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Rufina [12.5K]
3 years ago
13

Is groundwater contaminated by sewage from a ruptured septic tank can sometimes be naturally purified by flowing for a relativel

y short distance through a
_ _ __. WHATS THE ANSWER I NEED HELP!!
Chemistry
2 answers:
ANEK [815]3 years ago
8 0

Answer:

Sandstone aquifer

Explanation:

Groundwater is essentially water which is present under the earth's surface and stored in structures called aquifers.

Aquifers are structures that are made of unconsolidated materials (like sand, gravel, slit) or rocks which have the appropriate porosity and permeability for water to flow through. Sandstone aquifers are the most widespread and have the appropriate pore size to trap bacteria in contaminated groundwater. Hence, sewage from a ruptured septic tank can sometimes be naturally purified by flowing for a relatively short distance through a sandstone aquifer.

lora16 [44]3 years ago
6 0
<h2>Answer:</h2>

An Aquifer

<h2>Explanation:</h2>

An aquifer is an underground layer of water-bearing permeable rock or unconsolidated materials like gravel, sand, or silt. The most common aquifers are unconsolidated sands and gravels, permeable sedimentary rocks such as sandstones or limestones and heavily fractured volcanic and crystalline rocks. All of these aquifers can be used for the purification of groundwater contaminated by sewage from a ruptured septic tank.

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Answer:

C_2H_6O

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114.6~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=2.6~mol~of~CO_2

70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O

Now, in 1 mol of CO2 we have 1  mol of C and in 1 mol of H_2O we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to <u>calculate the grams of C and O</u> and then do the <u>substraction</u> form the initial amount, so:

2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C

3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H

Total~grams=~31.25~+~7.82=39.08~g

grams~of~O=60.00~g-~39.08~g=20.92~g~of~O

Now we can <u>convert the grams</u> of O to moles, so:

20.92~g~of~O\frac{1~mol~O}{16~g~O}=1.30~mol~O

The next step is to divide all the mol values by the <u>smallest one</u>:

O=\frac{1.30~mol~O}{1.30~mol~O}=~1

C=\frac{2.6~mol~C}{1.30~mol~O}=~2

H=\frac{7.82~mol~H}{1.30~mol~O}=6

Therefore the formula is C_2H_6O

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