Answer:
a reference book about weather
Explanation:
Answer:

Explanation:
Data:
50/50 ethylene glycol (EG):water
V = 4.70 gal
ρ(EG) = 1.11 g/mL
ρ(water) = 0.988 g/mL
Calculations:
The formula for the boiling point elevation ΔTb is

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.
1. Moles of EG

2. Kilograms of water

3. Molal concentration of EG

4. Increase in boiling point

5. Boiling point
Answer:
Check the explanation
Explanation:
When talking about our universe there are 5 d orbitals. The element of first transition series moves away from the universal principles of Hund's rule and Aufbav's principle. So in order to attain stability these elements tend to form half or full filled orbitals.
In our universe the ground state electronic configuration of sixth transition metal, Iron (Fe) : [Ar] 
and the electronic configuration of seventh transition metal, Cobalt (Co) : [Ar] 
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In universe L there are seven orbitals.
Ground state electronic configuration of sixth and seven transition element.
Sixth transition metal: [Ar] ![3d^{7} 4s^1 or [X] 3d^{7} 4s^1](https://tex.z-dn.net/?f=3d%5E%7B7%7D%204s%5E1%20or%20%5BX%5D%203d%5E%7B7%7D%204s%5E1)
Seventh transition metal: [Ar] ![3d^{7} 4s^{2}or [X] 3d^{7} 4s^{2}](https://tex.z-dn.net/?f=3d%5E%7B7%7D%204s%5E%7B2%7Dor%20%5BX%5D%203d%5E%7B7%7D%204s%5E%7B2%7D)
heterogeneous mixture, because, the layers are called phases and a heterogeneous mixture has two or more phases. the oil phase is less dense then water so it floats on top, etc.