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Mazyrski [523]
3 years ago
7

9.

Physics
1 answer:
Ghella [55]3 years ago
4 0

Given parameters:

Mass of the body  = 200g

Force on the body  = 10N

Unknown parameters:

Acceleration produced by the force  = ?

To solve this problem we must first define force in terms of mass and acceleration. This is possible due to the Newton's first law of motion.

  Force  = mass x acceleration

Here the unknown is acceleration and we can easily solve for it.

But we must take the mass to kilogram in order for it to cancel out.

        1000g  = 1 kg

        200g  = x kg =   \frac{200}{1000}   = 0.2kg

Now input the parameters and solve;

         10  = 0.2 x acceleration

   Acceleration  = \frac{10}{0.2}   = 50m/s²

The acceleration produced by the body is 50m/s²

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At 600.0 k the rate constant is 6.1× 10–8 s–1. what is the value of the rate constant at 785.0 k?
photoshop1234 [79]
Missing details. Complete text is:"The following reaction has an activation energy of 262 kJ/mol:
C4H8(g) --> 2C2h4(g)
At 600.0 K the rate constant is 6.1× 10–8 s–1. What is the value of the rate constant at 785.0 K?"
To solve the exercise, we can use Arrhenius equation:
\ln( \frac{K_2}{K_1} ) =  \frac{Ea}{R} ( \frac{1}{T_1}- \frac{1}{T_2}  )
where K are the reaction rates, Ea is the activation energy, R=8.314 J/mol*K and T are the temperatures. Using T1=600 K and T2=785 K, and Ea=262 kJ/mol = 262000 J/mol, on the right side of the equation we have
\frac{Ea}{R}( \frac{1}{T_1}- \frac{1}{T_2}  )=12.38
And so
\ln( \frac{K_2}{K_1})=12.38
And using K_1=6.1\cdot 10^{-8} s^{-1} , we find K2:
K_2=K_1 e^{12.38}=0.0145 s^{-1}


5 0
3 years ago
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