True
Carbon monoxide is a primary pollutant which no odor results from incomplete combustion of fuel. The man sources are gasoline and burning of biomass.
Depending on the source of emission, pollutants can be classified into two groups that is primary and secondary pollutants.
A primary pollutant is emitted in the atmosphere directly from a source. It can be either natural sch as volcanic eruptions, sandstorms or man-made that is due to industrial and vehicle emissions. Examples of primary pollutants are nitrogen oxides, carbon monoxide and particulate matter.
Secondary pollutant is due to interactions between primary and secondary pollutants. These can be chemical or physical interactions. Examples are photo-chemical oxidants and secondary particulate matter.
Therefore, carbon monoxide CO is a primary pollutant.
Answer:
[H2] = 0.012 M
[N2] = 0.019 M
[H2O] = 0.057 M
Explanation:
The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.
2NO(g) + 2H2(g) ⇒ N2(g) + 2H2O(g)
i mol 0.10 0.050 0.10
c mol -0.038 -0.038 +0019 +0.038
e mol 0.062 0.012 00.019 0.057
Since the volume of the vessel is 1.0 L, the concentrations in molarity are:
[NO] = 0.062 M
[H2] = 0.012 M
[N2] = 0.019 M
[H2O] = 0.057 M
Answer:
1. Comparative
2. Independent variable
3. The pH is the Dependent variable
<u>Answer:</u> The solubility of oxygen at 682 torr is
<u>Explanation:</u>
To calculate the molar solubility, we use the equation given by Henry's law, which is:
Or,
where,
are the initial concentration and partial pressure of oxygen gas
are the final concentration and partial pressure of oxygen gas
We are given:
Conversion factor used: 1 atm = 760 torr
Putting values in above equation, we get:
Hence, the solubility of oxygen gas at 628 torr is