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Mrrafil [7]
3 years ago
9

An archer shot a 0.06 kg arrow at a target. The arrow accelerated at 5,000 m/s/s to reach a speed of 50.0 m/s as it left the bow

. During this ACCELERATION, what was the net force on the arrow to the nearest newton? *
1 point
A. 3 N
B. 833 N
C. 300 N
D. None of the above
Chemistry
1 answer:
nordsb [41]3 years ago
8 0
<h3><u>Answer;</u></h3>

C. 300 N

<h3><u>Explanation</u>;</h3>

From Newton second's Law of motion, the resultant force is directly proportional to the rate of change in momentum.

Therefore;

F = ma ; where F is the resultant force, m is the mass and a is the acceleration.

Therefore;

F = 0.06 kg × 5,000 m/s/s

  <u>= 300 N</u>

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the equilibrium concentration of H₂(g) at 700°C =  0.00193 mol/L

0.00193 mol/L

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numbers of moles of H₂S = 0.59 moles

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2H₂S    ⇄   2H₂(g)  +  S₂(g)

The initial concentration of H₂S =

The initial concentration of H₂S =

= 0.1966 mol/L

The ICE table is shown be as :

                           2H₂S                ⇄         2H₂(g)        +        S₂(g)

Initial                    0.9166                           0                           0

Change                 -2 x                               +2 x                      + x

Equilibrium          (0.9166 - 2x)                   2x                         x

     

(since 2x < 0.1966 if solved through quadratic equation)

The equilibrium concentration for H₂(g) = 2x

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= 0.00193 mol/L

Thus, the equilibrium concentration of H₂(g) at 700°C =  0.00193 mol/L

To know more about equilibrium concentration

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