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3 years ago

a car travelling at 50km/h from rest covers a distance of 10km in 40minutes. Calculate the acceleration

Physics

1 answer:

Margarita [4]3 years ago

7 0

Answer:

$9.67\cdot 10^{-3}m/s^2$

Explanation:

We can solve the problem by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the car in this problem:

u = 0 (it starts from rest)

$v=50 km/h \cdot \frac{1000}{3600}=13.9 m/s$ is the final velocity

s = 10 km = 10 000 m is the displacement

Solving for a, we find:

$a=\frac{v^2-u^2}{2s}=\frac{13.9^2}{2(10000)}=9.67\cdot 10^{-3}m/s^2$

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3 years ago

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Answer:

Se obtienen 2,27 gramos de metanol.

Explanation:

La reacción entre monóxido de carbono e hidrógeno para producir metanol es la siguiente:

CO + 2H₂ → CH₃OH

Para encontrar el reactivo limitante y el reactivo en exceso, debemos calcular el número de moles de CO y H₂:

$\eta_{CO} = \frac{m}{M}$

En donde:

m: es la masa

M: es el peso molecular

$\eta_{CO} = \frac{m}{M_{CO}} = \frac{2,0 g}{28,01 g/mol} = 0,071 moles$

$\eta_{H_{2}} = \frac{2,0 g}{2,02 g/mol} = 0,99 moles$

Dado que la relación estequiométrica entre CO y H₂ es 1:2, el número de moles de hidrógeno gaseoso que reaccionan con el monóxido de carbono es:

$\eta_{H_{2}} = \frac{2}{1}*0,071 = 0,142 moles$

Entonces, se necesitan 0,142 moles de H₂ para reaccionar con 0,071 moles de CO y debido a que se tienen más moles de H₂ (0,99 moles) entonces el reactivo limitante es CO y el reactivo en exceso es H₂.

Ahora podemos encontar la masa de metanol obtenida usando el reactivo limitante (CO) y sabiendo que la realcion estequiométrica entre CO y CH₃OH es 1:1.

$\eta_{CH_{3}OH} = \eta_{CO} = 0,071 moles$