I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research. "Test a third group that does not receive the supplement" would be the procedure that would have made the <span>outcome of this study more valid. I hope the answer helps you.</span>
2PbO(s) + O₂(g) ⇄ 2PbO₂(s)
Then Δngas = -1
<h3>
What is Δngas?</h3>
The number of moles of gas that move from the reactant side to the product side is denoted by the symbol ∆n or delta n in this equation.
Once more, n represents the growth in the number of gaseous molecules the equilibrium equation can represent. When there are exactly the same number of gaseous molecules in the system, n = 0, Kp = Kc, and both equilibrium constants are dimensionless.
<h3>
Definition of equilibrium</h3>
When a chemical reaction does not completely transform all reactants into products, equilibrium occurs. Many chemical processes eventually reach a state of balance or dynamic equilibrium where both reactants and products are present.
Learn more about Equilibrium
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Answer: E
How much NH₃ can be produced from the reaction below:
N₂ + 3H₂ - 2NH₃
The stoichiometric ratio of the reactants = 1:3
Given
74.2g of N₂, and Molar mass = 14g/mole
Mole of N₂ = 74.2/14=5.3mols of N₂,
and 14mols of H₂
From this given values and comparing with the stoichiometric ratio, H₂ will be the limiting reagent while N₂ is the excess reactant.
i.e, for every 14mols of H₂, we need 4.67mols of N₂ to react with it to produce 9.33mols of NH₃ as shown (vice versa)
From this we have 9.33mols of NH₃ produced
Avogadro constant, we have n = no of particles = 6.022x10²³ molecules contained in every mole of an element.
For a 9.33mols of NH3, we have 9.33x6.022x10²³molecules in NH3
5.62x10²⁴molecules of NH₃
Answer:
The valid quantum numbers are l=0, l=-2 and l= 2.
Explanation:
Given that,
n = 3 electron shell
Suppose, the valid quantum numbers are
l = 3
m = 3
l = 0
m = –2
l = –1
m = 2
We know that,
The value of n = 3
Principle quantum number :
Then the principal quantum number is 3. Which is shows the M shell.
So, n = 3
Azimuthal quantum number :
The azimuthal quantum number is l.
Magnetic quantum number :
The magnetic quantum number is
Hence, The valid quantum numbers are l=0, l=-2 and l= 2.