The iodine clock experiment consists of the following three reactions, where reaction 1 is slow relative to reactions 2 and 3. Predict what would happen if reaction 2 was slow relative to reaction 1 (assume that reaction 3 is still fast). Reaction 1:3I^- (aq) + S_2O_8^2- (aq) right arrow I_3^-, (aq) + 2SO_4^2- (aq) slow
Reaction 2: l_3^-(aq) + 2S_2O_3^2- (aq) right arrow 3I^- (aq) + S_4O_6^2- (aq) fast
Reaction 3: l_3^- (aq) + starch (aq) right arrow 3I^ starch (bluish black) fast
A. A color change would appear immediately because I_3^+, would be removed slowly from solution, and would therefore be able to react with starch.
B. The reaction would turn blue slower because I_3^-, is removed at a slower rate.
C. The rate of color change would indicate the rate of reaction 3 instead of reaction 1.
D. Reaction 3 would not occur.

C. 1s22s22p63s23p43d104s24p5

Ch4 and ph3 are both covalent since they are composed both of non-metals.    ch4 is non- polar because of  london forces and has a geometry of tetrahedral  ph3 , on the other hand, is umbrella shaped polar and has dipole-dipole intermolecular force (imf) . since ph3 is polar, it has a stronger imf   

Answer two atoms of oxygen;

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