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1.99 M hydrofluoric acid, HF (Ka = 7.2  10–4), and 3.00 M hydrocyanic acid, HCN (Ka = 6.2  10–10). Determine the [CN–] at equilibrium.?

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  • 1.99 M hydrofluoric acid, HF (Ka = 7.2  10–4), and 3.00 M hydrocyanic acid, HCN (Ka = 6.2  10–10). Determine the [CN–] at equilibrium.?


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Anonymous74967
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