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4 years ago

The acidity of normal rain water is due to ______. a) SO2 b) CO2 c) NO2 d) NO

Chemistry

2 answers:

Viktor [21]4 years ago

6 0

The correct answer choice is CO₂ (B)

Rainwater naturally contains acids derived from carbon dioxide in the air and reacts with water vapor to form weak acids

H₂O + CO₂ ---> H₂CO₃

<h3><em>Further Explanation</em></h3>

Rainwater can cause damage to buildings, animals or plants due to acid rain

Acid rain occurs because other gases in the air mix with toxic gases to form acids

Rainwater naturally contains acids derived from carbon dioxide in the air and reacts with water vapor to form weak acids

H₂O + CO₂ ---> H₂CO₃

However, due to exhaust gases from industry, motor vehicles, or power plants that use fossil fuels containing sulfur to produce sulfur dioxide gas which will produce sulfuric acid

Sulfur combustion from fossil fuels

S + O₂ ----> SO₂

SO₂ produced from this exhaust gas reacts with O₂ oxygen which oxidizes SO₂ to SO₃

SO₂ + O₂ ----> SO₃

SO3 formed reacts with water vapor in the air to produce sulfuric acid as acid rain

SO₃ + H₂O ---> H₂SO₄

To reduce the pollution from this fossil fuel, various studies are now carried out as substitute energy that is more environmentally friendly such as solar energy and so on

<h3><em>Learn more </em></h3>

Acid rain is a secondary pollutant

brainly.com/question/1324440

The impact of from acid rain

brainly.com/question/8208861

Air pollution

brainly.com/question/1232950

Keywords: acid rain, air pollution, SO₃, H₂SO₄, fossil fuel

Gemiola [76]4 years ago

6 0

Answer:

SO₂

Explanation:

Acid rain is due to SO₂.

SO₂ + ½ O₂ + H2O = H₂SO₄

H₂SO₄= 2H⁺ + SO₄²⁻

Harmful pollutant in the air such as sulfur dioxide creates acid rain. SO₂ reacts with water vapor and oxygen to form sulfuric acid, which makes the rain acidic. H₂SO₄ releases H⁺ which decreases the pH of rain water. At acidic pH rainwater is harmful to both living beings and architectural objects.

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The density of a sample of metal is calculated using these three different sets of data: 1.2 g/mL, 1.4 g/mL, and 1.1 g/mL. If th

Iteru [2.4K]

Accurate data means the data experimentally obtained are close to the true value. Precise data means the data obtained are close to one another. In this case, the data are close to the true value which is 1.2 and the data are relatively close to one another. Hence the set is both accurate and precise.

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3 years ago

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate t

Sedbober [7]

Answer:

Approximately $6.30\times 10^{-3}\;\rm mol$ .

Explanation:

The gallium here is likely to be produced from a $\rm NaGaO_2\, (aq)$ solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?

Note the Roman Numeral " $\mathtt{(III)}$ " next to $\rm Ga$ . This numeral indicates that the oxidation state of the gallium in this solution is equal to $+3$ . In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of $\rm Ga\, (s)$ .

As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium $\mathtt{(III)}$ solution.

How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.

$t = \rm 80.0\; min = 80.0\; min \times 60\;s \cdot min^{-1} = 4800\; s$ .

$Q = I \cdot t = \rm 0.380 \; A \times 4800 \; s = 1.824\times 10^3\; C$ .

Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.

$\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} \cr &= \rm \dfrac{1.824\times 10^3\; C}{96485.332\; C \cdot mol^{-1}}\cr &\approx \rm 1.89\times 10^{-2}\; mol \end{aligned}$ .

It takes three moles of electrons to deposit one mole of gallium atoms $\rm Ga\, (s)$ . As a result, $\rm 1.89\times 10^{-2}\; mol$ of electrons would deposit $\displaystyle \rm \frac{1}{3}\times 1.89\times 10^{-2}\; mol \approx 6.30\times 10^{-3}\; mol$ of gallium atoms $\rm Ga\, (s)$ .

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3 years ago

The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa

Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of $MgCl_2$ (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of $MgCl_2$ (solute) = 3.37 g

First we have to calculate the mass of solute.

$\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2$

$\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g$

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg$