All categories

3 years ago

When fracking liquid waste is left in pools on the surface, _ can evaporate into the air and contribute to pollution

2 answers:

8 0

When fracking liquid waste is left in pools on the surface, Toxic Waste or Toxic Chemicals can evaporate into the air and contribute to pollution. A new study shows that these spills have left surface waters in the area carrying radium, selenium, thallium, lead, and other toxic chemicals that can continue for years at unsafe levels.

maks197457 [2]3 years ago

5 0

Answer:

The answer is "toxic chemicals".

Explanation:

Hydraulic fracture uses compounds such as sand to prevent fractures from closing, chemical additives, and several hundred chemical products, some very toxic and carcinogenic, that perform the function of preventing gas and oil from becoming contaminated. They are also used to prevent corrosion. By depositing fracking liquid waste in the pools, the most volatile products tend to evaporate contributing to air pollution. Some of the elements contained in this wastewater are selenium, radio, and lead.

Have a nice day!

You might be interested in

Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

Marizza181 [45]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

We are given:

$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = $\frac{173.008}{1}\times 2.25=432.52J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

3 0

3 years ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

3 years ago

Rutherford proposed that atoms have positive nuclei. Which statement correctly describes the nucleus of an atom?

DIA [1.3K]

I'm not 100% sure but I'm leaning towards D. :)

8 0

3 years ago

Read 2 more answers

Classify methanal according to the position of the c=o group

seraphim [82]

Answer : Methanal also known as Formaldehyde $CH_{2}O$ is a chemical Aldehyde which contain ( -CHO) group.

Explanation :

In organic chemistry, a carbonyl group is a functional group which contain a carbon atom double-bonded to an oxygen atom i.e, ( C=O).

If carbonyl group is present in a compound then it can be a carboxylic (RCOOH), aldehyde (RCHO), ketone (RCOR'), ester ((RCOOR') or amide (RCONR'R") group.

Here are some functional groups naming according to the IUPAC rules and image also attached,

Carboxylic acid → (RCOOH) → ( name end in 'OIC ACID' )

Aldehyde → (RCOH) → ( name end in 'AL' )

Ketone → (RCOR') → ( name end in 'ONE' )

Ester → (RCOOR') → ( name end in 'ATE' )

Amide → (RCONR'R") → ( name end in 'AMIDE' )

In an aldehyde, atleast one hydrogen atom must be attached to the carbonyl carbon. For an aldehyde, remove ( -e) from alkane name and add ( -al) at the end of the compound.

Methanal is the IUPAC name for Formaldehyde.

3 0

3 years ago

Select The on that most applys Will mark brainliest

sdas [7]

Answer:

A and C

Explanation:

7 0

3 years ago