All categories

3 years ago

Smog usually forms from gound-level ozone and what other human-made pollutant?

2 answers:

4 0

This happens when pollutants emitted by cars, power plants, industrial boilers, refineries, chemical plants, and other sources chemically react in the presence of sunlight. Ozone at ground level is a harmful air pollutant, because of its effects on people and the environment, and it is the main ingredient in “smog."

arsen [322]3 years ago

3 0

The word smog is created from the words "smoke" and "fog". It contains chemicals such carbon monoxide, sulfur dioxide, and various nitrogen oxides. It is created mostly from the burning of fossil fuels, such as gasoline, which is used commonly as fuel components for many different vehicles. Having known that such chemicals are found in exhaust fumes from cars, it is safe to say that the answer could either be smoke, or carbon dioxide(depending on teacher expectations).

You might be interested in

hydrogen gas and oxygen gas react to form liquid hydrogen peroxide. which starement is true about the reaction?

Strike441 [17]

The two elements mixed together to form a chemical bond and reacted as a chemical change

4 0

3 years ago

1. Which of the combinations in the lab activity had indications that a chemical change occured? Defend your argument with evide

ArbitrLikvidat [17]

Answer:The green growing on the penny of copper and the rust forming on the nail of iron are chemical changes. Boiling away salt water, scraping iron filings from a mixture of sand with a magnet, and breaking a rock with a hammer, are physical changes.

Explanation:

4 0

3 years ago

The metal tantalum becomes superconducting at temperatures below 4.483 K. Calculate the temperature at which tantalum becomes su

masha68 [24]

Answer:

The correct answer is "-268.667°C".

Explanation:

Given:

Temperature,

= 4.483 K (below)

Now,

The formula of temperature conversion will be:

⇒ $T(^{\circ} C)=T(K)-273.15$

By putting the values, we get

⇒ $=4.483-273.15$

⇒ $=-268.667^{\circ} C$

Thus the above is the correct answer.

3 0

3 years ago

A 6.000L tank at 19.2°C is filled with 18.0g of carbon monoxide gas and 10.6g of chlorine pentafluoride gas. You can assume both

Jobisdone [24]

Answer:

Total pressure: 2.89 atm

Mole fraction CO: 0.88

Partial pressure CO: 2.56 atm

Mole fraction ClF₅: 0.12

Partial pressure ClF₅: 0.33 atm

Explanation:

We should apply the Ideal Gases Law to solve this:

P . V = n . R . T

We need n, which is the total moles for the mixture

Total moles = Moles of CO + Moles of ClF₅

Moles of CO = mass of CO / molar mass CO → 18 g/28 g/mol = 0.643 mol

Moles of ClF₅ = mass of ClF₅ / molar mass ClF₅ → 10.6g/ 130.45 g/m = 0.0812 mol

0.643 mol + 0.0812 mol → 0.724 moles in the mixture

So we have the total moles so with the formula we would know the total pressure.

P . 6L = 0.724 mol . 0.082L.atm/mol.K . 292.2K

P = ( 0.724 mol . 0.082L.atm/mol.K . 292.2K) / 6L

P = 2.89 atm

Mole fraction is defined as the quotient between the moles of gas over total moles, and it is equal to partial pressure of that gas over total pressure

Moles of gas X /Total moles = Partial pressure of gas X/Total pressure

(Moles of gas X / Total moles) . Total pressure = Partial pressure of gas X

Mole fraction CO = 0.643 / 0.724 = 0.88

Partial pressure CO = 0.88 . 2.89 atm → 2.56 atm

Mole fraction ClF₅ = 0.0812 / 0.724 = 0.12

Partial pressure ClF₅ = 0.12 . 2.89 atm → 0.33 atm

5 0

3 years ago

At constant pressure, which of these systems do work on the surroundings? A ( s ) + B ( s ) ⟶ C ( g ) A(s)+B(s)⟶C(g) 2 A ( g ) +

Tju [1.3M]

Correct question:

At constant pressure, which of these systems do work on the surroundings?

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

(d) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

Answer:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

Explanation:

Work done by a system on the surroundings at a constant pressure is given as;

W = -PΔV

Where;

ΔV is gas expansion, that is final volume of the gas minus initial volume of the gas must be greater than zero.

Part (a)

A ( s ) + B ( s ) ⟶ C ( g )

ΔV = 1 - (0) = 1 (expansion)

Part (b)

2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

ΔV = 5 - ( 2+ 2) = 1 (expansion)

Part (c)

A ( g ) + B ( g ) ⟶ C ( g )

ΔV = 1 - ( 1 + 1) = -1 (compression)

Part (d)

2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

ΔV = 3 - ( 4) = -1 (compression)

Thus, systems where there is gas expansion are in part (a) and part (b). The correct answers are:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

4 0

3 years ago