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12 March, 03:46

How many moles of solute particles are present in 1 ml of aqueous 0.020 m (nh4) 2co3?

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  1. 12 March, 07:37
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    Vs = 1.0 mL = 0.001 L

    c ((NH4) 2CO3) = 0.02 M

    n ((NH4) 2CO3) = ?

    For the purpose, here we will use the next equation:

    c=n/V ⇒ n=cxV

    n ((NH4) 2CO3) = 0.02M x 0.001L

    n ((NH4) 2CO3) = 2x10⁻⁵ mole of (NH4) 2CO3 is presented in the solution
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