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lesya692 [45]
3 years ago
12

The ksp of lead(ii) carbonate, pbco3, is 7.40 × 10-14. calculate the solubility of this compound in g/l.

Chemistry
1 answer:
Dennis_Churaev [7]3 years ago
5 0
The  molar  solubility  is  the  Pb ions  present. PbCO3  dissociate  to  form  Pb  ions  and CO3  ions
Ksp  7.40x10^-14 is equal to  (Pb  ions) (CO3 ions)
Pb ions  correspond  to  CO3  ions
(pb ions)( Pb ions) is  equal to 7.40x10^-14
therefore  Pb  ions is square  root  of 7.4x10^14 which  is 2.27x10^-7m
In  
g/l   is (2.27x10^-7) x 267(RFM  of PbCO3) which  is  6.06x10^-5 g/l
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Would aluminum be as useful as food wrapping if it had a much lower melting point? Explain your answer.
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4 0
3 years ago
all my points! Calculate the change in free energy if 1.12 moles of Nal is dissolved in water at 25.0°C.​
Vaselesa [24]

Answer:

Free energy change is -7.500248 kJ/mol

Explanation:

If we have the enthalpy change and entropy change, we can find the free energy change using the equation,

ΔG = ΔH - TΔS

Temperature in K = 25°C + 273 = 298 K

ΔH for NaI = -7.50 kJ/mol

ΔS for NaI = 74 J/K mol = 0.074 kJ/K mol

Plugin the above values in the equation, we will get,

ΔG  =  -7.5 kJ/mol - 0.074 kJ/K mol / 298 K

    = -7.5 kJ/mol - 0.000248 kJ/mol

  = -7.500248 kJ/mol

7 0
3 years ago
Read 2 more answers
A solution is prepared by dissolving 36.7845 grams of disodium ethylenediaminetetraacetic acid dihydate in enough water to prepa
Igoryamba

<u>Answer:</u> The molar concentration of ethylenediminediacetic-dihydrate and sodium ions in solution is 0.1976 M and 0.3952 M respectively.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

We are given:

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Concentration of ethylenediminediacetic-dihydrate in solution = (1\times 0.1976)=0.1976M

Concentration of sodium ions in solution = (2\times 0.1976)=0.3952M

Hence, the molar concentration of ethylenediminediacetic-dihydrate and sodium ions in solution is 0.1976 M and 0.3952 M respectively.

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