I believe it’s the last option
Answer:
![AU^{3+} : [Rn] 5f^3](https://tex.z-dn.net/?f=AU%5E%7B3%2B%7D%20%3A%20%5BRn%5D%205f%5E3)
Explanation:
Writing electronic configuration of any element you should know atomic number of that element ,
and also electrons are filling according to their energy level and first electron is filled in the lower energy orbital
and it follows n+1 rule if n+1 is same for two orbital electron will go first in the lowest value of n.
writing electronic configuration of ion can be done like first for their neutral atom and then add or remove electron it will make things easy because there are also some eception case their you may do wrong.
![AU : [Rn] 5f^3 6d^1 7s^2](https://tex.z-dn.net/?f=AU%20%3A%20%5BRn%5D%205f%5E3%206d%5E1%207s%5E2)
remove three electron from outer most shell of AU
Answer:
Cooling a substance causes molecules to slow down and get slightly closer together, occupying a smaller volume that results in an increase in density. Hot water is less dense and will float on room-temperature water. <u>Cold water is more dense and will sink in room-temperature water.</u>
<u></u>
The number of Ml of C₅H₈ that can be made from 366 ml C₅H₁₂ is 314.7 ml of C₅H₈
<u><em>calculation</em></u>
step 1: write the equation for formation of C₅H₈
C₅H₁₂ → C₅H₈ + 2 H₂
Step 2: find the mass of C₅H₁₂
mass = density × volume
= 0.620 g/ml × 366 ml =226.92 g
Step 3: find moles Of C₅H₁₂
moles = mass÷ molar mass
from periodic table the molar mass of C₅H₁₂ = (12 x5) +( 1 x12) = 72 g/mol
moles = 226.92 g÷ 72 g/mol =3.152 moles
Step 4: use the mole ratio to determine the moles of C₅H₈
C₅H₁₂:C₅H₈ is 1:1 from equation above
Therefore the moles of C₅H₈ is also = 3.152 moles
Step 5: find the mass of C₅H₈
mass = moles x molar mass
from periodic table the molar mass of C₅H₈ = (12 x5) +( 1 x8) = 68 g/mol
= 3.152 moles x 68 g/mol = 214.34 g
Step 6: find Ml of C₅H₈
=mass / density
= 214.34 g/0.681 g/ml = 314.7 ml
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
