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:
They contain of atoms
Explanation:
That's because atomic weights or masses of each atom of each element are proportional to each other, the same number of atoms of each element will give masses that are also proportional to each other. If you start with 20 oxygen atoms, you will need 40 hydrogen atoms to make the water and you will get 20 molecules of water.
A phosphorus atom and a chlorine atom are likely to form a covalent bond because both elements are nonmetals.
Answer:
63 mL
Explanation:
To find the amount of mililiters in an amount of liters, we must multiply by the amount of liters by 1000.
0.063 L × 1000 = 63 mL
The molar mass of the gas is 77.20 gm/mole.
Explanation:
The data given is:
P = 3.29 atm, V= 4.60 L T= 375 K mass of the gas = 37.96 grams
Using the ideal Gas Law will give the number of moles of the gas. The formula is
PV= nRT (where R = Universal Gas Constant 0.08206 L.atm/ K mole
Also number of moles is not given so applying the formula
n= mass ÷ molar mass of one mole of the gas.
n = m ÷ x ( x molar mass) ( m mass given)
Now putting the values in Ideal Gas Law equation
PV = m ÷ x RT
3.29 × 4.60 = 37.96/x × 0.08206 × 375
15.134 = 1168.1241 ÷ x
15.134x = 1168.1241
x = 1168.1241 ÷ 15.13
x = 77.20 gm/mol
If all the units in the formula are put will get cancel only grams/mole will be there. Molecular weight is given by gm/mole.
