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diamong [38]
4 years ago
14

How many significant digits are there in the number 204.0920?

Chemistry
2 answers:
mihalych1998 [28]4 years ago
6 0
204.0920-It have 7 significant digits
Nikitich [7]4 years ago
5 0

Answer: 7 significant digits

Explanation:

Significant  digits.: The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant  digits.

All non-zero numbers are always significant. For example: 654, 6.54 and 65.4 all have three significant  digits.

All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant digits.

All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant digits.

All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant digits

Thus 204.0920 has 7 significant digits.

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Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,
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Answer:

Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 )   N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)  ; Kc = 1.54e - 31

2)   N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)  ; Kc = 2.16e - 24

   upon reversing  ( 2 )  equation

     N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)   Kc = 1/2.16e - 24  

    now adding 1 and reversed equation (2)

       N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)

      N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)

   we get ,

                  N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)  Kc = 1.54e-31 × 1/2.61e - 24

       equilibrium constant of equation (3) is -

            Kc = 1.54e - 31 / 2.61e - 24

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