Tag: limestone

Questions Related to limestone

Mg and Zn have following resemblance :

limestone uses of group 2 compounds group 2 industrial inorganic chemistry chemistry

  1. MgO and ZnO are amphoteric.

  2. $MgCO _3$ and $ZnCO _3$ both on heating give corresponding oxide.

  3. both are d-block elements.

  4. both are used to prevent corrosion.

Show answer
Correct Option: B,D
Explanation:

A) MgO is basic oxide and ZnO is acidic oxide.

B) $MgCO _3$ and $ZnCO _3$ both on heating give corresponding oxide(MgO and ZnO) carbon dioxide.

C) Mg is s block and Zn is d block element,both are not d block

D) both are used to prevent corrosion(Corrosion is the gradual destruction of 
materials (usually metals) by chemical reaction with their environment. In the most common use of the word, this means electrochemical oxidation of metals in reaction with an oxidant such as oxygen).

Hence options  B & D are correct.

State True or False.
Tricalcium phosphate is used as a nutritional supplement and occurs naturally in cow milk.

limestone uses of group 2 compounds group 2 industrial inorganic chemistry chemistry

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Tricalcium phosphate is used as a nutritional supplement and occurs naturally in cow milk.
Thus, the given statement is true.
However commonly used supplements include calcium carbonate and calcium citrate.
Calcium carbonate should be taken with food and calcium citrate should be taken without food.

Kieserite is not an ore of:

limestone uses of group 2 compounds group 2 industrial inorganic chemistry chemistry

  1. Cu

  2. Fe

  3. Mg

  4. Al

Show answer
Correct Option: A,B,D
Explanation:

Kieserite is the magnesium sulphate.


Kieserite is not the ore of Cu,Fe & Al.Its the ore of Mg.

Hence options A,B & D are correct.

Lithopone is mixture of :

limestone uses of group 2 compounds group 2 industrial inorganic chemistry chemistry

  1. $ZnCO _3, BaCO _3$

  2. $ZnS, Na _2SO _4$

  3. $ZnSO _4, BaSO _4$

  4. $ZnS, BaSO _4$

Show answer
Correct Option: D
Explanation:

Lithopone, is Pigment White 5, is a mixture of inorganic compounds barium sulfate and zinc sulfide $(BaSO _4 , ZnS)$, is a widely used as a white pigment powder. 

Match the column I with column II and mark the appropriate choice.

Column I Column II
(A) $Na^+$ (i) Chlorophyll
(B) $K^+$ (ii) Bones and teeth
(C)  $Ca^{2+}$ (iii)  Regulating flow of water across cell membrane
(D) $Mg^{2+}$ (iv) Activation of enzyme in this cell fluids
limestone uses of group 2 compounds group 2 industrial inorganic chemistry chemistry

  1. $(A)\rightarrow (i), (B)\rightarrow (iii), (C)\rightarrow (ii), (D)\rightarrow (iv)$

  2. $(A)\rightarrow (iv), (B)\rightarrow (iii), (C)\rightarrow (ii), (D)\rightarrow (i)$

  3. $(A)\rightarrow (i), (B)\rightarrow (ii), (C)\rightarrow (iii), (D)\rightarrow (iv)$

  4. $(A)\rightarrow (iii), (B)\rightarrow (iv), (C)\rightarrow (ii), (D)\rightarrow (i)$

Show answer
Correct Option: D
Explanation:

Uses of alkaline earth metals:

$Na^+$ regulates the flow of water across the cell membrane by the process of diffusion.
$K^+$ helps in activation of enzyme in this cell fluids.
$Ca^{2+}$is present in our bones and teeth.
$Mg^{2+}$ is bound as the central atom of the porphyrin ring of the green plant pigment, chlorophyll.

Nuclear attraction is often the deciding control factor for the association of natural molecules to
a given metal ion. Which one of the following represents the correct order of stability of the ions?
$[Be(H _2O) _4]^{2+} , [Mg(H _2O) _4]^{2+} , Ca(H _2O) _4]^{2+} and          Sr(H _2O)^4]^{2+}$

limestone uses of group 2 compounds group 2 industrial inorganic chemistry chemistry

  1. $[Be(H _2O) _4]^{2+} > Sr(H _2O) _4]^{2+}] > [Mg(H _2O) _4]^{2+} > [Ca(H _2O)^4]^{2+}$

  2. $[Ca(H _2O) _4]^{2+} > [Mg(H _2O) _4]^{2+} > [Be(H _2O) _4]^{2+} > Sr(H _2O) _4]^{2+}$

  3. $[Sr(H _2O) _4]^{2+} > [Ca(H _2O) _4]^{2+} > [Mg(H _2O) _4]^{2+} > [Be(H _2O) _4]^{2+}$

  4. $[Be(H _2O) _4]^{2+} > [Mg(H _2O) _4]^2 > [Ca(H _2O) _4]^{2+} > Sr(H _2O) _4]^{2+}$

Show answer
Correct Option: D
Explanation:
The degree of hydration and the amount of hydration energy decreases as the size of the ion increases from   Be^{2+} to Sr^{2+}. 

                      $Be^{2+}       <      Mg^{2+}     <    Ca^{2+}    <     Sr^{2+}$

Hydration   -2494        - 1921          -1577         -1443 

Ergy (kJ moI$^{-1})$ 

Thus, stability of hydrated ion is 

$[Be(H _2O) _4]^{2+} >[Mg(H _2O _4]^{2+} > [Ca(H _2O) _4]^{2+} > [Sr(H _2O) _4]^{2+}$

Based on the following analytical data, answer the given question.
A mineral, which can be represented by the formula $Mg _xBa _y(CO _3) _2$, was analyzed as described below:

A sample of the mineral was dissolved in excess hydrochloric acid and the solution made up to $100 cm^3$ with water. During the process, $48 cm^3$ of carbon dioxide, measured at $25^{\circ}C$ and 1-atmosphere pressure, were evolved.
A $25.0 cm^3$ portion of the resulting solution required $25.0 cm^3$ of EDTA solution of concentration $0.02 \ mol / dm^3$ to reach an end-point. A further $25.0 cm^3$ portion gave a precipitate of barium sulphate of mass 0.058 g on treatment with excess dilute sulphuric acid. You may assume that group-2 metal ions form 1:1 complexes with EDTA. Molar volume of any gas at $25^{\circ}C$ and 1 atmosphere pressure $ = 24 dm^3$).

The formula of the mineral is: 

limestone uses of group 2 compounds group 2 industrial inorganic chemistry chemistry

  1. $MgBa(CO _3) _4$

  2. $MgBa(CO _3) _2$

  3. $MgBa(CO _3) _3$

  4. $Mg _2Ba(CO _3) _4$

Show answer
Correct Option: B
Explanation:

The formula of the mineral is $MgBa(CO _3) _2$.

$48 cm^3$ of carbon dioxide corresponds to 0.002 moles.
Thus, $MgBa(CO _3) _2$ reacts with excess HCl to give 0.002 moles of carbon dioxide.

So $100.0cm^3$ solution will give 0.002 moles of carbon dioxide.
$25.0cm^3$ solution will give 0.0005 moles of carbon dioxide.

A $25.0cm^3$ portion of the resulting solution required $25.0cm^3$ of EDTA solution of concentration $0.02mol.dm^{-3}$ to reach an end-point.

Thus $25.0cm^3$ portion of the resulting solution contains $0.02 \times \frac {25.0}{1000}=0.0005 mol$ of metal ions.

A further $25.0cm^3$ portion gave a precipitate of barium sulphate of mass 0.058 g on treatment with excess dilute sulphuric acid.

0.058 g of barium sulphate corresponds to $\frac {0.058 g}{233.43  g/mol}=0.00025\  mol$.

$\therefore $  Moles of $Ba^{2+}$ ion $=$ Moles of $Mg^{2+}$ ion $=0.00025\ mol$  

So, formula of the compound is $MgBa(CO _3) _2$.