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3 years ago

5

Sulfur dioxide is a main component of: photochemical smog, thermal inversion, industrial smog.

2 answers:

Marrrta [24]3 years ago

4 0

Answer: industrial smog.

Explanation:

The industrial smog typically exists in the urban industrial areas where the large scale burning of the fossil fuels may lead to generation of huge amount of smoke and sulfur dioxide gas liberated into the atmosphere. These gaseous products gets mixed with the droplets of fog already present in the atmosphere. This lead to the formation of industrial smog. The industrial smog forms a thick blanket of smog or haze nearly close to the ground.

DedPeter [7]3 years ago

3 0

The correct answer is industrial smog. This type of smog exists in coal power plants which creates smoke and sulfur dioxide which may mix with fog creating a thick blanket of haze. Sulfur dioxide is one primary component of an industrial smog.

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A student reacted NiS2 (MM = 122.83 g/mol) with O2 (MM=32.00 g/mol) to make SO2 (MM=64.07 g/mol) according to this balanced equa

Ilya [14]

Answer:

Percentage yield = 0.49 × 10² %

Explanation:

Given data:

Actual yield of SO₂ = 4.309 ×10² g

Theoretical yield of SO₂ = 8.78 ×10² g

Percentage yield = ?

Solution:

Chemical equation:

2NiS₂ + 5O₂ → 2NiO + 4SO₂

Percentage yield:

Percentage yield = actual yield / theoretical yield × 100

Percentage yield = 4.309 ×10² g / 8.78 ×10² g × 100

Percentage yield = 0.491 × 100

Percentage yield = 49.1%

In scientific notation:

Percentage yield = 0.49 × 10² %

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3 years ago

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The kilogram is the si

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A piece of high-density Styrofoam measuring 24.0 cm by 36.0 cm by 5.0 cm floats when placed in a tub of water. When a 1.5 kg boo

zheka24 [161]

Explanation:

The given data is as follows.

Width of Styrofoam = 24.0 cm

Length of Styrofoam = 36.0 cm

Height of Styrofoam = 5.0 cm

Therefore, volume of the Styrofoam will be calculated as follows.

Volume = length × width × height

= (36.0 × 24.0 × 5.0) $cm^{3}$

= 4320 $cm^{3}$

or, = $4.32 \times 10^{3} cm^{3}$

As Styrofoam partially sinks at 3.0 cm and total height of Styrofoam is 5.0 cm. Hence, height of Styrofoam above the water is (5.0 - 3 cm) = 2 cm.

So, volume of water displaced is as follows.

24.0 cm × 36.0 cm × 2.0 cm

= $1.73 \times 10^{3} cm^{3}$

Hence, mass of displaced water is as follows.

mass = density × volume

= $1.00 g/cm^{3} \times 1.73 \times 10^{3} cm^{3}$

= $1.73 \times 10^{3} g$

Since, book is placed on the Styrofoam. Therefore, mass of water displaced is also equal to the following.

Mass of water displaced = mass of book + mass of Styrofoam

$1.73 \times 10^{3} g$ = 1500 g + mass of Styrofoam

(1730 - 1500) g = mass of Styrofoam

mass of Styrofoam = 230 g

Therefore, calculate the density of Styrofoam as follows.

Density = $\frac{mass}{volume}$

= $\frac{230}{4.32 \times 10^{3} cm^{3}}$

= $53.24 \times 10^{-3} g cm^{-3}$

Thus, we can conclude that the density of Styrofoam is $53.24 \times 10^{-3} g cm^{-3}$ .

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