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ANTONII [103]
2 years ago
5

Which of the following air pollutants causes a reddish-brown haze in the sky? Sulfur dioxide Carbon monoxide Nitrogen dioxide Ca

rbon dioxide
Chemistry
2 answers:
Zanzabum2 years ago
8 0

The answer would most likely he sulfur dioxide

love history [14]2 years ago
3 0

Answer:

The answer is nitrogen dioxide

Explanation:

The term "photochemical smog" refers to a mist formed by a mixture of pollutants (NOx and volatile hydrocarbons) with others such as ozone, peroxyacyl, hydroxyl radicals, etc., which are formed by the reactions that occur with the light of the sun to affect the NO. This mixture dyes the atmosphere of a reddish brown color, full of very dangerous components for living beings.

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How many moles of hydrogen gas will be produced if 5.00 moles of zinc reactions with an excess amount of sulfuric acid
harkovskaia [24]

Answer:

Moles of Hydrogen produced is 5 moles

Explanation:

The balanced Chemical equation for reaction between zinc and sulfuric acid is :

Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)

This equation tells that ; when  1 mole of Zn react with 1 mole of sulfuric acid, it produces 1 mole of zinc sulfate and 1 mole of hydrogen.

Since sulfuric acid is in excess so Zinc is the limiting reagent

(Limiting reagent : Substance which get consumed when the reaction completes, limiting reagent helps in predicting the amount of products formed)

Limiting reagent (Zn) will decide the amount of Hydrogen produced

Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)

1\ mole\ zinc\rightarrow 1\ mole\ H_{2}

So,

5\ mole\ zinc\rightarrow 5\ mole\ H_{2}

Hence moles of Hydrogen produced is 5 moles

3 0
3 years ago
What are bacteria? Chapter 7 lesson 2 AYU
Maru [420]

Answer:

a group of unicellular microorganisms

7 0
3 years ago
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Artemon [7]
The answer to this problem is 11.6m
3 0
3 years ago
The compound dioxane, which is used as a solvent in various industrial processes is composed of C,H, and O atoms. Combustion of
Lady_Fox [76]

Answer:

The correct formula of dioxane is C₄H₈O₂.

Explanation:

Given data:

mass of dioxane= 2.23 g

mass of water = 1.802 g

mass of carbon dioxide = 4.401 g

molar mass of dioxane = 88.1 g / mol

Molecular formula of dioxane = ?

Solution:

percentage of carbon = (4.401 g/2.23 g ) × (12 /44) × 100

                                     =  (1.98 × 0.273) × 100 = 54.1

percentage of hydrogen =  (1.802 g/ 2.23 g) × (2.016 /18.016) × 100

                                         = (0.81 × 0.112) × 100 = 9.072

percentage of oxygen = 100 - (54.1 + 9.072)

                                     = 100 - 63.172 = 36.828

Now we will determine the number of grams atoms of carbon, hydrogen and oxygen.

No. of gram atoms of carbon = 54.1 /12 = 4.51

No. of gram atoms of hydrogen = 9.072 / 1.008 = 9

No. of gram atoms of oxygen = 36.828 / 16 = 2.302

Atomic ratio:

    C :H :O               4.51/ 2.302    :   9 / 2.302   :  2.302 /2.302

    C :H :O                2 : 4 : 1

Molecular formula:

   Molecular formula = n × (empirical formula)

   n = molar mass of compound / empirical formula mass

   empirical formula mass= 2 × 12 + 4 × 1.008  + 1 × 16

    empirical formula mass= 24+ 4.032 +16 = 44.032

                 n = 88.1 / 44.032 = 2

        Molecular formula = n × (empirical formula)

        Molecular formula = 2 × (C₂H₄O)

         Molecular formula = C₄H₈O₂

8 0
3 years ago
The vapor pressure of pure water at 25 °c is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by
maxonik [38]

The vapour pressure of the solution is 23.4 torr.

Use <em>Raoult’s Law</em> to calculate the vapour pressure:  

<em>p</em>₁ = χ₁<em>p</em>₁°  

where  

χ₁ = the mole fraction of the solvent  

<em>p</em>₁ and <em>p</em>₁° are the vapour pressures of the solution and of the pure solvent  

The formula for vapour pressure lowering Δ<em>p</em> is  

Δ<em>p</em> = <em>p</em>₁° - <em>p</em>₁  

Δ<em>p</em> = <em>p</em>₁° - χ₁<em>p</em>₁° = p₁°(1 – χ₁) = χ₂<em>p</em>₁°  

where χ₂ is the mole fraction of the solute.  

<em>Step 1</em>. Calculate the <em>mole fraction of glucose </em>

<em>n</em>₂ = 18.0 g glu × (1 moL glu/180.0 g glu) = 0.1000 mol glu  

<em>n</em>₁ = 95.0 g H_2O × (1 mol H_2O/18.02 g H_2O) = 5.272 mol H_2O  

χ₂ = <em>n</em>₂/(<em>n</em>₁ + n₂) = 0.1000/(0.1000 + 5.272) = 0.1000/5.372 = 0.018 62  

<em>Step 2</em>. Calculate the <em>vapour pressure lowering</em>  

Δ<em>p</em> = χ₂<em>p</em>₁° = 0.018 62 × 23.8 torr = 0.4430 torr  

<em>Step 3</em>. Calculate the <em>vapour pressure</em>  

<em>p₁</em> = <em>p</em>₁° - Δ<em>p</em> = 23.8 torr – 0.4430 torr = 23.4 torr

3 0
3 years ago
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