Answer:
In the n = 3 energy level
Explanation:
There's is no further explanation for this.
All the electrons in an energy level are distribuited according to the period in the periodic table they are.
So, if we have an atom in period 1, like Hydrogen (H), that atom would only have 1 level energy (n = 1) and in that level, we only have the sub level 1s.
Electrons in the 3d sublevel, are found mostly in all the transition metals of period 3, and it can go from 1 to 10 electrons. To be with the 3d sub level it's neccesary that the energy level to be 3.
energy levels beyond that, like n = 4, we have electrons occupying the 3d sub level, so, primordly, the 3d is found only in energy level 3.
Hope this helps
Answer:
The minimum rate of fresh air in the room is 176 moles/min
Explanation:
High exposure of CO₂ has health effects as headaches, increased heart rate, elevated blood pressure, coma, asphyxia, convulsions, etc.
0,500 mole% of CO₂ in air means 0,500 moles of CO₂ per 100 moles of air
As the rate of sublimation of CO₂ is 0,880, the minimum rate of fresh air in the room must be:
X = <em>176 moles of Air/min</em>
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I hope it helps!
<h3><u>Answer;</u></h3>
Empirical formula = C₂H₃O
Molecular formula = C₁₄H₂₁O₇
<h3><u>Explanation</u>;</h3>
Empirical formula
Moles of;
Carbon = 55.8 /12 = 4.65 moles
Hydrogen = 7.04/ 1 = 7.04 moles
Oxygen = 37.16/ 16 = 2.3225 moles
We then get the mole ratio;
4.65/2.3225 = 2.0
7.04/2.3225 = 3.0
2.3225/2.3225 = 1.0
Therefore;
The empirical formula = <u>C₂H₃O</u>
Molecular formula;
(C2H3O)n = 301.35 g
(12 ×2 + 3× 1 + 16×1)n = 301.35
43n = 301.35
n = 7
Therefore;
Molecular formula = (C2H3O)7
<u> = C₁₄H₂₁O₇</u>
We convert the masses of our reactants to moles and use the stoichiometric coefficients to determine which one of our reactants will be limiting.
Dividing the mass of each reactant by its molar mass:
(10 g C2H6)(30.069 g/mol) = 0.3326 mol C2H6
(10 g O2)(31.999 g/mol) = 0.3125 mol O2.
Every 2 moles of C2H6 react with 7 moles of O2. So the number of moles of O2 needed to react completely with 0.3326 mol C2H6 would be (0.3326)(7/2) = 1.164 mol O2. That is far more than the number of moles of O2 that we are given: 0.3125 moles. Thus, O2 is our limiting reactant.
Since O2 is the limiting reactant, its quantity will determine how much of each product is formed. We are asked to find the number of grams (the mass) of H2O produced. The molar ratio between H2O and O2 per the balanced equation is 6:7. That is, for every 6 moles of H2O that is produced, 7 moles of O2 is used up (intuitively, then, the number of moles of H2O produced should be less than the number of moles of O2 consumed).
So, the number of moles of H2O produced would be (0.3125 mol O2)(6 mol H2O/7 mol O2) = 0.2679 mol H2O. We multiply by the molar mass of H2O to convert moles to mass: (0.2679 mol H2O)(18.0153 g/mol) = 4.826 g H2O.
Given 10 grams of C2H6 and 10 grams of O2, 4.826 g of H2O are produced.
