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andreyandreev [35.5K]
3 years ago
8

Atmospheric pressure decreases as altitude increases. in other words, there is more air pushing down on you at sea level, and th

ere is less air pressure pushing down on you when you are on a mountain. if pentane (c5h12), hexane (c6h14), and hexanol (c6h13oh) are heated evenly at different altitudes, rank them according to the order in which you would expect them to begin boiling. 1 = boils first ; 4 = boils last hexane at sea level pentane at high altitude hexanol at sea level hexane at high altitude
Physics
1 answer:
Nutka1998 [239]3 years ago
3 0
At sea level, the size amid the 2 alkanes lets for pentane to simmer at a lower temperature than hexane. Phenol has a higher boiling point due to hydrogen bonding High altitude would have the same order while low pressure only cuts the temperature at which a solvent boils. Boiling has to do with molecular size, the occurrence/nonappearance of hydrogen bonds, and other steric issues.
So the answer would be pentane high altitude, hexane high altitude, hexane sea level, hexanol sea level. In order of boil first to boil last. This is clarified because altitude has a better effect on vapor pressure (and hence boiling points) than inter-molecular forces.
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A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
jeka94

Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

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To solve this, we need to understand the principles of electromotive forces and Lenz’ Law; changing the magnetic field conditions around anything with this potential causes an induced current in the wire that resists this change. This principle is known as Lenz’ Law, and can be described using equations that are specific to certain situations. For this, we need the two that are useful here:

e = -N•dI/dt; dI = ABcos(theta)

where “e” describes the electromotive force, “N” describes the number of loops in the coil, “dI” describes the change in magnetic flux, “dt” describes the change in time, “A” describes the area vector of the coil (this points perpendicular to the loops, intersecting it in open space), “B” describes the magnetic field vector, and theta describes the angle between the area and mag vectors.

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