The volume of CO2 at STP =124.298 L
<h3>Further explanation</h3>
Given
Reaction
4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O
701,52 g of KMnO4
Required
volume of CO2 at STP
Solution
mol KMnO4 (MW=158,034 g/mol) :
mol = mass : MW
mol = 701.52 : 158.034
mol = 4.439
mol CO2 from equation : 5/4 x mol KMnO4 = 5/4 x 4.439 = 5.549
At STP 1 mol = 22.4 L, so for 5.549 moles :
=5.549 x 22.4
=124.298 L
Answer:
c. reduces the concentration of the hazardous material in the air.
Explanation:
Pollution can be defined as the physical degradation or contamination of the environment through an emission of harmful, poisonous and toxic chemical substances.
Particulate population is a form of pollution that is responsible for the degradation of the environment.
Particulate matter is also referred to as particle pollution or atmospheric aerosol particles and it can be defined as a complex microscopic mixture of liquid droplets and solid particles that are suspended in air. Other forms of particle pollution includes space debris and marine debris.
Some examples of particulate pollution are dusts, soot, dirt, smoke, etc.
Basically, various anthropogenic activities such as construction and agriculture are primary sources of particulate matter because they're capable of causing particle pollution on their own. The other sources of particle pollution is the secondary source which includes factories, cars, trucks, etc.
Vapor dispersion can be defined as a process which is typically used for removing particle pollutants from the atmosphere through the use of vapor or steam.
Hence, vapor dispersion when adopted, reduces the concentration of the hazardous material such as soot, dusts, smoke, etc., in the air.
Given that the gas inside the balloon is an ideal gas, then the ideal gas equation can be used to describe the system. I think the correct answer from the choices listed above is option B. If the balloon is heatedto a temperature of 15.5T while it is placed under a high pressure of 15.5P, then the volume of the balloon stays the same since the change in temperature corresponds to same degree of change with pressure.
