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3 years ago

What minimum concentration of clâ is required to begin to precipitate pbcl2? for pbcl2, ksp=1.17Ã10â5?

1 answer:

5 0

Missing in your question : [Pb2+] = 0.085 M

So from this balanced equation:
PbCl2 ↔ Pb+2 + 2Cl-

when Ksp = [Pb+2][Cl]^2
when we have Ksp = 1.17x10^-5 & [Pb+2] = 0.085 M so by substitution:
1.17x10^-5 = 0.085 [Cl-]^2

∴[Cl]^2 = (1.17x10^-5) / 0.085 = 1.38x10^-4
∴[Cl-] = √0.0002
= 0.0117 M
So 0.0117 M is the minimum concentration requires precipitating PbCl2

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Alexxandr [17]

Answer:

testable questions are answer through observation or an experiment that provides evidence that the questions connects to the scientific concepts rather the opinion feelings

therefore the question can be tested through observation or experiment

7 0

3 years ago

(06.01 MC)

skelet666 [1.2K]

Answer:

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2 years ago

A sulfuric acid solution containing 571.3 g of h2so4 per liter of aqueous solution has a density of 1.329 g/cm3. Part a calculat

loris [4]

Mass percentage of a solution is the amount of solute present in 100 g of the solution.

Given data:

Mass of solute H2SO4 = 571.3 g

Volume of the solution = 1 lit = 1000 ml

Density of solution = 1.329 g/cm3 = 1.329 g/ml

Calculations:

Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g

Therefore we have:

571.3 g of H2SO4 in 1329 g of the solution

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Mass percentage of H2SO4 (%w/w) is 42.99 %

3 0

3 years ago

Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose "average" formula is C₁₂H₂₆.

padilas [110]

The calculated enthalpy of formation of kerosene is 365.4 kJ and heat produced is 78650.3 kJ

For this, we need the normal enthalpy of formation given below

ΔH∘f(CO2)=−393.5kJ/molΔH∘f(H2O)\s=−241.8kJ/molΔH∘f(O2)=0kJ/mol

We shall now determine the enthalpy of kerosene formation:

H rxn = 24 mol H f (CO2) + 26 mol H f (H2O) + 2 mol H f (C12H26) + 37 mol H f (O2) + 1.50 104 kJ = 9444 kJ + 6286.8 kJ + 1500 kJ 2 mol H f (C12H26) = 730.8 kJ H f (C12H26) = 365.4 kJ

Kerosene has a density of 0.74 g/mL.

Kerosene volume (V) equals 0.63 gallons, or 0.63 x 3785.4, or 2384. 8 mL.

We shall now calculate the mass (m) of kerosene:

ρ=mVm\s=ρ×Vm\s=0.749g/mL×2384.mLm\s=1786.2g

We shall now discover the heat that 1786 generated.

Two grams of kerosene

Kerosene's molar mass is 170.33 g/mol.

The mass of two moles of kerosene is equal to 2*170.33*340.66g.

1.50104kJ of heat are generated by 340.66 g of kerosene.

1786 produced heat.

Kerosene 2 grams = 1.50 104 kJ 340.66 1786.2 g = 78650.3 kJ

Learn more about enthalpy here-

brainly.com/question/13996238

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5 0

1 year ago

Balance the following equations. Do not include the states of matter.<br><br> (a) C + O2 → CO

krek1111 [17]

Answer:

C + O2 → CO2

Explanation:

C + O2 → CO ----------------- (1)

from equ (1) on reactant side, C has 1 mole, O has 2 moles

from equ (1) on product side, C has 1 mole, O has 1 mole

Thus, to balance the equation, O should have 2 moles

C + O2 → CO2

7 0

3 years ago

Read 2 more answers