Manganese (II) chloride has a solubility of 5.75 M at 0 degrees C. If 125 mL of saturated MnCl2 is evaporated to dryness, what mass of MnCl2 will be left?

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Manganese (II) chloride has a solubility of 5.75 M at 0 degrees C. If 125 mL of saturated MnCl2 is evaporated to dryness, what mass of MnCl2 will be left?

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Answer 1 · 2018-05-01T06:32:00

And so the solubility of $M n C l_{2}$ $MnCl_2$ is $5.75 \cdot m o l \cdot L^{- 1}$ $5.75*mol*L^-1$ ..I am not sure if I believe that....I get a mass of approx. $90 \cdot g$ $90*g$ ...

Explanation:

We gots .................

$125 \cdot m L \times 10^{- 3} \cdot L \cdot m L^{- 1} \times 5.75 \cdot m o l \cdot L^{- 1} = 0.719 \cdot m o l$ $125*mLxx10^-3*L*mL^-1xx5.75*mol*L^-1=0.719*mol$

...the which represents a mass of ....... $125.84 \cdot g \cdot m o l^{- 1} \times 0.719 \cdot m o l = ? ? \cdot g$ $125.84*g*mol^-1xx0.719*mol=??*g$

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Manganese (II) chloride has a solubility of 5.75 M at 0 degrees C. If 125 mL of saturated MnCl2 is evaporated to dryness, what mass of MnCl2 will be left?

1 Answer

Explanation:

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