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2 years ago

How many moles in 28.0 grams of Oxygen

Chemistry

1 answer:

anastassius [24]2 years ago

3 0

Answer:

1.) 28.0 grams of oxygen28 grams (1 mole/16 grams per mole)=1.75 moles oxygen2.)5.0 moles of Iron5 moles(55.845 grams/1 mole)=279.225 grams Iron3.) 452 grams Argon452 grams(1 mole/39.948 grams)=11.315 moles Argon4.) 16.5 moles Hydrogen16.5 moles(1.01 grams/1 mole)=16.665 grams Hydrogen

Explanation:

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Sodium sulfate is slowly added to a solution containing 0.0500 M Ca 2 + ( aq ) and 0.0390 M Ag + ( aq ) . What will be the conce

levacccp [35]

The given question is incomplete. The complete question is as follows.

Sodium sulfate is slowly added to a solution containing 0.0500 M $Ca^{2+}(aq)$ and 0.0390 M $Ag^{+}(aq)$ . What will be the concentration of $Ca^{2+}$ (aq) when $Ag_{2}SO_{4}(s)$ begins to precipitate? What percentage of the $Ca^{2+}(aq)$ can be separated from the Ag(aq) by selective precipitation?

Explanation:

The given reaction is as follows.

$Ag_{2}SO_{4} \rightleftharpoons 2Ag^{+} + SO^{2-}_{4}$

$[Ag^{+}]$ = 0.0390 M

When $Ag_{2}SO_{4}$ precipitates then expression for $K_{sp}$ will be as follows.

$K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]$

$1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]$

$[SO^{2-}_{4}]$ = 0.00788 M

Now, equation for dissociation of calcium sulfate is as follows.

$CaSO_{4} \rightleftharpoons Ca^{2+} + SO^{2-}_{4}$

$K_{sp} = [Ca^{2+}][SO^{2-}_{4}]$

$4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788$

$[Ca^{2+}]$ = 0.00625 M

Now, we will calculate the percentage of $Ca^{2+}$ remaining in the solution as follows.

$\frac{0.00625}{0.05} \times 100$

= 12.5%

And, the percentage of $Ca^{2+}$ that can be separated is as follows.

100 - 12.5

= 87.5%

Thus, we can conclude that 87.5% will be the concentration of $Ca^{2+}(aq)$ when $Ag_{2}SO_{4}(s)$ begins to precipitate.

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