Harrietking5571 Harrietking5571
  • 01-03-2018
  • Chemistry
contestada

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

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zelinsky
zelinsky zelinsky
  • 10-03-2018
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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