Mole word problems
the recommended daily allowance of selenium in your diet is 6.97x10-7 mole. how many atoms of selenium is this?

a bracelet containing 0.200 mol metal is 75% gold. how many atoms of gold are in the bracelet?

which has more atoms, 10.0 g c or 10.0 g of ca? how many atoms does each contain?
a mixture contains 0.250 mol fe and 1.20 g c. what is the total number of atoms in the mixture?

air contains several gases. argon makes up 0.934% of the air. if a person takes a breath that contains 0.600 g of air, calculate the number of argon atoms inhaled.

a chemist needs 0.015 mol of iodine crystals, i2, for an experiment. what mass of iodine crystals should the student obtain?

bonus problem
one atom of an unknown element has a mass of 6.66x10-23 g. what is the identity of this element?

Q: The recommended daily allowance of selenium in your diet is 6.97x10-7 mole. How many atoms of selenium is this?

                  4.19 × 10¹⁷ Se Atoms

Solution:

                     As we know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of particles (atoms) contained by 6.97 × 10⁻⁷ moles of Se, we will use following relation,

           Moles  =  Number of Atoms ÷ 6.022 × 10²³ Atoms.mol⁻¹

Solving for Number of Atoms,

           Number of Atoms  =  Moles × 6.022 × 10²³ Atoms.mol⁻¹

Putting values,

           Number of Atoms  =  6.97 × 10⁻⁷ mol × 6.022 × 10²³ Atoms.mol⁻¹

           Number of Atoms  =  4.19 × 10¹⁷ Atoms

Q: A bracelet containing 0.200 mol metal is 75% gold. How many atoms of gold are in the bracelet?

                  9.033 × 10²² Atoms of Gold

Solution:

                     As we know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of particles (atoms) contained by 0.200 moles of Au, we will use following relation,

           Moles  =  Number of Atoms ÷ 6.022 × 10²³ Atoms.mol⁻¹

Solving for Number of Atoms,

           Number of Atoms  =  Moles × 6.022 × 10²³ Atoms.mol⁻¹

Putting values,

           Number of Atoms  =  0.200 mol × 6.022 × 10²³ Atoms.mol⁻¹

           Number of Atoms  =  1.20 × 10²³ Atoms

For 75%,

           Number of Atoms  =  1.20 × 10²³ Atoms × 0.75

           Number of Atoms  =  9.033 × 10²² Atoms of Gold

Q: Which has more atoms, 10.0 g C or 10.0 g of Ca? How many atoms does each contain?

               Carbon has greater number of Atoms.

Solution:

Data Given:

                 Mass of Carbon =  10 g

                 M.Mass of Carbon  =  12.01 g.mol⁻¹

                 Mass of Calcium =  10 g

                 M.Mass of Calcium  =  40.08 g.mol⁻¹

Step 1: Calculate Moles of Carbon as,

                 Moles  =  Mass ÷ M.Mass

Putting values,

                 Moles  =  10 g ÷ 12.01 g.mol⁻¹

                 Moles  =  0.8326 mol

Step 2: Calculate number of Carbon atoms,

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Carbon atoms can be written as,

                 Moles  =  Number of C Atoms ÷ 6.022 × 10²³ Atoms.mol⁻¹

Solving for Number of C atoms,

                 Number of C atoms  =  Moles × 6.022 × 10²³ Atoms.mol⁻¹

Putting value of moles,

                 Number of C atoms  =  0.8326 mol × 6.022 × 10²³ Atoms.mol⁻¹

                 Number of C atoms =  5.01 × 10²³ C atoms

Step 3: Calculate Moles of Calcium as,

                 Moles  =  Mass ÷ M.Mass

Putting values,

                 Moles  =  10 g ÷ 40.08 g.mol⁻¹

                 Moles  =  0.25 mol

Step 4: Calculate number of Calcium atoms,

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Calcium atoms can be written as,

                 Moles  =  Number of Ca Atoms ÷ 6.022 × 10²³ Atoms.mol⁻¹

Solving for Number of Ca atoms,

                 Number of Ca atoms  =  Moles × 6.022 × 10²³ Atoms.mol⁻¹

Putting value of moles,

                 Number of Ca atoms  =  0.25 mol × 6.022 × 10²³ Atoms.mol⁻¹

                 Number of Ca atoms =  1.50 × 10²³ C atoms

Q: A mixture contains 0.250 mol Fe and 1.20 g C. What is the total number of atoms in the mixture?

                 2.10 × 10²³ Atoms in Mixture

Solution:

Atoms in Fe:

                  As we know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of particles (atoms) contained by 0.25 moles of Fe, we will use following relation,

           Moles  =  Number of Atoms ÷ 6.022 × 10²³ Atoms.mol⁻¹

Solving for Number of Atoms,

           Number of Atoms =  Moles × 6.022 × 10²³ Atoms.mol⁻¹

Putting values,

           Number of Molecules  =  0.25 mol × 6.022 × 10²³ Atoms.mol⁻¹

           Number of Molecules  =  1.50 × 10²³ Atoms of Fe

Atoms in C:

Step 1: Calculate Moles of Carbon as,

                 Moles  =  Mass ÷ M.Mass

Putting values,

                 Moles  =  1.20g ÷ 12.01 g.mol⁻¹

                 Moles  =  0.0999 mol

Step 2: Calculate number of Carbon atoms,

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Carbon atoms can be written as,

                 Moles  =  Number of C Atoms ÷ 6.022 × 10²³ Atoms.mol⁻¹

Solving for Number of C atoms,

                 Number of C atoms  =  Moles × 6.022 × 10²³ Atoms.mol⁻¹

                 Number of C atoms  =  0.0999 × 6.022 × 10²³ Atoms.mol⁻¹

                 Number of C atoms  =  6.01 × 10²² C Atoms

Total Number of Atoms:

  =  6.01 × 10²² C Atoms + 1.50 × 10²³ Atoms of Fe

  =  2.10 × 10²³ Atoms in Mixture

Remaining answers are attached.

Carbon based or cells?

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.