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Morgarella [4.7K]
3 years ago
8

1. Consider the decomposition reaction of sodium chlorate. There are 100 grams of

Chemistry
1 answer:
IRINA_888 [86]3 years ago
6 0
A. NaCl(s) and O2(g)

B. 2NaClO3(s) —> 2NaCl(s) + 3O2(g)

C. moles NaClO3 = 100 g / 106.44 g/mol = 0.939 mol NaClO3

D. 0.939 mol NaCl (because the NaClO3 and NaCl are in a 1 to 1 ratio)

E. grams NaCl = 0.939 mol • 58.44 g/mol = 54.9 g NaCl

F. moles of O2 = 0.939 mol NaClO3 • (3 mol O2 / 2 mol NaClO3) = 1.41 mol O2

G. grams of O2 = 1.41 mol • 32 g/mol = 45.1 g O2

H. Percent yield = 10/45.1 • 100% = 22.2% yield
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Answer:

C) Q < K, reaction will make more products

Explanation:

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∴ Kc = 0.425 = [ SF6 ] / [ F2 ]³

∴ Q = [ SF6 ] / [ F2 ]³

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5 0
2 years ago
Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,
7nadin3 [17]

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)

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      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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3 years ago
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krek1111 [17]

Answer:

Sodium fluoride (NaF) does indeed follow the octet rule without any violations.

3 0
1 year ago
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umka2103 [35]
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3 years ago
Use the following equation to answer the questions below.
sertanlavr [38]

Answer:

Explanation:

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