is found to be 24.9 mmgh at 25C. what is the mole fraction of solute in this solution?

Gekata [30.6K]

Answer:

Mole fraction of solute is 0.0462

Explanation:

To solve this we use the colligative property of lowering vapor pressure.

First of all, we search for vapor pressure of pure water at 25°C = 23.8 Torr

Now, we convert the Torr to mmHg. Ratio is 1:1, so 23.8 Torr is 23.8 mmHg.

Formula for lowering vapor pressure is:

ΔP = P° . Xm

Where ΔP = P' (Vapor pressure of solution) - P° (Vapor pressure of pure solvent)

Xm = mole fraction

24.9 mmHg - 23.8 mmHg = 23mmHg . Xm

Xm = (24.9 mmHg - 23.8 mmHg) / 23mmHg

Xm = 0.0462

8 0

2 years ago

Identify which one is the reducing agent in this reaction

Romashka-Z-Leto [24]

Answer:

Na.

Explanation:

The oxidation-reduction reaction contains a reductant and an oxidant (oxidizing agent).

An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction. Also known as the electron acceptor, the oxidizing agent is normally in one of its higher possible oxidation states because it will gain electrons and be reduced.

A reducing agent (also called a reductant or reducer) is an element (such as calcium) or compound that loses (or "donates") an electron to another chemical species in a redox chemical reaction.

For the reaction:

2Na + S → Na₂S.

Na is oxidized to Na⁺ in (Na₂S) (loses 1 electron). "reducing agent".

S is reduced to S²⁻ in (Na₂S) (gains 2 electrons). "oxidizing agent".

6 0

3 years ago