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3 years ago

A horse runs 1 km in 15 s. What is the average speed

Chemistry

1 answer:

Reptile [31]3 years ago

6 0

The average speed of the horse for the entire motion is 66.67 m/s.

The given parameters;

<em>distance traveled by the horse, d = 1 km = 1,000 m</em>
<em>time of motion of the horse, t = 15 s</em>

The average speed of the horse is calculated by dividing the total distance traveled by the horse by the time of motion.

The average speed of the horse is calculated as follows;

$average\ speed = \frac{distance}{time} \\\\average\ speed = \frac{1000 \ m}{15 \ s} = 66.67 \ m/s$

Thus, the average speed of the horse for the entire motion is 66.67 m/s.

Learn more here: brainly.com/question/17289046

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Problem PageQuestion Iron(III) oxide and hydrogen react to form iron and water, like this: (s)(g)(s)(g) At a certain temperature

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Complete question

The complete question is shown on the first uploaded image

Answer:

The value is $K_c = 2.69 *10^{-5}$

Explanation:

From the question we are told that

The equation is

$Fe_20_3_{(s)}+3H_{(g)}\to2Fe_{(s)}+3H_2O_{(g)}$

Generally the equilibrium is mathematically represented as

$K_c = \frac{[H_2O]^2}{[H_2]^3}$

Here $[H_2O]$ is the concentration of water vapor which is mathematically represented as

$[H_2O ] = \frac{n_w}{V_s }$

Here $V_s$ is the volume of the solution given as 8.9 L

$n_w$ is the number of moles of water vapor which is mathematically represented as

$n_w = \frac{m_w}{Z_w}$

Here $m_w$ is the mass of water given as 2.00 g

and $Z_w$ is the molar mass of water with value 18 g/mol

$n_w = \frac{2}{18}$

=> $n_w = 0.11 \ mol$

$[H_2O ] = \frac{0.11}{8.9 }$

=> $[H_2O ] = 0.01236 \ M$

Also

$[H]$ is the concentration of hydrogen gas which is mathematically represented as

$[H ] = \frac{n_v}{V_s }$

Here $V_s$ is the volume of the solution given as 8.9 L

$n_v$ is the number of moles of hydrogen gas which is mathematically represented as

$n_v = \frac{m_v}{Z_v}$

Here $m_w$ is the mass of water given as 4.77 g

and $Z_v$ is the molar mass of water with value 2 g/mol

$n_w = \frac{4.77}{2}$

=> $n_w = 2.385 \ mol$

$[H_2O ] = \frac{2.385}{8.9 }$

=> $[H_2O ] = 0.265 \ M$

$K_c = \frac{( 0.01236 )^3}{ (0.265 )^2}$

=> $K_c = 2.69 *10^{-5}$

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