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Lelu [443]
3 years ago
7

HURRY PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Chemistry
2 answers:
aivan3 [116]3 years ago
8 0

Answer:


When <em>a scientist on Earth drops a hammer and a feather at the same time an astronaut on the moon drops a hammer and a feather, the result</em>  expected is that <em>the hammer hits the ground before the feather on Earth, and the hammer and feather hit at the same time on the moon (option D).</em>


Explanation:


In the abscence of atmosphere (vacuum), the objects fall in free fall. This is, the only force acting on the objects is the gravitational pull, which is directed vertlcally downward.


Under such absecence of air, the equations that rules the motion are:


  • V = Vo + gt
  • d = Vo + gt² / 2
  • Vf² = Vo² + 2gd

As you see, all those equations are independent of the mass and shape of the object. This explains why <em>when an astronaut on the moon drops a hammer and a feather at the same time</em>, <em>the hammer and feather hit at the same time on the moon</em>, a space body where the gravitational attraction is so small (approximately 1/6 of the gravitational acceleration on Earth) that does not retain atmosphere.


On the other hand, the air (atmosphere) present in Earth  will exert a considerable drag force on the feather (given its shape and small mass), slowing it down, whereas, the effect of the air on the hammer is almost neglectable. In general and as an approximation, the motion of the heavy bodies that fall near the surface is ruled by the free fall equations shown above, so, <em>the result </em>that is<em> expected  when a scientist on Earth drops a hammer and a feather at the same time is that the hammer hits the ground before the feather</em>.

Ludmilka [50]3 years ago
3 0

Answer

Correct Answer is D(D. The hammer hits the ground before the feather on Earth, and the hammer and feather hit at the same time on the moon.)

Explanation:

We know that, if an object drop from certain height with initial velocity u and it hits the ground with velocity v then time taken by the body to touch the ground will b given by following equation

t=\frac{(v-u)}{g}.........(1)

Here, g is gravitational acceleration

From equation(1), it is clear that time taken by the falling body to touch the ground does not depend on its mass but it only depends on initial velocity, final velocity and gravitational acceleration.

When a feather and hammer are dropped at the same time on earth  then hammer will touch the ground first because the feather is very light that, air pressure will act on it uniformly and it will counteract the acceleration due to gravity.

On moon there is no air so no pressure will be there on feather so both feather and hammer will hit at the same time on the moon.

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Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for t
ratelena [41]

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\Delta G^0 _{rxn} = 207.6\ kJ/mol

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Explanation:

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2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}

temperature = 298.15K

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\Delta G^0 _{rxn} = 2 \times \Delta G_f^0  \ N_2O_{(g)} - ( 2 \times  \Delta G_f^0  \ N_2{(g)} +   \Delta G_f^0  \ O_{2(g)})

\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times  0 +   0)

\Delta G^0 _{rxn} = 207.6\ kJ/mol

The equilibrium constant determined from the partial pressure denoted as K_p can be expressed as :

K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}

K_p = \dfrac{1}{ (22.20)}

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\Delta G = \Delta G^0 _{rxn} + RT \ lnK

where;

R = gas constant = 8.314 × 10⁻³ kJ

\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15  \ ln(0.045)

\Delta G =207.6 + 2.4788191 \times \ ln(0.045)

\Delta G =207.6+ (-7.687048037)

\Delta G = 199.912952  kJ

ΔG ≅ 199.91 kJ

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