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Dmitrij [34]
3 years ago
5

Students in a science class learn that groundwater is an important source of drinking water for many communities. They construct

a model to study how leaking chemicals from a landfill can affect groundwater.
What is the most effective way for the students to see how leaking chemicals from a landfill can enter a well?


A.Tilt the tank towards the well and drain some of the water out of the tank.

B.Drop food color into the well at the well's opening.

C.Tear some of the colored tissue paper and put the paper pieces into the well at the well's opening.

D.Pump water out of the well to see how stained water from the tissue paper moves toward the well.
Chemistry
1 answer:
sveticcg [70]3 years ago
5 0

Groundwater is also one of our most important sources of water for irrigation from septic tanks and toxic chemicals from underground storage tanks and leaky landfills ... Drinking contaminated groundwater can have serious health effects. If there is a leak, these contaminants can eventually make their way down

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_NH3 + _ 02 →_NO + _ H2O<br> Balanced
vovikov84 [41]

Answer:

4NH3+5O2=4N0+6H2O

Explanation:

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3 years ago
The mass of a single uranium atom is 4.70x10^-22 grams. How many uranium atoms would there be in 111 milligrams of uranium?
emmainna [20.7K]

2.4 x 10²² atoms

<h3>Further explanation</h3>

Atomic mass is the average atomic mass of all its isotopes  

In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu  

So the atomic mass obtained is the mass of the atom relative to the 12th carbon atom  

mass single Uranium atom=4.7 x 10⁻²² g

then for 111 mg=0.111 g

\tt \dfrac{0.111}{4.7\times 10^{-22}}=2.4\times 10^{22}

5 0
3 years ago
A liquid dissolved in a gas results in a solution that is _____.
SVETLANKA909090 [29]

The correct answer is C


6 0
3 years ago
Read 2 more answers
Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms
nirvana33 [79]

Answer:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }

With further simplification, we have:

P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]

Then, we have:

P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]

Therefore,

PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]

Using the expansion:

(1-x)^{-1} = 1 + x + x^{2} + ....

Therefore,

PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]

Thus:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]           equation (1)

Using the virial equation of state:

P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]

Thus:

PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]     equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol[/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0
3 years ago
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