Answer: The last electron will be filled in first orbital of 3p sub-shell.
Explanation: Filling of electrons in orbitals is done by using Hund's Rule.
Hund's rule states that the electron will be singly occupied in the orbital of the sub-shell before any orbital is doubly occupied.
For filling up of the electrons in Sulfur atom having 16 electrons. First 10 electrons will completely fill according to Aufbau's Rule in 1s, 2s and 2p sub-shells and last 6 electrons are the valence electrons which will be filled in the order of 3s and then 3p.
3s sub-shell will be fully filled and the orbitals of 3p sub-shell will be first singly occupied and then pairing will take place. Hence, the last electron will be filled in the first orbital of 3p-sub-shell.
Answer:
129,600kg/day
Explanation:
The river is flowing at 30.0
1 m³ is equivalent to 1000L
flowrate of river = 30*1000 =30,000L/s
Convert L/s to litre per day by multiplying by 24*60*60
flowrate of river = 30,000 * 24*60*60 L/day
= 2,592,000,000L/day
if the river contains 50mg of salt in 1L of solution
lets find how many mg of salt Y is contained in 2,592,000,000L/day
by cross multiplying we have
Y=
Y= 129,600,000,000 mg/day
convert this value to kg/day by dividing by 1 million
Y= 129,600,000,000/1000000
Y= 129,600kg/day
D. Cycloalkene
________________________
Answer:
All the options are correct except "has 7 electrons"
Explanation:
1. 842g of NaOH will form 547.3 g of Al(OH)₃
2. The yield is 93.55%
<u>Explanation:</u>
3NaOH + Al → Al(OH)₃ + 3Na
1.
Molar mass of NaOH = 40 g/mol
Molar mass of Al = 27 g/mol
Molar mass of Al(OH)₃ = 78 g/mol
According to the balanced equation:
3 moles of NaOH requires 1 mole of Al to form 1 mole of Al(OH)₃
The ratio of NaOH : Al : Al(OH)₃ = 3 : 1 : 1
3 X 40 g of NaOH reacts with 27 g of Al to form 78 g of Al(OH)₃
120 g of NaOH + 27g of Al → 78 g of Al(OH)₃
120g of NaOH form 78g of Al(OH)₃
1g of NaOH will form 
g of Al(OH)₃
842g of NaOH will form 
of Al(OH)₃
= 547.3 g of Al(OH)₃
Therefore, 842g of NaOH will form 547.3 g of Al(OH)₃
2. Only 512 g of Al(OH)₃ is formed
Yield % = ?
Therefore, the yield is 93.55%
