Hello!
To find electron configuration for Idoine we need to understand the following steps:
- Finding the Atom's Atomic Number (tells us the specific number of electrons)
- Determining the Charge of the Atom
- Understanding the orbitals (Set S [Contains 2], P [Contains 3, Holds 6], D [Contains 5, Holds 10], F [Contains 7, Holds 14], and there are some theoritical ones.) [Overall the sets go SPDFGHIK
- Understanding notations in configuartion. The notations display the number of electrons in the atom and set.
In this case, for Iodine. If we follow these rules we can see that the electron configuration is [Kr] 4d^10 5s^2 5p^5. We use Krytpon in front because that is the last full and stable noble gas before this particular element. Atoms are just trying to be stable so the goal is to achieve that full shell.
Number of moles = volume / (molar volume)
Molar volume at stp = 22.4 dm^3
Volume = no of moles × molar volume
= 0.987 × 22.4
= 22.1088 dm^3
= 22108.8 cm^3
Hope it helped!
Hence, concentration of base is 1.17 M
Answer:
1. B
2. D. the form of a substance changes but not its identity
3. C
4. D
∆H ° rxn =-2855.56 kJ
<h3>Further explanation</h3>
Given
ΔHf CO₂ = -393.5 kJ/mol
ΔHf H₂O = -241.82 kJ/mol
ΔHf C₂H₆ = - 84.68 kJ/mol
Reaction
2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(g)
Required
ΔHrxn=
Solution
<em>∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants) </em>
∆H ° rxn = (4.-393.5+6.-241.82)-(2.-84.68)
∆H ° rxn = (-1574-1450.92)-(-169.36)
∆H ° rxn =-3024.92+169.36
∆H ° rxn =-2855.56 kJ