Suppose you start with 1.80 g of mgco3·5h2o, after complete dehydration how many grams of the sample do you have remaining?

The  number of  grams of the sample that remained  after  hydration is 0.87  grams

                calculation
calculate the molar  mass of MgCO3. 5 H2O  which is

24 + 12+ (3 x16) + ( 5  x18)  =  174 g/mol

we  well  know  after  hydration  water evaporate leaving behind   MgCO3
therefore the  mass  remained  is  for  MgCO3

find  the   molar  mass of  MgCO3 =  24 +12 +( 3 x16)  = 84  g/mol

the  mass  remained is therefore

 =  molar  mass of MgCO3 /  molar mass of MgCO3.5H2O  x 1.80 g

=  84g/mol / 174 g/mol  x1.8g = 0.87  grams

Idk the answer sorry

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.