Answer:
Partial pressure N₂ . (Partial pressure H₂O)² / (Partial pressure H₂)² . (Partial pressure NO)² = Kp
Explanation:
The reaction is:
2NO + 2H₂ → N₂ + 2H₂O
The expression for Kp (pressure equilibrium constant) would be:
Partial pressure N₂ . (Partial pressure H₂O)² / (Partial pressure H₂)² . (Partial pressure NO)²
There is another expression for Kp, where you work with Kc (equilibrium constant)
Kp = Kc (R.T)^Δn
where R is the Ideal Gases constant
T° is absolute temperature
Δn = moles of gases formed - moles of gases, I had initially
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
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!
What's wrong with this setup is the substrate on which you have positioned
the drop is "dirty and unclean" meaning it is not being dampened by
the solution. This action can be corrected by comprehensively cleaning the
substrate where the drop will be positioned.
Answer:
The ground state configuration is the lowest energy, most stable arrangement. An excited state configuration is a higher energy arrangement (it requires energy input to create an excited state). Valence electrons are the electrons utilised for bonding.
or the
FIGURE 5.9 The arrow shows a second way of remembering the order in which sublevels fill. Table 5.2 shows the electron configurations of the elements with atomic numbers 1 through 18.
Element Atomic number Electron configuration
sulfur 16 1s22s22p63s23p4
chlorine 17 1s22s22p63s23p5
argon 18 1s22s22p63s23p6
or the
Two electrons
Two electrons fill the 1s orbital, and the third electron then fills the 2s orbital. Its electron configuration is 1s22s1.
Explanation:
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