<u>Answer:</u> The electronic configuration of the elements are written below.
<u>Explanation:</u>
Electronic configuration is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom is determined by the atomic number of that atom.
For the given options:
- <u>Option a:</u> Carbon (C)
Carbon is the 6th element of the periodic table. The number of electrons in carbon atom are 6.
The electronic configuration of carbon is 
- <u>Option b:</u> Phosphorus (P)
Phosphorus is the 15th element of the periodic table. The number of electrons in phosphorus atom are 15.
The electronic configuration of phosphorus is 
- <u>Option c:</u> Vanadium (V)
Vanadium is the 23rd element of the periodic table. The number of electrons in vanadium atom are 23.
The electronic configuration of vanadium is 
- <u>Option d:</u> Antimony (Sb)
Antimony is the 51st element of the periodic table. The number of electrons in antimony atom are 51.
The electronic configuration of antimony is 
- <u>Option e:</u> Samarium (Sm)
Samarium is the 62nd element of the periodic table. The number of electrons in samarium atom are 62.
The electronic configuration of samarium is 
Hence, the electronic configuration of the elements are written above.
protons and electrons are both always the atomic number which is 9 in this case.
For neutrons you subtract the atomic number (9) from the weight of the atom (18.998) some teachers will want you to round to the nearest whole (19). We do this because the number of protons is the atomic number so if you subtract the protons from the whole weight of the atom you would have the electrons and neutrons left. Since electrons weigh so little we don't have to subtract them. Weighing neutrons and electrons would be like weighing an elephant (neutrons) and then putting one marshmallow on the scale (electron).
The balanced equation for the reaction is as follows;
2H₂S + SO₂ —> 2H₂O + 3S
Stoichiometry of H₂S to SO₂ is 2:1
Limiting reactant is fully used up in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of H₂S moles - 8.0 g / 34 g/mol = 0.24 mol of H₂S
Number of SO₂ moles = 12.0 g / 64 g/mol = 0.188 mol of SO₂
According to molar ratio of 2:1
If we assume H₂S to be the limiting reactant
2 mol of H₂S reacts with 1 mol of SO₂
Therefore 0.24 mol of H₂S requires - 1/2 x 0.24 = 0.12 mol of SO₂
But 0.188 mol of SO₂ is present therefore SO₂ is in excess and H₂S is the limiting reactant.
H₂S is the limiting reactant
Amount of S produced depends on amount of H₂S present
Stoichiometry of H₂S to S is 2:3
2 mol of H₂S forms 3 mol of S
Therefore 0.24 mol of H₂S forms - 3/2 x 0.24 mol = 0.36 mol of S
Mass of S produced = 0.36 mol x 32 g/mol = 11.5 g of S is produced
Answer:
0.2402 M
Explanation:
0.945 L sol contains = 0.227 moles
1 L sol will contain = (0.227/0.945)*1 = 0.2402 moles
Molarity = 0.2402 moles/L or M
You can find this on google easily