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yan [13]
3 years ago
11

In the past mining companies were not made responsible for the clean-up of any of the mine sites. As a result, abandoned mines h

ad a severe impact on the environment in Ontario. Which of these is not a result of these mining practices?
1) Chemical reactions producing sulphuric acid decrease soil and water PH in the area.
2) Acid leaching dissolves metals found in the soil thus allowing them to enter the water system.
3) Heavy metals which are left tailing ponds can dissolve and enter the water system.
4) Old and inefficient mining smokestacks contaminate the soils around abandoned mine sites.
Chemistry
1 answer:
Andre45 [30]3 years ago
4 0

Answer:

Old and inefficient mining smokestacks contaminate the soils around abandoned mine sites.

Explanation:

A smokestack, is a very tall channel commonly used in many instances to release gases produced by combustion processes directly into the air. These high towers are aimed at dispersing the gaseous pollutants over a wider area thereby minimizing their impact.

Old and inefficient smokestack do not contaminate the soil since they are very high towers that discharge gases directly into the atmosphere. Hence they are not part of the sources of soil contamination in abandoned mines.

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A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of
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Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH}  =0.1900gO

4. Compute the moles of oxygen by using its molar mass:

n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1

6. Search for the closest whole number (in this case multiply by 2):

C_3H_6O_2

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

M=12*3+1*6+16*2=74g/mol

Which is about three times in the molecular formula, for that reason, the actual formula is:

C_9H_{18}O_6

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

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Yes; the graph shows a strong positive linear association.

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