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Ivahew [28]
3 years ago
15

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

Chemistry
1 answer:
Dima020 [189]3 years ago
6 0
Vs = 1.0 mL = 0.001 L
c((NH4)2CO3) = <span>0.02 M
n(</span>(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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