Tag: limestone

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

Mg and Zn have following resemblance :

  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.
Reveal answer Fill a bubble to check yourself
B,D Correct answer
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.

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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
  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. $(A)\rightarrow (iii), (B)\rightarrow (iv), (C)\rightarrow (ii), (D)\rightarrow (i)$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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?
$[Be(H _2O) _4]^{2+} , [Mg(H _2O) _4]^{2+} , Ca(H _2O) _4]^{2+} and          Sr(H _2O)^4]^{2+}$

  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. $[Be(H _2O) _4]^{2+} > [Mg(H _2O) _4]^2 > [Ca(H _2O) _4]^{2+} > Sr(H _2O) _4]^{2+}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 

$[Be(H _2O) _4]^{2+} >[Mg(H _2O _4]^{2+} > [Ca(H _2O) _4]^{2+} > [Sr(H _2O) _4]^{2+}$
Multiple choice limestone uses of group 2 compounds group 2 industrial inorganic chemistry chemistry

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: 

  1. $MgBa(CO _3) _4$
  2. $MgBa(CO _3) _2$
  3. $MgBa(CO _3) _3$
  4. $Mg _2Ba(CO _3) _4$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $MgBa(CO _3) _2$.