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Pani-rosa [81]
3 years ago
6

The second-order decomposition of hi has a rate constant of 1.80 x 10−3 m−1 s−1. How much hi remains after 45.6 s if the initial

concentration of hi is 3.81 m?
Physics
1 answer:
zlopas [31]3 years ago
6 0

Answer:

2.9 M

Explanation:

The concentration-time equation for a second order reaction is:

1/[A] = kt + 1/[A°]

Where,

A = concentration remaining at time, t

A° = initial concentration

k = rate constant

1/[A] = (1.80 x 10^-3) * (45.6) + 1/3.81

1/[A] = 0.345

= 1/0.345

= 2.9 M.

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Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

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Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

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As it just touches the 15 mm high roof, the final velocity will be zero. So,

v_f=0\ m/s.

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Change in gravitational potential energy = Final PE - Initial PE

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\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J                    [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

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