Your limiting reagent is NaOH and your excess is H2SO4 with 0.025 mol in excess.
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.
<span>In chemical reaction, atoms
always are rearranged. They are never destroyed or created because it would
contradict the law of conservation of energy. The law of conservation of energy
states that a matter cannot be created nor destroyed. It can only be formed in
one way or the other. Also it cannot be neutralized. Resulting in one cannot be
part of a chemical reaction. An example of this is the reaction HCl + NaOH
-> NaCl + H2O. the reactants and products are not created, they are just
formed into another substance.</span>
Answer : The molality of solution is, 5.69 mole/L
Explanation :
The relation between the molarity, molality and the density of the solution is,
![d=M[\frac{1}{m}+\frac{M_b}{1000}]](https://tex.z-dn.net/?f=d%3DM%5B%5Cfrac%7B1%7D%7Bm%7D%2B%5Cfrac%7BM_b%7D%7B1000%7D%5D)
where,
d = density of solution = 1.25 g/mL
m = molality of solution = ?
M = molarity of solution = 4.57 M
= 98 g/mole
Now put all the given values in the above formula, we get
![1.25g/ml=4.57M[\frac{1}{m}+\frac{98g/mole}{1000}]](https://tex.z-dn.net/?f=1.25g%2Fml%3D4.57M%5B%5Cfrac%7B1%7D%7Bm%7D%2B%5Cfrac%7B98g%2Fmole%7D%7B1000%7D%5D)

Therefore, the molality of solution is, 5.69 mole/L