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Ludmilka [50]
3 years ago
12

"Reduce, reuse, recycle." Give examples of how one or more of

Chemistry
1 answer:
Anarel [89]3 years ago
6 0

Answer:

Examples of how the ideas can be applied to the issues and practices of hydraulic fracturing for the acquisition of shale gas are;

Reuse; The produced water in obtained from oil and gas well production are reused for fracking, drilling, and if the water is good enough, it can be used for farming

Reduce; The use of recycled brine and water in drilling and fracking process reduces the application of freshwater in the those processes and reduces pollution of natural water sources

Recycle; Recycling involves creating products from waste. In the hydraulic fracturing process approximately 13 percent of the water produced and the flowback water are recycled to be used more than once thereby reducing the net consumption of freshwater

Explanation:

In hydraulic fracturing, also known informally as fracking, is the drilling method used in oil and gas well development process that makes use of water sand and chemical injection through the well bore to open and widen cracks in the bedrock formations in the areas around the wellbore.

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10. If 3.5 kJ of energy are added to a 28.2 g sample of iron at 20°C, what
igor_vitrenko [27]

Answer:

569K

Explanation:

Q = 3.5kJ = 3500J

mass = 28.2g

∅1 = 20°C = 20 + 273 = 293K

∅2 = x

c = 0.449

Q = mc∆∅

3500 = 28.2×0.449×∆∅

3500 = 12.6618×∆∅

∆∅ = 3500/12.6618

∆∅ = 276.4220

∅2 - ∅1 = 276.4220

∅2 = 276.4220 + ∅1

∅2 = 276.4220 + 293

∅2 = 569.4220K

∅2 = 569K

3 0
3 years ago
VA 19.75-g sample was heated by 12.35 calories. The specific heat of the sample is 0.125 cal/g°C. What was the initial temperatu
SOVA2 [1]

Answer:

31.9 °C  

Explanation:

The formula for the heat q absorbed by an object is

q = mCΔT where ΔT = (T₂ - T₁)

Data:

q = 12.35 cal

m = 19.75 g

C = 0.125 cal°C⁻¹g⁻¹

T₂ = 37.0 °C

Calculations

(a) Calculate ΔT

q = mCΔT

12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT

12.35 = 2.406ΔT °C⁻¹  

ΔT  = 12.35/(2.406 °C⁻¹) = 5.13 °C

(b) Calculate T₂

ΔT = T₂ - T₁

T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C

The original temperature was 31.9 °C.

 

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