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sergiy2304 [10]
3 years ago
9

Selina drops a golf ball down a deep well. The ball falls with only the force of gravity acting on it. Without any air resistanc

e, as the golf ball falls into the well, its speed is:
A. constant
B. increasing
C. decreasing
D. directional
Chemistry
1 answer:
ss7ja [257]3 years ago
8 0

<u>Answer:</u>    A. constant

Without any air resistance, as the golf ball falls into the well, its speed is constant.

<u>Explanation:</u>

Given that when the ball is released,<em>there is no air resistance acting on ball. So the ball is freely falling. </em>

The only force acting on the freely falling body is the gravitational force pulling the body towards the centre of the earth.Here the only acceleration acting on the body is the <em>acceleration due to gravity.</em>

The acceleration due to gravity is 9.8 m/s^2. so through out the journey of the ball,the acceleration is 9.8 m/s^2. <em>So as the acceleration is constant, the speed is also constant.</em>

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Consider the second-order reaction:
kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

Explanation:

2HI(g)\rightarrow H_2(g)+I_2(g)&#10;

Rate Law: k[HI]^2&#10;

Rate constant of the reaction = k = 6.4\times 10^{-9} L/mol s

Order of the reaction = 2

Initial rate of reaction = R=1.6\times 10^{-7} Mol/L s

Initial concentration of HI =[A_o]

1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2

[A_o]=5 mol/L

Final concentration of HI after t = [A]

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Integrated rate law for second order kinetics is given by:

\frac{1}{[A]}=kt+\frac{1}{[A_o]}

\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}

[A]=0.00345 mol/L

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

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