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3 years ago

PLEASE HELP ASAP!

Chemistry

1 answer:

hoa [83]3 years ago

3 0

Answer:

A. 145.2 g NH3

B. 0.76 g H2

C. 1.41 x 10^22 molecules NH3

Explanation:

A. 1 mol N2 -> 2 mol NH3

4.27 mol N2 -> x

x= (4.27 mol N2 * 2 mol NH3)/1 mol N2 x= 8.54 mol NH3

1 mol N2 -> 17 g

8.54 mol N2 -> x x= 145.2 g NH3

2 g H2 -> 34 g NH3

x -> 13.01 g NH3

x= 0.76 g H2

2 g H2 -> 34 g NH3

0.0235 g H2 -> x

x= 0.39 g NH3

0.39 g NH3 (1 mol NH3/17 g NH3)(6.023 x 10^23 molecules/1 mol NH3) =

1.41 x 10^22 molecules NH3

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Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

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$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

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