Answer:
4 orbits in the fourth period.
19 electrons in the atom from group 1 and fourth period.
Explanation:
Potassium has 19 electrons distributed in its 4 orbits
The balanced chemical equation is given as:
2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)
We are given the yield of CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.
Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3
% yield = .775 = actual yield / 264.66
actual yield = 205.11 g CH3CH2OCH2CH3
First, we need to calculate the principal quantum number n for this electron, using the equation:
E = (-13.60 eV) / (n x n)
where E is the energy that is used to bound the electron (here, E = - 0.544 eV).
- 0.544 eV = (-13.60 eV) / (n x n)
n x n = (- 13.60 eV) / (- 0.544 eV)
n x n = 25
n = 5
The orbital radius that is equal to the radius of a hydrogen atom is calculated using the equation:
r = 0.053 nm x n x n
r = 0.053 nm x 5 x 5
r = 0.053 nm x 25
r = 1.325 nm
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
