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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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Number 51 is 0.00150 ml to l
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Answer:

The answer to your question is

51.- 1.59 x 10⁻⁶

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Explanation:

51.- Convert 0.00159 ml to liters

Use a rule of three

                                 1000 ml -------------- 1 l

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Reaction

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8 0
3 years ago
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Answer:

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Formula CaF

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Symbol H S

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