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Romashka-Z-Leto [24]

3 years ago

11

How can the major components of air be separated

1 answer:

WARRIOR [948]3 years ago

4 0

Air can be separated into its components by means of distillation in special units. So-called air fractionating plants employ a thermal process known as cryogenic rectification to separate the individual components from one another in order to produce high-purity nitrogen, oxygen and argon in liquid and gaseous form.

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Calculate the molar mass of each of the following:

Allushta [10]

Explanation:

Molar mass

The mass present in one mole of a specific species .

The molar mass of a compound , can easily be calculated as the sum of the all the individual atom multiplied by the number of total atoms .

(a) S₈

Molar mass of of the atoms are -

sulfur, S = 32 g/mol.

Molar mass of S₈ = 8 * 32 g/mol. = 256 g/mol.

(b) C₂H₁₂

Molar mass of of the atoms are -

Hydrogen , H = 1 g/mol

Carbon , C = 12 g/mol

Molar mass of C₂H₁₂ = ( 2 * 12 ) + (12 * 1 ) = 36 g /mol

(c) Sc₂(SO₄)₃

Molar mass of of the atoms are -

sulfur, S = 32 g/mol.

oxygen , O = 16 g/mol.

scandium , Sc = 45 g/mol.

Molar mass of Sc₂(SO₄)₃ = (2 * 45 ) + ( 3 *32 ) + ( 12 * 16 ) = 378 g /mol

(d) CH₃COCH₃ (acetone)

Molar mass of of the atoms are -

Carbon , C = 12 g/mol

oxygen , O = 16 g/mol.

Hydrogen , H = 1 g/mol

Molar mass of CH₃COCH₃ (acetone) = (3 * 12 ) + ( 1 * 16 ) + ( 6 * 1 ) = 58g/mol

(e) C₆H₁₂O₆ (glucose)

Molar mass of of the atoms are -

Carbon , C = 12 g/mol

oxygen , O = 16 g/mol.

Hydrogen , H = 1 g/mol

Molar mass of C₆H₁₂O₆ (glucose) = ( 6 * 12 ) + ( 12 * 1 ) + ( 6 * 16 ) = 108g/mol.

6 0

3 years ago

When a cell is in a solution where the concentration of solute is the same in the cell as in the solution, the solution is calle

Elodia [21]

The answer to this question would be: isotonic

An isotonic solution has a same tonicity/concentration to the cell. This kind of solution will not cause osmosis to occur so it wouldn't shrink or swell the cells. This is why an isotonic solution is safer for the cells. The normal saline solution is isotonic to human cells and used frequently in medicine to replace body fluid.

7 0

4 years ago

Solid chromium (III) reacts with oxygen gas to form solid Cr2O3. What is this type of reaction?

mafiozo [28]

Answer:

.081 g of O2

Explanation:

4Cr + 3O2 -----> 2Cr2O3

.175 g Cr x [1 mole / 52.0 g] x [2 moles Cr2O3 / 4 moles Cr] x [152 g / 1 mole] = .256 g of Cr2O3

.175 g Cr x [1 mole / 52.0 g] x [3 moles O2 / 4 moles Cr] x [32 g / 1 mole] = .081 g of O2

5 0

3 years ago

Weak noncovalent interactions:__________ a. do not include ionic interactions b. always involve water. c. can have a large cumul

valentina_108 [34]

Answer:

c. can have a large cumulative effect

Explanation:

Noncovalent interactions between molecules are weaker than covalent interactions. Noncovalent interactions between molecules are of various types which include van der Waals forces, hydrogen bonding, and electrostatic interactions or ionic bonding.

van der Waals forces are weak interactions found in all molecules. They include dipole-dipole interactions - formed due to the differences in the electronegativity of atoms - and the London dispersion forces.

Hydrogen Bonds results when electrons are shared between hydrogen and a strongly electronegative atoms like fluorine, nitrogen, oxygen. The hydrogen acquires a partial positive charge while the electronegative atom acquires a partial negative. This results in attraction between hydrogen and neighboring electronegative molecules.

Ionic bonds result due to the attraction between groups with opposite electrical charges, for example in common salt between sodium and chloride ions.

Even though these noncovalent interactions are weak, cumulatively, they exert strong effect. For example, the high boiling point of water and the crystal structure of ice are due to hydrogen bonding.

7 0

4 years ago

Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an

Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 0

4 years ago