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castortr0y [4]
3 years ago
14

How many molecules are in 30 liters of methane (CH4) at STP?

Chemistry
1 answer:
mylen [45]3 years ago
4 0

Hello!

How many molecules are in 30 liters of methane (CH4) at STP ?

We have the following information:

Knowing that by Avogadro's Law for each mole of substance we have 6.02 * 10²³ molecules, it is known that in STP (Standard Temperature and Pressure) one mole of any gas equals 22.4 L, then:

22.4 L ----------------- 6.02*10²³ molecules

30 L -------------------- y molecules

\dfrac{22.4}{30} = \dfrac{6.02*10^{23}}{y}

multiply the means by the extremes

22.4*y = 30*6.02*10^{23}

22.4*y = 1.806*10^{25}

y = \dfrac{1.806*10^{25}}{22.4}

\boxed{\boxed{y = 8.0625*10^{23}\:molecules}}\:\:\:\:\:\:\bf\green{\checkmark}

Answer:

8.0625*10²³ molecules of methane

________________________  

\bf\green{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

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Draw a complete structural formula and a condensed structural formula for(a) three compounds of formula C3H8O (b) five compounds
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Explanation:

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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

         

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