Answer:
Explanation:
The air 9% mole% methane have an average molecular weight of:
9%×16,04g/mol + 91%×29g/mol = 27,8g/mol
And a flow of 700000g/h÷27,8g/mol = 25180 mol/h
In the reactor where methane solution and air are mixed:
In = Out
Air balance:
91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)
Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>
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The air in the product gas is
95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>
43058 mol air×29g/mol <em>1249 kg air</em>
Percent of oxygen is:
=<em>0,231 kg O₂/ kg air</em>
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I hope it helps!
Answer:
Increasing the surface area of a reactant increases the frequency of collisions and increases the reaction rate. Several smaller particles have more surface area than one large particle. The more surface area that is available for particles to collide, the faster the reaction will occur.
Explanation:
:)
Answer:
False
Explanation:
Few organisms were not preserved as well as others, and new fossils are found every day