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ch4aika [34]
3 years ago
8

0.05 grams od cobalt ii oxalate will dissolve in one liter of solution at 25 degree celcius. Calculate the ksp for cobalt ii oxa

late at 25 degree celcius
Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
5 0

Answer:

The equilibrium expression is:

CoC2O4(s)⇌Co2+(aq)+C2O2−4(aq)

For this reaction:

Ksp = [Co2+][C2O2−4]=1.96×10−8

Explanation:

Batteries will not clot if cobalt ions are removed from its cells. Some blood collection tubes contain salts of the oxalate ion,

C2O2−4

, for this purpose. At sufficiently high concentrations, the calcium

and oxalate ions form solid, CoC2O4·H2O (which also contains water bound in the solid). The concentration of Co2+ in a sample of blood serum is 2.2 × 10–3M. What concentration of

C2O2−4

ion must be established before CoC2O4·H2O begins to precipitate.

CoC2O4 does not appear in this expression because it is a solid. Water does not appear because it is the solvent.

Solid CoC2O4 does not begin to form until Q equals Ksp. Because we know Ksp and [Co2+], we can solve for the concentration of

C2O2−4

that is necessary to produce the first trace of solid:

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If a gas sample has a pressure of 30.7 kPa at 0.00*C, by how much does the temperature have to decrease to lower the pressure to
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                      252.68 K  or   -20.46 °C

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                    According to Gay-Lussac's Law, "Pressure and Temperature at given volume are directly proportional to each other".

Mathematically,

                                              P₁ / T₁  =  P₂ / T₂   ---- (1)

Data Given:

                  P₁  =  30.7 kPa

                  T₁  =  0.00 °C  =  273.15 K

                  P₂  =  28.4 kPa

                  T₂  =  <u>???</u>

Solving equation for T₂,

                  T₂  =  P₂ T₁ / P₁

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                  T₂  =  28.4 kPa × 273.15 K / 30.7 kPa

                  T₂  =  252.68 K  or   -20.46 °C

8 0
3 years ago
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