20) (17 points) Consider the following reaction: equation #2 MnCl5-2 (aq)  Mn+3 (aq) + 5 Cl-1 (aq)
(red) (yellow) (colorless)

a) At equilibrium, [ Mn+3 ] = 1.500 M, [ Cl-1 ] = 6.000 M, and [ MnCl5-2 ] = 2.750 M. Calculate the value of the equilibrium constant.

b) In another situation, [ Mn+3 ] = 5.000 M and [ MnCl5-2 ] = 2.000 M.
Calculate [ Cl-1 ].

Aqueous lead (II) ions react with aqueous chloride ions forming solid lead (II) chloride. The equation for this precipitation reaction is:

Pb+2 (aq) + 2 Cl-1 (aq)  PbCl2 (s)

Lead (II) ions are added to the solution that has equation #2.

c) Of the three chemicals in equation #2, which one of those chemicals will have its concentration directly affected by addition of lead (II) ions?

d) For that chemical in your answer to part ‘c’, is it a reactant or a product in equation #2?

e) For that chemical in your answer to part ‘c’, will its concentration increase or decrease with addition of lead (II) ions?

f) Based upon your answer from part ‘e’, which way will the equilibrium in equation #2 shift with addition of lead (II) ions?

g) Based upon your answer from part ‘f’, will the red color intensify and the yellow color fade, or will the red color fade and the yellow color intensify?

h) The solution that has equation #2 is placed in an ice water bath to lower the temperature of the equilibrium. One observes that the red color intensifies and the yellow color fades. Is this an exothermic reaction or an endothermic reaction?

Please answer c-h

Your answer is a just took the test

Answers are 1,2,4 its correct on edgenuity.

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

Ahigh energy (high speed) particle (alpha or beta or neutron) are shot at the target;