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3 years ago

Why are naturally occurring substances, such as particulates or carbon dioxide, sometimes considered pollutants?

Chemistry

2 answers:

iogann1982 [59]3 years ago

6 0

Answer:

Because it is in such high concentrations that it cannot be compensated with its natural elimination process.

Explanation:

Substances like carbon dioxide are produced naturally by living beings in the breathing process. This carbon dioxide is taken by plants for their photosynthesis process, but currently, the carbon dioxide is also produced by the burning of organic materials in industries. The quantities that are produced are exceeding the capacity of the plants to transform it, which causes gas to accumulate in the atmosphere damaging the ozone layer

____ [38]3 years ago

5 0

Because despite being natural they still harm the ozone layer and cause breathing problems, even suffocation.

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How to write a chemical formula ?

Kaylis [27]

Answer:

To write a chemical equation, the reactants should be written on the left, and the products should be written on the right and the coefficients should be written next to the symbols of entities to indicate the number of moles of a substance produced used in the chemical reaction.

8 0

2 years ago

Read 2 more answers

How many moles are in 4.818 x 1024 chloride ions

erastovalidia [21]

Answer:

<h3>The answer is 8.00 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{4.818 \times {10}^{24} }{6.02 \times {10}^{23} } \\ = 8.00332225...$

We have the final answer as

<h3>8.00 moles</h3>

Hope this helps you

4 0

3 years ago

Where and when are acute-phase proteins produced?

mrs_skeptik [129]

Answer:

The answer should be C. Primarily in the liver in response to inflammation :)

Have an amazing day!!

Please rate and mark brainliest!!

3 0

2 years ago

A gas sample has volume 3.00 dm3 at 101

gavmur [86]

Answer:

V₂ = 21.3 dm³

Explanation:

Given data:

Initial volume of gas = 3.00 dm³

Initial pressure = 101 Kpa

Final pressure = 14.2 Kpa

Final volume = ?

Solution;

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

101 Kpa × 3.00 dm³ = 14.2 Kpa × V₂

V₂ = 303 Kpa. dm³/ 14.2 Kpa

V₂ = 21.3 dm³

Explanation:

3 0

3 years ago

a. If 42.5 g of CH3OH reacts with 22.8 L of O2 at 27°C and a pressure of 2.00 atm, calculate the number of grams of water vapor

Korvikt [17]

Answer:

The mass of water vapor is 44.46 grams

The volume of water is 30.37 L

Explanation:

Step 1: Data given

Mass of CH3OH =42.5 grams

Molar mass CH3OH = 32.04 g/mol

Volume of O2 = 22.8 L

Pressure = 2.00 atm

Step 2: The balanced equation

2CH3OH + 3O2 → 2CO2 + 4H2O

Step 3: Calculate moles CH3OH

Moles CH3OH = mass CH3OH / molar mass CH3OH

Moles CH3OH = 42.5 grams / 32.04 g/mol

Moles CH3OH = 1.326 moles

Step 4: Calculate moles O2

p*V = n*R*T

⇒with p = the pressure = 2.00 atm

⇒with V = the volume of O2 = 22.8 L

⇒with n = the moles of O2 = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*Atm/mol*K

⇒with T = the temperature = 27 °C = 300 K

n = (p*V) / (R*T)

n = (2.00 * 22.8) / (0.08206*300)

n = 1.85 moles

Step 5: Calculate the limiting reactant

For 2 moles CH3OH we need 3 moles O2 to produce 2 moles CO2 and 4 H2O

O2 is the limiting reactant. It will completely be consumed ( 1.85 moles). CH3OH is in excess. There will react 2/3*1.85 = 1.233 moles. There will remain 1.326 - 1.233 = 0.093 moles

Step 6: Calculate moles products

For 2 moles CH3OH we need 3 moles O2 to produce 2 moles CO2 and 4 H2O

For 1.85 moles O2 we'll have 1.233 moles CO2 and 2.467 moles H2O

Step 7: Calculate mass H2O

Mass H2O = moles H2O * molar mass H2O

Mass H2O = 2.467 moles * 18.02 g/mol

Mass H2O = 44.46 grams

Step 8: Calculate volume H2O

p*V = n*R*T

⇒with p = the pressure = 2.00 atm

⇒with V = the volume of H2O = TO BE DETERMINED

⇒with n = the moles of H2O = 2.467 moles

⇒with R = the gas constant = 0.08206 L*Atm/mol*K

⇒with T = the temperature = 27 °C = 300 K

V = (n*R*T)/p

V = (2.467 * 0.08206 * 300) / 2.00

V = 30.37 L

The mass of water vapor is 44.46 grams

The volume of water is 30.37 L

3 0

3 years ago