use dimensional analysis to find how many moles of nitrate ions are in 40.0 mL of a 0.75 M aqueous solution of aluminum nitrate

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(0.75 moles Al(NO3)3/L Al(NO3)3) x 0.040 L Al(NO3)3 = ??moles Al(NO3)3. Note that L Al(NO3)3 in the numerator cancel with L Al(NO3)3 in the denominator to leave moles Al(NO3)3. Now you need a factor to convert moles Al(NO3)3 to moles NO3^-.