Chemistry, 16.07.2019 09:30 manarsadi6

Achemical engineer determines the mass percent of iron in an ore sample by converting the fe to fe2+ in acid and then titrating the fe2+ with mno4−. a 1.3909−g sample was dissolved in acid and then titrated with 21.63 ml of 0.04462 m kmno4. the balanced equation is

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A chemical engineer determines the mass percent of iron in an ore sample by converting the Fe to $Fe^{2+}$ in acid and then titrating the $Fe^{2+}$ with Mn $O^{4-}$ .
A 1.3909 g sample was dissolved in acid and then titrated with 21.63 mL of 0.03709 M KMn $O^{4-}$ . The balanced equation is given below.
8 $H^{+}$ (aq) + 5 $Fe^{2+}$ (aq) + Mn $O^{4-}$ (aq) → 5 $Fe^{3+}$ (aq) + $Mn^{+2}$ +(aq) + 4 $H_{2}O$ (l)
Calculate the mass percent of iron in the ore.
______%

Calculation : Given :- weight of sample - 1.3909 g,
Moles of KMn $O^{4-}$ - 0.04462 m
Volume of KMn $O^{4-}$ - 21.63 mL

On solving we get,
(21.63 mL KMn $O^{4-}$ / 1.3909g sample) x (1L / 1000mL) x (0.04462mol KMn $O^{4-}$ / L KMn $O^{4-}$ ) x (5 mol Fe2+ / 1 mol KMn $O^{4-}$ ) x (55.845g $Fe^{2+}$ / mol $Fe^{2+}$ ) x 100% = 19.37%

So the concentration of the ore is 19.37%

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.

Answer from: Quest

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

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Achemical engineer determines the mass percent of iron in an ore sample by converting the fe to fe2+...