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!
Good idea!maybe I should try that
The mass of 254 mL of water is 254 g. Since the density of water is 1g/mL, we can simply multiply the density 1g/mL by 254 mL of water and get 254 g as our answer. Since mL is in the numerator and denominator, mL cancels out and we are left with g only.
Answer:
1.72x10⁻⁵ g
Explanation:
To solve this problem we use the PV=nRT equation, where:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 25 °C ⇒ (25+273.16) = 298.16 K
And we <u>solve for n</u>:
- 1 atm * 5.7x10⁶ L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
Finally we <u>convert moles of helium to grams</u>, using its <em>molar mass</em>:
- 4.29x10⁻⁶ mol * 4 g/mol = 1.72x10⁻⁵ g
