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storchak [24]
4 years ago
14

HELP! CHEMISTRY QUESTIONS!

Chemistry
1 answer:
Gennadij [26K]4 years ago
4 0
E= m× C2
... than:
m=E/c 2
m=1,366•10^3/3•10 A 8 2
m=1,518•10^14 kg
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150 mL of 0.15 M Na2SO4 is mixed with an equal volume of 0.050 M Agno3. Select all the options that correctly show the steps use
Lera25 [3.4K]

Answer:

No precipitate will be formed.

Explanation:

Given that ;

150 mL of 0.15 M Na_2SO_4 is mixed with an equal volume of 0.050 M AgNO_3.

Then;

[Ag^+] = \frac{0.05*150}{(150+150)}

[Ag^+] = \frac{7.5}{(300.0)}

[Ag^+] = 0.025 \ M

[SO_4^{2-}] = \frac{0.15*150.0}{150.0+150.0}

[SO_4^{2-}] = \frac{22.5}{300.0}

[SO_4^{2-}] = 0.075 \ M

For Ag_2SO_4 ; [Ag^+]^2 [SO_4^{2-}]

=(0.025)^2 *(0.075)

=4.6875*10^{-5}

The Ksp of Ag_2SO_4 = 1.5*10^{-5} ;

Therefore since 4.6875*10^{-5} is < Ksp of Ag_2SO_4 ; Then no precipitate will be formed.

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Explanation:

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