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Alborosie
3 years ago
6

A kinetics experiment is set up is set up to collect the gas that is generated when a sample of chalk consisting primarily of so

lid CaCO3 is added to a solution of ethanoic acid, CH3COOH. The rate of reaction between CaCO3 and CH3COOH is determined by measuring the volume of gas generated at 25 degrees C and 1 atm as a functionof time. Which of the following experimental conditions is most likely to increase gas production?
A) Decreasing the amount of ethnoic acid solution used in the experiment
B) Decreasing the concentration of the ethnoic acid solution used in the experiment
C) Decreasing the temperature at which the experiment is performed
D) Decreasing the partical size of the CaC03 by grinding it into a fine powder
Chemistry
1 answer:
inn [45]3 years ago
4 0

Answer:

Correct option -D

Explanation:

Here, A kinetics experiment is set up is set up to collect the gas that is generated from the cacarbonate and methanoic acid.

Among the given conditions, decreasing the particle size of calcium carbonate only increases the production of gas.

Smaller particles of reactant increases the surface area then followed by rate of reaction will be increases it leads to increases the production of gas.

Therefore, the suitable experimental condition most likely to increase the gas production is-

Decreasing the particle size of the CaCO_{3} by grinding it into a fine powder.

Hence, correct option -D.

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The volume of the 39% acid solution that Roger needs to use to make a mixture of 86 mL of a 36% acid solution is 74.27 mL.

The mixture of the acids solutions is given by:

CV = C_{1}V_{1} + C_{2}V_{2}    (1)      

Where:

C: is the concentration if the mixture = 36%  

V: is the total volume of the mixture = 86 mL

C₁: is the concentration of acid 1 = 39%

V₁: is the volume if acid 1 =?

C₂: is the concentration of acid 2 = 17%

V₂: is the volume of acid 2

The sum of V₁ and V₂ must be equal to V, so:

V = V_{1} + V_{2}

V_{2} = V - V_{1}  (2)

By entering equation (2) into (1), we have:

CV = C_{1}V_{1} + C_{2}(V - V_{1})

36\%*86 mL = 39\%*V_{1} + 17\%(86 mL - V_{1})

Changing the percent values to decimal ones:

0.36*86 mL = 0.39*V_{1} + 0.17(86 mL - V_{1})  

Now, by solving the above equation for V₁:

V_{1} = 74.27 mL  

Therefore, the volume of the 39% acid solution is 74.27 mL.

       

To learn more about mixture and solutions, go here: brainly.com/question/6358654?referrer=searchResults

I hope it helps you!  

         

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