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:
1.67g/cm3
Explanation:
The formula for density is 
. The m variable stands for mass and the v variable stands for volume.
The mass of the brown sugar is 10.0g and the volume is 6.0cm3, so we can plug those values into the equation.
Rounded to 3 significant figures, the density of the block of brown sugar is 1.67 g/cm3. If the mass is in grams and the volume is in cm3, the unit for the final answer is (grams per centimetres cubed).
The orbital hybridization of the central carbon atom in CSe2 is sp.
In chemical bonding, atomic orbitals may be combined to form appropriate hybrid orbitals suitable for bonding. The orbitals that combine during hybridization must be close enough in energy.
In the compound Cse2, carbon is the central atom bonded to two selenium atoms. The carbon atom in CSe2 is sp hybridized.
Learn more about orbital hybridization: brainly.com/question/1869903
Answer:
25.8
Explanation:
Let's write the reaction between magnesium-phosphide and potassium:
Mg3P2 + K = Mg + K3P
And now let's balance this equation:
Mg3P2+6K=3Mg+2K3P
We see that the ratio of magnesium-phosphide and potassium is 1:6, which means that for every mole of magnesium-phosphide there need to be 6 moles of potassium.
Since we have 4.3 moles of Mg3P2, there need to be 6 • 4.3 = 25.8 moles of potassium.
