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0.9 M Al(NO3)3 = 0.9 M *3 M (NO3-) = 2.700 M nitrate ion's (=NO3-)

(Mark the bold numbers: 3above !! because they correspond to e. o.)

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14y ago
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10y ago

Aluminium nitrate is Al(NO3)3. Therefore, for every molecule of aluminium nitrate, there are 3 molecules of nitrates. 0.0525M x 3 is 0.1575M nitrate.

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11y ago

I am assuming that the ratio is in aluminium nitrate, Al(NO3)3. So the ratio of aluminium to nitrate will be 1 : 3.

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8y ago

The concentration of ammonium ion is 32,47 g/L.

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Q: What is the molarity of nitrate ion in a 0.0525 M solution of aluminum nitrate?
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