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OverLord2011 [107]
3 years ago
12

When balancing a chemical reaction what are the small numbers that we do not change?

Chemistry
1 answer:
Phoenix [80]3 years ago
6 0

Answer:

Subscripts

Explanation:

Example:

H2O

2 is the subscript and does not change  

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Verdich [7]

The amount of physical material in an object

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Which of the following samples will have the greatest volume at STP?
Tamiku [17]

Answer:

Option b. 22 g of He will have the greatest volume at STP

Explanation:

In order to determine the volume, we apply the Ideal Gases Law equation:

P . V = n . R . T

V = n . R . T / P

R, T and P are the same in all the situation we must define n (number of moles).

The one that has the greatest number of moles will have the greatest volume at STP

22 g of Ne . 1mol / 20.1 g = 1.09 moles of Ne

22g of He . 1mol / 4 g = 5.5 moles of He

22 g of O₂ . 1mol / 32g = 0.68 moles of O₂

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3 years ago
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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Frequency, f = v / λ

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What elements were formed in Nuclear fusion?
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Answer:

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