Mass of Al wire before reaction = 3.96 g Mass of Al wire after reaction = 3.65 g
Mass of Al lost = 0.31 g
Moles of Al lost = 0.011 moles
Mass of Pb + filter paper = 4.26 g
Mass of filter paper = 0.92 g
Mass of Pb = 3.34 g
Moles of Pb formed = 0.016 moles
Recall that there were 100 mL of solution. 0.016 moles of Pb were removed from the solution.

What was the [Pb+2] for the saturated solution of PbCl2?
Options: 0.16, 0.016, or 0.0016

Recall that there was 100 mL of solution. 0.016 moles of Pb were removed from the solution.
Also recall the original equilibrium: PbCl2(s) Pb+2(aq) + 2(Cl-)(aq)

Since there are 2 Cl ions formed for every Pb ion, what was the [Cl-] for the saturated solution of PbCl2?
Options: 0.32 M, 0.032 M, or 0.0032 M

A. is your answer.  hope this  

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is  2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry  problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that     10 dm = 1 m  ⇔ 10³ dm³ = 1 m³then we use the fact that     10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting:     now we have to find the moles of hydrochloric  acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by    molar mass = (mass ca) + (mass c) + 3(mass o)    = 40 + 12 + 3(16)    = 100 g/molconvert the given 5 g to moles:     mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis    mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃)    = 0.05 mol hcl * 2    = 0.1 mol hcl  we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get     volume = moles / concentration    volume = 0.1 mol / (0.002 mol/cm³)    volume = 50 cm³

H tricky one let me think and get back to u