<span>Several
important pollutants are produced by fossil fuel combustion: carbon
monoxide, nitrogen oxides, sulfur oxides, and hydrocarbons. In addition,
total suspended particulates contribute to air pollution, and nitrogen
oxides and hydrocarbons can combine in the atmosphere to form
tropospheric ozone, the major constituent of smog.
Carbon monoxide is a gas formed as a by-product during the incomplete
combustion of all fossil fuels. Exposure to carbon monoxide can cause
headaches and place additional stress on people with heart disease. Cars
and trucks are the primary source of carbon monoxide emissions.
Two oxides of nitrogen--nitrogen dioxide and nitric oxide--are formed in
combustion. Nitrogen oxides appear as yellowish-brown clouds over many
city skylines. They can irritate the lungs, cause bronchitis and
pneumonia, and decrease resistance to respiratory infections. They also
lead to the formation of smog. The transportation sector is responsible
for close to half of the US emissions of nitrogen oxides; power plants
produce most of the rest.
Sulfur oxides are produced by the oxidization of the available sulfur in
a fuel. Utilities that use coal to generate electricity produce
two-thirds of the nation's sulfur dioxide emissions. Nitrogen oxides and
sulfur oxides are important constituents of acid rain. These gases
combine with water vapor in clouds to form sulfuric and nitric acids,
which become part of rain and snow. As the acids accumulate, lakes and
rivers become too acidic for plant and animal life. Acid rain also
affects crops and buildings.
Hydrocarbons are a broad class of pollutants made up of hundreds of
specific compounds containing carbon and hydrogen. The simplest
hydrocarbon, methane, does not readily react with nitrogen oxides to
form smog, but most other hydrocarbons do. Hydrocarbons are emitted from
human-made sources such as auto and truck exhaust, evaporation of
gasoline and solvents, and petroleum refining.
The white haze that can be seen over many cities is tropospheric ozone,
or smog. This gas is not emitted directly into the air; rather, it is
formed when ozone precursors mainly nonmethane hydrocarbons and nitrogen
oxides react in the presence of heat and sunlight. Human exposure to
ozone can produce shortness of breath and, over time, permanent lung
damage. Research shows that ozone may be harmful at levels even lower
than the current federal air standard. In addition, it can reduce crop
yields.
Finally, fossil fuel use also produces particulates, including dust,
soot, smoke, and other suspended matter, which are respiratory
irritants. In addition, particulates may contribute to acid rain
formation.
Also, water and land pollution.
</span>
A.Force because your adding pressure to what you are pushing or pulling.
Answer:
Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)
Explanation:
<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.
Calculating Percent Composition of NaHCO₃:
1: Calculating Molar Masses of all elements present in NaHCO₃:
a) Na = 22.99 g/mol
b) H = 1.01 g/mol
c) C = 12.01 g/mol
d) O₃ = 16.0 × 3 = 48 g/mol
2: Calculating Molecular Mass of NaHCO₃:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O₃ = 48 g/mol
----------------------------------
Total 84.01 g/mol
3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:
For Na:
= 22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 27.36 %
For H:
= 1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 1.20 %
For C:
= 12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 14.29 % ≈ 14.30 %
For O:
= 48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 57.13 % ≈ 57.14 %
For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:
ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3
Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol
m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg
For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf
Tf = -1.59 celsius
You would need 8.1 <span>moles of LiOH</span>
