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masha68 [24]
4 years ago
13

Carbon dioxide is a colorless odorless gas why is it considered a pollutant?

Chemistry
2 answers:
WARRIOR [948]4 years ago
7 0

Carbon dioxide is a gas having chemical formula CO_{2}, a gas which exhaled by humans and inhaled by plants. It is also emitted from the sources such as cement production and combustion of fossil fuels. It is also known as greenhouse gas associated with acidification of ocean.

Carbon dioxide with other gases acts like a blanket of Earth which absorbs infrared radiations emitted from the sun and thus, preventing the escaping of these rays into the outer space which results in the gradual heating of Earth's surface and atmosphere i.e. increase in temperature of Earth. This process is known as Global Warming.

Hence, carbon dioxide is responsible for the increase in warming of the atmosphere i.e. option (D).



saw5 [17]4 years ago
7 0
Even though carbon dioxide is a colorless, odorless gas, it is still considered a pollutant because D. it increases the warming of the atmosphere.
Global warming is the result of the abundance of carbon dioxide on Earth, which is why ecologists and environmentalists do everything in their power to raise awareness of this problem.
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Lina20 [59]

Answer:

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Explanation:

6 0
3 years ago
Read 2 more answers
4. A 14.5-L balloon is in the air where it is 20.0 C at 0.980-atm. A wind passes over and the balloon's pressure becomes 740.0-m
jeka94

The temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.

<h3>What is Combined gas law?</h3>

Combined gas law put together both Boyle's Law, Charles's Law, and Gay-Lussac's Law. It states that "the ratio of the product of volume and pressure and the absolute temperature of a gas is equal to a constant.

It is expressed as;

P₁V₁/T₁ = P₂V₂/T₂

Given the data in the question;

  • Initial volume V₁ = 14.5L
  • Initial pressure P₁ = 0.980atm
  • Initial temperature T₁ = 20.0°C = 293.15K
  • Final volume V₂ = 14.3L
  • Final pressure P₂ = 740.mmHg = 0.973684atm
  • Final temperature T₂ = ?

We substitute our given values into the expression above.

P₁V₁/T₁ = P₂V₂/T₂

( 0.980atm × 14.5L )/293.15K = ( 0.973684atm × 14.3L )/T₂

14.21Latm / 293.15K = 13.92368Latm / T₂

14.21Latm × T₂ = 13.92368Latm × 293.15K

14.21Latm × T₂ = 4081.72679LatmK

T₂ = 4081.72679LatmK / 14.21Latm

T₂ = 287.24K

T₂ = 14.09°C

Therefore, the temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.

Learn more about the combined gas law here: brainly.com/question/25944795

#SPJ1

4 0
3 years ago
If 0.220 molmol of a nonvolatile nonelectrolyte are dissolved in 3.70 molmol of water, what is the vapor pressure PH2OPH2O of th
Masteriza [31]

Answer:

Vapor pressure for solution is 22.4 Torr

Explanation:

P° → Vapor pressure of pure solvent

P' → Vapor pressure of solution

Formula for lowering vapor pressure is: P° - P' = P°  . Xm

Xm is the mole fraction. Let's determine it:

0.220 mol of solute + 3.70 mol of water = 3.92 total moles

Mole fraction for solute → 0.220 / 3.92 = 0.056

23.8 Torr - P' = 23.8 Torr  . 0.056

P' = - ( 23.8 Torr  . 0.056 - 23.8 Torr ) → 22.4 Torr

4 0
3 years ago
Which method of finding slope do you prefer? Why?<br> (slope formula, table, other)
grigory [225]
I prefer using the slope formula because it’s easy to pick two numbers on a graph and solve for the slope with a simple equation.
5 0
4 years ago
A cylinder of compressed gas rolls off a boat and falls to the bottom of a lake. Eventually it rusts and the gas bubbles to the
igomit [66]

Answer:

0.228 L is the volume that the gas occupies after it is driedand stored at STP.

Explanation:

Pressure of the wet gas = P=695 Torr=\frac{695}{760}atm=0.914 atm

1 atm = 760 Torr

Pressure of water vapor = p = 0.0372 atm

Pressure of gas = P_1=P-p=0.914 atm-0.0372 atm = 0.8768 atm

Volume of wet gas = V_1=287 mL=0.287 L

1 mL = 0.001 L

Temperature of the wet gas = T_1=28.0^oC=28.0+273 K=301 K

Pressure of dry gas at STP = P_2=1 atm

Temperature of the dry gas at STP ,T_2= 273 K

Volume of dry gas at STP = V_2

Using combined gas law:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

\frac{0.8768 atm\times 0.287 L}{301 K}=\frac{1 atm\times V_2}{273 K}

V_2=0.228 L =228 mL

0.228 L is the volume that the gas occupies after it is driedand stored at STP.

4 0
3 years ago
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