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!
A space-filling model shows the relative amount of space each atom takes up. In other words, a space-filling model can show relative sizes of atoms. However, unlike ball-and-stick or structural models, space-filling models do not show bond lengths clearly. Bonds are not really like sticks in a ball-and-stick model.
This may help you
<span>You need to use some stoichiometry here. The only way to do that is if you're working in moles. Since you're given grams of Al, you can convert that moles by dividing by the molar mass.
Then from looking at the coefficients in your equation, you can see that for however many moles of Al react, the same numbers of moles of Fe will be produced, but only half as many moles of Al2O3 will be produced.
To go back to grams, multiply the moles of each product that you get by their molar masses!</span>
98 elements are naturally forming elements.
Answer: 98
B. false is the answer of this question
