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fenix001 [56]
3 years ago
11

Cherylanne mixes together Chemical A and Chemical B in a test tube, forming Chemical C. The mixture instantly turns cool to the

touch. She then adds Chemical D, and the reaction becomes hot and explodes. What can be known about this situation? Chemicals A and B form an endothermic reaction, and chemicals C and D form an exothermic reaction. Chemicals A and B form an exothermic reaction, and chemicals C and D form an endothermic reaction. Chemicals A and C form an endothermic reaction, and chemicals B and D form an exothermic reaction. Chemicals A and B form an exothermic reaction, and chemicals B and D form an endothermic reaction.
Chemistry
2 answers:
mina [271]3 years ago
8 0

Answer : Chemicals A and B form an endothermic reaction, and chemicals C and D form an exothermic reaction.

Explanation :

Endothermic reaction : When the system absorb heat from the surrounding then the surrounding become cool.

Exothermic reaction : when the system releases heat into the surrounding then the surrounding become hot.

According to the question,

when we mixed chemical A and chemical B together in a test tube to form chemical C, the mixture become cool. This means that the system is absorbing heat from the surrounding and thus the reaction is called a endothermic reaction.

And when we added chemical D in chemical C, the new mixture becomes hot and explodes. This means that the system is releasing heat into the surrounding and thus the reaction is called as exothermic reaction.


Juli2301 [7.4K]3 years ago
6 0

Answer:

for the lazy the answer is (A)

Explanation:

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rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is
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Complete Question

The  rate of a certain reaction is given by the following rate law:

            rate =  k [H_2][I_2]

rate Use this information to answer the questions below.

What is the reaction order in H_2?

What is the reaction order in I_2?

What is overall reaction order?

At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer:

The reaction order in H_2 is  n =  1

The reaction order in I_2 is  m = 1

The  overall reaction order z =  2

When the hydrogen is double the the initial rate is   rate_n  =  4.0*10^{-4} M/s

The rate constant is   k = 35.23 \  M^{-1} s^{-1}

Explanation:

From the question we are told that

   The rate law is  rate =  k [H_2][I_2]

   The rate of reaction is rate =  2.0 *10^{4} M /s

Let the reaction order for H_2 be  n and for I_2  be  m

From the given rate law the concentration of H_2 is raised to the power of 1 and this is same with I_2 so their reaction order is  n=m=1

   The overall reaction order is  

               z  = n +m

               z  =1 +1

               z  =2

At  rate =  2.0 *10^{4} M /s

        2.0*10^{4}  = k  [H_2] [I_2] ---(1)

= >    k  = \frac{2.0*10^{4}}{[H_2] [I_2]  }

given that the concentration of hydrogen is doubled we have that

            rate  = k [2H_2] [I_2] ----(2)

=>      k = \frac{rate_n  }{ [2H_2] [I_2]}

 So equating the two k

           \frac{2.0*10^{4}}{[H_2] [I_2]  } = \frac{rate_n  }{ [2H_2] [I_2]}

    =>    rate_n  =  4.0*10^{-4} M/s

So when

      rate_x =  52.0 M/s

        [H_2] = 1.8 M

         [I_2] =  0.82 \ M

We have

      52 .0 =  k(1.8)* (0.82)

     k = \frac{52 .0}{(1.8)* (0.82)}

      k = 35.23 M^{-2} s^{-1}

     

     

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Answer:

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