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mamaluj [8]
3 years ago
10

Why must an air traffic controller know a plane's velocity

Chemistry
2 answers:
Kitty [74]3 years ago
6 0
So they can avoid planes meeting up at the same time
aliina [53]3 years ago
3 0

Answer:

1. In order to prevent collision

2. To be able to control free flow of air traffic

Explanation:

The reason why it's paramount for air traffic controller to know the velocity of a plane is to enable them to avoid or prevent collision of planes coming in opposite direction. Knowing the velocity of the plane by the air controller would also give the air controller a good insight of how the air traffic should flow in oder to avoid accidents and disorderliness.

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The substance that is in the smaller amount is the solute. so yes, the answer is true.
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Oil platforms may be built in the middle of the ocean. How does an oil platform work?
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If an element is a reactant or product in a chemical reaction, the reaction must be an oxidaton-reduction reaction. Why is this
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Explanation:

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How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo
riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)

The expression used will be:

\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]

where,

\Delta H = heat released by the reaction = ?

m = mass of benzene = 125 g

c_{p,g} = specific heat of gaseous benzene = 1.06J/g^oC

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC

\Delta H_{vap} = enthalpy change for vaporization = 33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]

\Delta H=-67682.5J=-67.7kJ

Therefore, the energy removed must be, -67.7 kJ

4 0
3 years ago
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