Chemistry, 05.10.2019 01:30 bubster5820

To be considered acid rain, precipitation would have to
-contain a lot of carbon dioxide
-have a ph of less than 5.6
-contain chlorofluorocarbons
-occur in an industrial area

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Answer from: tinydesb

-occur in an industrial area

Explanation:

Precipitation in industrial area leads to the formation of acid rain. Acid rain is made up of precipitation that contains dissolved Nitrogen and Sulfur oxides. This form of precipitation has more acidity than usual and expected. Normal rain will have an acidity of about 5.0 to 5.5 on a pH scale according to EPA. For acid rains, the pH is usually lesser and between 4.2 to 4.4.

Most industrial rains use engines that utilize fossil fuels. Combustion of fossil fuels release carbon dioxide, water, nitrogen and sulfur oxides. These sulfur and nitrogen oxides when combined with rain water forms acid rains. Ordinary carbon dioxide forms a relatively weak carbonic acid which is not so powerful for an acid rain.

Answer from: paulandigi

The correct answer is B. Have a pH of less than 5.6

Explanation:

Acid rain is the result of water in precipitation mixing with acidic components or substances such as nitrogen oxide that are the result of natural processes such as volcanic activity or human activities such as industries. This type of precipitation contains more hydrogen than normal and therefore is more acidic, in these terms acid rain is any precipitation with a pH (Hydrogen concentration) lower than 5.6 as unpolluted precipitation pH is above 5.7 which means regular precipitation is slightly acid or neutral. According to this, to be considered acid rain, precipitation would have to have a pH of less than 5.6.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

Hairy thick gross thing? idk

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To be considered acid rain, precipitation would have to
-contain a lot of carbon dioxide
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