# 19.1 Occurrence, preparation, and properties of transition metals  (Page 12/27)

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Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A 2.5000-g sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 M Na 2 Cr 2 O 7 is required in the titration. What percentage of the ore sample was iron?

2.57%

How many cubic feet of air at a pressure of 760 torr and 0 °C is required per ton of Fe 2 O 3 to convert that Fe 2 O 3 into iron in a blast furnace? For this exercise, assume air is 19% oxygen by volume.

Find the potentials of the following electrochemical cell:

Cd | Cd 2+ , M = 0.10 ‖ Ni 2+ , M = 0.50 | Ni

0.167 V

A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?

The standard reduction potential for the reaction ${\left[\text{Co}{\left({\text{H}}_{2}\text{O}\right)}_{6}\right]}^{3+}\left(aq\right)+{\text{e}}^{-}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\left[{\text{Co}\left(\text{H}}_{2}\text{O}{\right)}_{6}\right]}^{2+}\left(aq\right)$ is about 1.8 V. The reduction potential for the reaction ${\left[\text{Co}{\left({\text{NH}}_{3}\right)}_{6}\right]}^{3+}\left(aq\right)+{\text{e}}^{-}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\left[\text{Co}{\left({\text{NH}}_{3}\right)}_{6}\right]}^{2+}\left(aq\right)$ is +0.1 V. Calculate the cell potentials to show whether the complex ions, [Co(H 2 O) 6 ] 2+ and/or [Co(NH 3 ) 6 ] 2+ , can be oxidized to the corresponding cobalt(III) complex by oxygen.

E ° = −0.6 V, E ° is negative so this reduction is not spontaneous. E ° = +1.1 V

Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)

(a) ${\text{MnCO}}_{3}\left(s\right)+\text{HI}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶$

(b) $\text{CoO}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶$

(c) $\text{La}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶$

(d) $\text{V}\left(s\right)+{\text{VCl}}_{4}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶$

(e) $\text{Co}\left(s\right)+{xs\text{F}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶$

(f) ${\text{CrO}}_{3}\left(s\right)+\text{CsOH}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶$

Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)

(a) $\text{Fe}\left(s\right)+{\text{H}}_{2}{\text{SO}}_{4}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶$

(b) ${\text{FeCl}}_{3}\left(aq\right)+\text{NaOH}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶$

(c) $\text{Mn}{\left(\text{OH}\right)}_{2}\left(s\right)+\text{HBr}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶$

(d) $\text{Cr}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶$

(e) ${\text{Mn}}_{2}{\text{O}}_{3}\left(s\right)+\text{HCl}\left(aq\right)⟶$

(f) $\text{Ti}\left(s\right)+xs{\text{F}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶$

(a) $\text{Fe}\left(s\right)+2{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{SO}}_{4}{}^{2-}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Fe}}^{2+}\left(aq\right)+{\text{SO}}_{4}{}^{2-}\left(aq\right)+{\text{H}}_{2}\left(g\right)+2{\text{H}}_{2}\text{O}\left(l\right);$ (b) ${\text{FeCl}}_{3}\left(aq\right)+{\text{3Na}}^{\text{+}}\left(aq\right)+{\text{3OH}}^{\text{−}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\text{Fe}{\left(\text{OH}\right)}_{3}\left(s\right)+{\text{3Na}}^{\text{+}}\left(aq\right)+{\text{3Cl}}^{\text{+}}\left(aq\right);$ (c) $\text{Mn}{\left(\text{OH}\right)}_{2}\left(s\right)+2{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{2Br}}^{\text{−}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Mn}}^{2+}\left(aq\right)+{\text{2Br}}^{\text{−}}\left(aq\right)+4{\text{H}}_{2}\text{O}\left(l\right);$ (d) $\text{4Cr}\left(s\right)+3{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{Cr}}_{2}{\text{O}}_{3}\left(s\right);$ (e) ${\text{Mn}}_{2}{\text{O}}_{3}\left(s\right)+6{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{6Cl}}^{\text{−}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{MnCl}}_{3}\left(s\right)+9{\text{H}}_{2}\text{O}\left(l\right);$ (f) $\text{Ti}\left(s\right)+xs{\text{F}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{TiF}}_{4}\left(g\right)$

Describe the electrolytic process for refining copper.

Predict the products of the following reactions and balance the equations.

(a) Zn is added to a solution of Cr 2 (SO 4 ) 3 in acid.

(b) FeCl 2 is added to a solution containing an excess of ${\text{Cr}}_{2}{\text{O}}_{7}{}^{2-}$ in hydrochloric acid.

(c) Cr 2+ is added to ${\text{Cr}}_{2}{\text{O}}_{7}{}^{2-}$ in acid solution.

(d) Mn is heated with CrO 3 .

(e) CrO is added to 2HNO 3 in water.

(f) FeCl 3 is added to an aqueous solution of NaOH.

(a)
$\begin{array}{}\\ \\ {\text{Cr}}_{2}{\left({\text{SO}}_{4}\right)}_{3}\left(aq\right)+\text{2Zn}\left(s\right)+{\text{2H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2Zn}}^{2+}\left(aq\right)+{\text{H}}_{2}\left(g\right)+{\text{2H}}_{2}\text{O}\left(l\right)+{\text{2Cr}}^{2+}\left(aq\right)+{\text{3SO}}_{4}{}^{2-}\left(aq\right);\end{array}$ (b) ${\text{4TiCl}}_{3}\left(s\right)+{\text{CrO}}_{4}{}^{2-}\left(aq\right)+{\text{8H}}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{4Ti}}^{4+}\left(aq\right)+\text{Cr}\left(s\right)+{\text{4H}}_{2}\text{O}\left(l\right)+{\text{12Cl}}^{\text{−}}\left(aq\right);$ (c) In acid solution between pH 2 and pH 6, ${\text{CrO}}_{4}{}^{2-}$ forms ${\text{HCrO}}_{4}{}^{\text{−}},$ which is in equilibrium with dichromate ion. The reaction is ${\text{2HCrO}}_{4}{}^{\text{−}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Cr}}_{2}{\text{O}}_{7}{}^{2-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right).$ At other acidic pHs, the reaction is ${\text{3Cr}}^{2+}\left(aq\right)+{\text{CrO}}_{4}{}^{2-}\left(aq\right)+{\text{8H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{4Cr}}^{3+}\left(aq\right)+{\text{12H}}_{2}\text{O}\left(l\right);$ (d) ${\text{8CrO}}_{3}\left(s\right)+\text{9Mn}\left(s\right)\phantom{\rule{0.2em}{0ex}}\stackrel{\text{Δ}}{⟶}\phantom{\rule{0.2em}{0ex}}{\text{4Cr}}_{2}{\text{O}}_{3}\left(s\right)+{\text{3Mn}}_{3}{\text{O}}_{4}\left(s\right);$ (e) $\text{CrO}\left(s\right)+{\text{2H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{2NO}}_{3}{}^{\text{−}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Cr}}^{2+}\left(aq\right)+{\text{2NO}}_{3}{}^{\text{−}}\left(aq\right)+{\text{3H}}_{2}\text{O}\left(l\right);$ (f) ${\text{CrCl}}_{3}\left(s\right)+\text{3NaOH}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Cr}{\left(\text{OH}\right)}_{3}\left(s\right)+{\text{3Na}}^{\text{+}}\left(aq\right)+{\text{3Cl}}^{\text{−}}\left(aq\right)$

What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid?

Predict the products of each of the following reactions and then balance the chemical equations.

(a) Fe is heated in an atmosphere of steam.

(b) NaOH is added to a solution of Fe(NO 3 ) 3 .

(c) FeSO 4 is added to an acidic solution of KMnO 4 .

(d) Fe is added to a dilute solution of H 2 SO 4 .

(e) A solution of Fe(NO 3 ) 2 and HNO 3 is allowed to stand in air.

(f) FeCO 3 is added to a solution of HClO 4 .

(g) Fe is heated in air.

(a) $\text{3Fe}\left(s\right)+{\text{4H}}_{2}\text{O}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Fe}}_{3}{\text{O}}_{4}\left(s\right)+{\text{4H}}_{2}\left(g\right);$ (b) $\text{3NaOH}\left(aq\right)+\text{Fe}{\left({\text{NO}}_{3}\right)}_{3}\left(aq\right)\phantom{\rule{0.2em}{0ex}}\stackrel{\phantom{\rule{0.5em}{0ex}}{\text{H}}_{2}\text{O}\phantom{\rule{0.5em}{0ex}}}{\to }\phantom{\rule{0.2em}{0ex}}\text{Fe}{\left(\text{OH}\right)}_{3}\left(s\right)+{\text{3Na}}^{\text{+}}\left(aq\right)+3{\text{NO}}_{3}{}^{\text{−}}\left(aq\right);$ (c) $\begin{array}{}\\ \\ \\ \text{MnO}{}^{4-}+5\text{Fe}{\text{2+}}^{}+8\text{H}{\text{+}}^{}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Mn}{}^{\text{2+}}+5{\text{Fe}}_{3}+4{\text{H}}_{2}\text{O}\end{array};$ (d) $\text{Fe}\left(s\right)+{\text{2H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{SO}}_{4}{}^{2-}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Fe}}^{2+}\left(aq\right)+{\text{SO}}_{4}{}^{2-}\left(aq\right)+{\text{H}}_{2}\left(g\right)+{\text{2H}}_{2}\text{O}\left(l\right);$ (e) ${\text{4Fe}}^{2+}\left(aq\right)+{\text{O}}_{2}\left(g\right)+{\text{4HNO}}_{3}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{4Fe}}^{3+}\left(aq\right)+{\text{2H}}_{2}\text{O}\left(l\right)+{\text{4NO}}_{3}{}^{\text{−}}\left(aq\right);$ (f) ${\text{FeCO}}_{3}\left(s\right)+{\text{2HClO}}_{4}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Fe}{\left({\text{ClO}}_{4}\right)}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)+{\text{CO}}_{2}\left(g\right);$ (g) $\text{3Fe}\left(s\right)+{\text{2O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}\stackrel{\phantom{\rule{0.2em}{0ex}}\text{Δ}\phantom{\rule{0.2em}{0ex}}}{⟶}\phantom{\rule{0.2em}{0ex}}{\text{Fe}}_{3}{\text{O}}_{4}\left(s\right)$

Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state.
$\text{Co}{\left({\text{NO}}_{3}\right)}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Co}}_{2}{\text{O}}_{3}\left(s\right)+{\text{NO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)$

Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility.

${\text{Ag}}^{\text{+}}\left(aq\right)+{\text{CN}}^{\text{−}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{AgCN}\left(s\right)$
$\begin{array}{l}{\text{Ag}}^{\text{+}}\left(aq\right)+2{\text{CN}}^{\text{−}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\left[\text{Ag}{\text{(CN)}}_{2}\right]}^{\text{−}}\left(aq\right)\\ \text{AgCN}\left(s\right)+{\text{CN}}^{\text{−}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\left[\text{Ag}{\text{(CN)}}_{2}\right]}^{\text{−}}\left(aq\right)\end{array}$

Predict which will be more stable, [CrO 4 ] 2− or [WO 4 ] 2− , and explain.

Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula M 3 O 4 are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format MO·M 2 O 3 , to permit estimation of the metal’s two oxidation states.)

(a) Sc 2 O 3

(b) TiO 2

(c) V 2 O 5

(d) CrO 3

(e) MnO 2

(f) Fe 3 O 4

(g) Co 3 O 4

(h) NiO

(i) Cu 2 O

(a) Sc 3+ ; (b) Ti 4+ ; (c) V 5+ ; (d) Cr 6+ ; (e) Mn 4+ ; (f) Fe 2+ and Fe 3+ ; (g) Co 2+ and Co 3+ ; (h) Ni 2+ ; (i) Cu +

what is a balanced equation 4 trioxonitrate (V)acid and sodium hydroxide?
proved ur Worth: If A is a of trioxonitrate(V)acid,HNO3' of unknown concentration .B is a standard solution of sodium hydroxide containing 4.00g per dm cube of solution.25cm cube portions solution B required an average of 24.00cm cube of solution A for neutralization,using 2 drops of methyl orange.
Marcel
calculate the concentration of solution B in moles per dm cube
Marcel
calculate the concentration of solution A and B in moles per DM cube
Marcel
finally calculate the concentration in g/dm cube of HNO3 in solution A (H=1,N=14,O=16,Na=23)
Marcel
wat is electrolysis?
it is the chemical decomposition of a substance when electric current is passed through it either in molten form or aqueous solution
Nuru
list the side effect of chemical industries
how do you ionise an atom
many ways ,but one of them is when the atom becomes heated to a certain temperature the surface electron becomes too energetic and leaves the atom because the attraction between the nucleus and the electron becomes overpowered by the energetic eletron
sunday
also hitting of two atoms can cause transfer of surface electrons
sunday
and when this transfers occur the atom becomes ionised
sunday
who is doing Cape chemistry tomorrow?
What is hybridization
the mix between different breeds of species in one
Jared
it is the blending of orbitals.
stanley
the mixing of orbital
caramel
are covalent bonds influenced by factors such as temperature and pressure?
what is catalyst used for mirror test
when an atom looses electron, what does it become?
it's oxidized and called an ion
Anora
thanks
Abdullahi
Now, I get it
Abdullahi
cation
Anora
can you give an example please, if you don't mind
Abdullahi
a positive ion,become positively charged/a cation.
Janis
sodium plus one is simple cation is exmpl
ajmal
Taking Sodium as example..... it carries a positive charge which means it is positively charged.....when it gains an electron, it is reduced cuz an electron is negatively charged.....also when an atom looses an electron, it becomes positively charged and when it gains, it becomes negatively charged.
Nuru
typically, ionization is the process where an atom looses or gains electron(s) to form ion(s) either a positively or negatively
Nuru
what is copper
just an element
Power
Cu
daniel
Why is water a single covalent bond?
nitrogen is a gas whereas phosphorus is solid .Explain.
can you explain what you are needing it now better than maybe I'm just not interpreting it what you're needing to know
Alex
cool nitrogen down to around negative 270 °F and it will be solid. now they are both solid
daniel
they are different elements and dats how they are pal.....check the periodic table
Nuru
Nitrogen is a diatomic molecule with relatively weak van de waals forces between the molecules. These forces are overcome when the solid melts or liquid evaporates. Phosphorus forms larger molecules consisting of four phosphorus atoms in a tetradedral shape. The intermolecular forces are stronger
Paul
whats a base
A base is a substance which will neutralize an acid to yield salt and water only
Zainab
base is a substance that produces OH(aq) ions in aqueous solution. Strong soluable bases are in water and are completely dislocated. Therefore weak base ionize slightly...
Roy
a base is a substance that neutralise and acid to form salt and water
Daksalma
write electrolysis of bright solution using either carbon or platinum and write the reaction at the anode or at the cathode
what is the H3O of a solution with the pH of 2.5
pH<7, therefore there are only H3O+HX3OX+particles in the solution. [H3O+]=10−pH=10−6.99=1.02⋅10−7[HX3OX+]=10−pH=10−6.99=1.02⋅10−7 When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking  and I'll give you another one if this is ki
Alex
Alex
When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking
Alex
sorry I don't know why that sent again
Alex
We have [H3O+]=10−pH=10−6.99=1.02⋅10−7[HX3OX+]=10−pH=10−6.99=1.02⋅10−7 and [OH−]=10−pOH=10−7.01=9.77⋅10−8[OHX−]=10−pOH=10−7.01=9.77⋅10−8.  Because of H3O++OH−⟶2H2OHX3OX++OHX−⟶2HX2O we are left with [H3O+]=1.02⋅10−7−9.77⋅10−8=4.6⋅10−9
Alex