Answer:
more drag is created because the air molecules are not moving out of the way of the airplane
Explanation:
Answer : The molal freezing point depression constant of X is 
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of X liquid (solvent) = 450.0 g
Molar mass of urea = 60 g/mole
Formula used :
where,
= change in freezing point
= freezing point of solution = 
= freezing point of liquid X= 
i = Van't Hoff factor = 1 (for non-electrolyte)
= molal freezing point depression constant of X = ?
m = molality
Now put all the given values in this formula, we get
![[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}](https://tex.z-dn.net/?f=%5B0.4-%28-0.5%29%5D%5EoC%3D1%5Ctimes%20k_f%5Ctimes%20%5Cfrac%7B5.90g%5Ctimes%201000%7D%7B60g%2Fmol%5Ctimes%20450.0g%7D)
Therefore, the molal freezing point depression constant of X is
Pure substance element I think
Well they would definitely rip up the flooring and replace it
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 4p²
Explanation:
This atom will likely have 4 electron shells denotation of – 2.8.8.4
Orbitals shells show the probability, in space around the nucleus, where to find an electron. It is important to note that the 3rd shell has an additional d orbital (-in addition to s and p). However, because the d orbital has a higher energy state than the 4s and 4p orbitals, the d orbital only fills up when these latter ones are completely filled. In this case, the 4p does not completely fill (hence we don't see the d orbital in the notation).
