Answer:
First oil well in the United States, built in 1859 by Edwin L. Drake, Titusville, Pennsylvania.
Answer:
b. precipitate is your answer
Answer:
-169°C to -104°C
Explanation:
Ethene, also known as ethylene exists in solid, liquid and gaseous states. Ethene is an aliens with condensed structural formula C2H4. Athens is a colourless gas. It is flammable and is also a sweet smelling gas in its pure form. It is the monomer in the production of polyethylene which is of great importance in the plastic industry. In agriculture, it is used to induce the ripening of fruits. It can be hydrated in order to produce ethanol.
The liquid range of ethene refers to the temperatures at which ethene is found in the liquid state of matter. It is actually the difference between the melting point and the boiling points of ethene. Hence the liquid range of ethene is -169°C to -104°C
Answer : The mass of sulfuric acid needed is 
.
Solution : Given,
pH = 8.94
Volume of solution = 380 ml = 
Molar mass of sulfuric acid = 98.079 g/mole
As we know,
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
![5.06=-log[OH^-]](https://tex.z-dn.net/?f=5.06%3D-log%5BOH%5E-%5D)
![[OH^-]=0.00000871=8.71\times 10^{-6}mole/L](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.00000871%3D8.71%5Ctimes%2010%5E%7B-6%7Dmole%2FL)
Now we have to calculate the moles of 
.
Formula used : 
![\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7D%5BOH%5E-%5D%3D%5Ctext%7B%20Concentration%20of%20%7D%5BOH%5E-%5D%5Ctimes%20Volume%5C%5C%5Ctext%7B%20Moles%20of%20%7D%5BOH%5E-%5D%3D%288.71%5Ctimes%2010%5E%7B-6%7Dmole%2FL%29%5Ctimes%20%28380%5Ctimes%2010%5E%7B-3%7DL%29%3D3309.8%5Ctimes%2010%5E%7B-9%7Dmoles)
For neutralization, equal number of moles of 
ions will neutralize same number of ions.
![\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7D%5BOH%5E-%5D%3D%5Ctext%7B%20Moles%20of%20%7D%5BH%5E%2B%5D%3D3309.8%5Ctimes%2010%5E%7B-9%7Dmoles)
As, 
From this reaction, we conclude that
2 moles of ion is given by the 1 mole of 
moles of ion is given by 
moles of
Now we have to calculate the mass of sulfuric acid.
Mass of sulfuric acid = Moles of × Molar mass of sulfuric acid
Mass of sulfuric acid = 
Therefore, the mass of sulfuric acid needed is .
