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Bumek [7]
4 years ago
9

A solution of saturated pbi2 is found to contain 2.7 ✕ 10−3 m iodide ion. calculate the ksp of pbi2

Chemistry
1 answer:
BigorU [14]4 years ago
4 0
When the compound PbI₂ dissolves, it dissociates as follows;
PbI₂ --> Pb²⁺ + 2I⁻
Molar solubility is the number of moles dissolved in 1 L of solution
A saturated solution is when the maximum amount of solute is dissolved in the solution.
Molar solubility of Iodide when solution is saturated is 2.7 x 10⁻³ mol/L, then solubility of Pb²⁺ is (2.7 x 10⁻³ mol/L) / 2 = 1.35 x 10⁻³ mol/L
ksp is the solubility product constant that can be calculated as follows;
ksp = [Pb²⁺][I⁻]
ksp = (1.35 x 10⁻³ mol/L) x (2.7 x 10⁻³ mol/L)²
      = 1.35 x 10⁻³ x 7.29 x 10⁻⁶
      = 9.8 x 10⁻⁹ 
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LIST TWO SUBSTANCE THAT CAN BE USED TO (i)REDUCE THE ACIDITY OF A SUBSTANCE
jenyasd209 [6]

Answer:

An acidic substance is any substance with a pH of less than 7, the smaller the number the more acidic it is. ... An acid is a substance that donates hydrogen ions. Because of this, when an acid is dissolved in water, the balance between hydrogen ions and hydroxide ions is shifted.

Explanation:

3 0
3 years ago
A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce
spin [16.1K]

Explanation:

The given data is as follows.

       Volume of lake = 15 \times 10^{6} m^{3} = 15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}

        Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = 5.6 \times 15 \times 10^{9} mg

                                                                    = 84 \times 10^{9} mg

                                                                    = 84 \times 10^{3} kg

Flow rate of river is 50 m^{3} sec^{-1}

Volume of water in 1 day = 50 \times 10^{3} \times 86400 liter

                                          = 432 \times 10^{7} liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are 2.9792 \times 10^{10} mg or 2.9792 \times 10^{4} kg

Flow rate of sewage = 0.7 m^{3} sec^{-1}

Volume of sewage water in 1 day = 6048 \times 10^{4} liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = 1.8144 \times 10^{10} mg or 1.8144 \times 10^{4}kg

Therefore, total concentration of lake after 1 day = \frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l

                                        = 6.8078 mg/l

                 k_{D} = 0.2 per day

       L_{o} = 6.8078

Hence, L_{liquid} = L_{o}(1 - e^{-k_{D}t}

             L_{liquid} = 6.8078 (1 - e^{-0.2 \times 1})  

                             = 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

                                                             = 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0
4 years ago
Match with O for organic and I for inorganic for each compound.
Vika [28.1K]

Answer:

Organic compounds: C₃H₆, and C₂H₂.

Inorganic compounds: NaS, SO₂, and HI.

Explanation:

Organic compounds are the molecules that have a carbon backbone with hydrogen atoms in its structure.

While, inorganic molecules are composed of other elements. They can contain hydrogen or carbon, but if they have both, they are organic.

<em>So:</em>

<em>Organic compounds are C₃H₆, and C₂H₂.</em>

<em>Inorganic compounds are NaS, SO₂, and HI.</em>

5 0
3 years ago
The density of acetic acid is 1.05 g/mL. What is the volume of 327 g of acetic acid
Eva8 [605]

Answer:

311.43\ \text{mL}

Explanation:

\rho = Density of acetic acid = 1.05\ \text{g/mL}

m = Mass of acetic acid = 327\ \text{g}

V = Volume

Density is given by

\rho=\dfrac{m}{V}\\\Rightarrow V=\dfrac{m}{\rho}\\\Rightarrow V=\dfrac{327}{1.05}\\\Rightarrow V=311.43\ \text{mL}

The volume of acetic acid is 311.43\ \text{mL}.

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3 years ago
Compared to the charges of a proton the charge of an electron has
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4 because electrons and protons have the same magnitude and electron is - and proton is +.
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