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Radda [10]
3 years ago
6

6.0L of a gas exert a pressure of 2.5 atm. When the pressure is increased to 10.0atm what is the new volume of the gas

Chemistry
1 answer:
jeka57 [31]3 years ago
7 0

Answer:

<h2>1.5 L</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{6 \times 2.5}{10}  =  \frac{15}{10}  =  \frac{3}{2}  \\

We have the final answer as

<h3>1.5 L</h3>

Hope this helps you

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The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichl
nevsk [136]

Complete Question

A student is extracting caffeine from water with dichloromethane. The K value is 4.6. If the student starts with a total of 40 mg of caffeine in 2 mL of water and extracts once with 6 mL of dichloromethane

The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichloromethane each time. How much caffeine will be in each dichloromethane extract?

Answer:

The mass of  caffeine extracted is  P =  39.8 \ mg

Explanation:

From the question above  we are told that

    The  K value is  K =  4.6

     The  mass of the caffeine is  m  = 40 mg

      The  volume of water is  V  = 2 mL

      The volume  of caffeine is  v_c =  2 mL

     The number of times the extraction was done is  n =  3

Generally the mass of  caffeine that will be  extracted is  

           P =  m  *  [\frac{V}{K *  v_c + V} ]^3

substituting values  

           P =  40   *  [\frac{2}{4.6 *  2 + 2} ]^3

           P =  39.8 \ mg

6 0
3 years ago
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Answer:

669.48 kJ

Explanation:

According to the question, we are required to determine the heat change involved.

We know that, heat change is given by the formula;

Heat change = Mass × change in temperature × Specific heat

In this case;

Change in temperature = Final temp - initial temp

                                       = 99.7°C - 20°C

                                       = 79.7° C

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Specific heat of water is 4.2 J/g°C

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