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Flura [38]
1 year ago
10

A quantity of an ideal gas is compressed to half its initial volume. the process may be adiabatic, isothermal or isobaric. the g

reatest amount of work is required if the process is?
Chemistry
1 answer:
JulsSmile [24]1 year ago
3 0

the greatest amount of work is required if the process is adiabatic.The correct option is adiabatic.

The process in  which heat is constant is called adiabatic process.

The The process in  which temperature is constant is called isothermal process.

The process in  which pressure is constant is called isobaric process.

The P-V diagram for adiabatic , isothermal and isobaric process is given below.

Work done in process = area encloses by P-V diagram axis . Since area under the curve is maximum for adiabatic process which is shown in the above diagram. So, work done by the gas will be maximum for adiabatic process.

learn more about adiabatic process.

brainly.com/question/17192213

#SPJ4

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Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank
Masja [62]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration equilibrium constant is K_c  = 14.39

Explanation:

The chemical equation for this decomposition of ammonia is

                2 NH_3  ↔   N_2 + 3 H_2

The initial concentration of ammonia is mathematically represented a

          [NH_3] =  \frac{n_1}{V_1}  = \frac{29}{75}

          [NH_3] = 0.387  \  M

The initial concentration of nitrogen gas  is mathematically represented a

         [N_2] =  \frac{n_2}{V_2}

         [N_2] =  0.173  \  M

So  looking at the equation

   Initially (Before reaction)

      NH_3 = 0.387 \ M

      N_2  =  0 \  M

      H_2 =  0 \ M

During reaction(this is gotten from the reaction equation )

        NH_3 = -2 x(this implies that it losses two moles of concentration )

         N_2 = + x  (this implies that it gains 1 moles)

         H_2  =  +3 x(this implies that it gains 3 moles)

Note : x denotes concentration

At equilibrium

        NH_3 = 0.387 -2x

       N_2 =  x

        H_2  =  3 x

Now since

     [NH_3] = 0.387  \  M

     x= 0.387  \  M    

H_2  =  3 * 0.173    

H_2  =  0.519 \ M    

NH_3 = 0.387 -2(0.173)

NH_3 = 0.041 \ M

Now the equilibrium constant is

           K_c  =  \frac{[N_2][H_2]^3}{[NH_3]^2}

substituting values

           K_c  =  \frac{(0.173) (0.519)^3}{(0.041)^2}

           K_c  = 14.39

         

3 0
3 years ago
What was Anton van Lee
solniwko [45]

I would say the answer is A.

5 0
3 years ago
What is the volume of 0.250 m hydrochloric acid required to react completely with 20.0 ml of 0.250 m ca(oh)2?
Soloha48 [4]
The balanced equation for the above neutralisation reaction is as follows;
Ca(OH)₂ + 2HCl ----> CaCl₂ + 2H₂O
Stoichiometry of Ca(OH)₂ to HCl is 1:2
number of Ca(OH)₂ moles reacted - 0.250 mol/L x 20.0 x 10⁻³ L = 5.00 x 10⁻³ mol
according to molar ratio of 1:2
number of HCl moles required = 2 x number of Ca(OH)₂ moles reacted
number of HCl moles = 5.00 x 10⁻³ x 2 = 10.0 x 10⁻³ mol
molarity of HCl solution - 0.250 M
there are 0.250 mol  in volume of 1 L
therefore 10.0 x 10⁻³ mol in - 10.0 x 10⁻³ mol  / 0.250 mol/L = 40.0 mL 
40.0 mL of 0.250 M HCl is required
5 0
3 years ago
How many grams of Mg(NO3)2 would be produced? please show how you got the answer​
Len [333]

Answer:

148 grams of relative atomic mass

Explanation:

magnesium atomic mass : 24

nitrogen : 14

oxygen : 16

24 × 1

14 × 1 × 2

16 × 3 × 2

24 + 28 + 80 = 148 grams

7 0
3 years ago
A substance in which light can travel through such as air,glass, or water
Stels [109]

Answer:

The correct answer is - transparent medium.

Explanation:

A transparent substance or medium is the substance that allows light to pass through it. Light moves through these substances as they do not absorb the light and do not reflect too.

The example of such substances is glass, air or water. These substances allow light to pass through them.

Thus, The correct answer is - transparent medium.

6 0
3 years ago
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