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SCORPION-xisa [38]
3 years ago
15

Sulfuric acid is a component of acid rain formed when gaseous sulfur dioxide pollutant reacts with gaseous oxygen and liquid wat

er to form aqueous sulfuric acid. part a write a balanced chemical equation for this reaction. express your answer as a chemical equation. identify all of the phases in your answer.
Chemistry
2 answers:
Vesna [10]3 years ago
6 0
2SO₂(g) + O₂(g) + 2H₂O(aq) → 2H₂SO₄(aq)
Igoryamba3 years ago
6 0

Answer:

The chemical reaction is :

2SO_2(g)+O_2(g)+2H_2O(l)\rightarrow 2H_2SO_4(aq)

Explanation:

When sulfur dioxide gas reacts with oxygen gas and liquid water it gives aqueous solution of sulfuric acid.

This is reason behind the acidic nature of the rain water during acid rain.The chemical reaction is given as:

2SO_2(g)+O_2(g)+2H_2O(l)\rightarrow 2H_2SO_4(aq)

According to stoichiometry, 2 moles of sulfur dioxide gas reacts with 1 mole of oxygen gas and 2 moles of water to give 2 moles of aqueous sulfuric acid.

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4 0
2 years ago
A certain mass of nitrogen gas occupies a volume of 7.56 L at a pressure of 5.08 atm. At what pressure will the volume of this s
faust18 [17]

Answer:

P₂ = 3.86 atm .

Explanation:

We assume that during this change , the temperature of the gas remains constant .

So the gas will obey Boyle's law .

P₁ V₁ = P₂V₂

5.08 x 7.56 = P₂ x 9.94

P₂ = 3.86 atm .

8 0
3 years ago
A student heats a sample of hydrate once, and the mass of the sample and the evaporating dish is 16.428 g. After a second heatin
jolli1 [7]

Answer:

12.371 g

Explanation:

Given :

m_{evaporating\ dish}=1.135\ g

m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g

m_{evaporating\ dish}+m_{First\ heated\ sample}=16.428\ g

m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g

Mass of salt hydrate:

m_{evaporating\ dish}=1.135\ g

m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g

m_{Hydrate\ sample}=25.637-m_{evaporating\ dish}\ g=25.637-1.135\ g=24.502\ g

Mass of salt anhydrous:

m_{evaporating\ dish}=1.135\ g

m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g

m_{Second\ heated\ sample}=m_{salt\ anhydrous}=13.266-m_{evaporating\ dish}\ g=13.266-1.135\ g=12.131\ g

Mass of water:

m_{water}=m_{Hydrate\ sample}-m_{salt\ anhydrous}=24.502-12.131\ g=12.371\ g

m_{water}=12.371\ g

4 0
3 years ago
Boiling point of a solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water. kB= 0.512 c/m.
Nookie1986 [14]

100.133 degree celsius is the boiling point of the solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water.

Explanation:

Balanced eaquation for the reaction

CaCl2 + 2H20 ⇒ Ca(OH)2 + HCl

given:

mass of CaCl2 = 15.2 grams

mass of the solution = 57 grams

Kb (molal elevation constant) = 0.512 c/m

i = vont hoff factor is 1 as 1 mole of the substance is given as product.

Molality is calculated as:

molality = \frac{grams of solute}{grams of solution}

              = \frac{15.2}{57}

               = 0.26 M

Boiling point is calculated as:

ΔT = i x Kb x M

     = 1 x 0.512 x 0.26

      = 0.133 degrees

The boiling point of the solution will be:

100 degrees + 0.133 degrees (100 degrees is the boiling point of water)

= 100.133 degree celcius is the boiling point of mixture formed.

4 0
3 years ago
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