228728
$\mathrm{l}\left(\mathrm{FeO} \cdot \mathrm{Cr}_2 \mathrm{O}_3\right)+\mathrm{mNa}_2 \mathrm{CO}_3+\mathrm{nO}_2 \times \mathrm{xNa}_2 \mathrm{CrO}_4+$ $\mathrm{yFe}_2 \mathrm{O}_3+\mathrm{zCO}_2$ $1, m, n, x, y$ and $z$ respectively are
1 $4,8,7,8,2$ and 8
2 $4,8,7,8,8$ and 2
3 $8,8,8,2,4$ and 7
4 2, 4,7,8, 8 and 8
Explanation:
The given reaction is : $\begin{aligned} & \mathrm{l}\left(\mathrm{FeO} \cdot \mathrm{Cr}_2 \mathrm{O}_3\right)+\mathrm{mNa}_2 \mathrm{CO}_3+\mathrm{nO}_2 \longrightarrow \mathrm{xNa}_2 \mathrm{CrO}_4+ \\ & \mathrm{yFe} \mathrm{O}_3+\mathrm{zCO}_2 \\ & \text { after balancing the equation we get - } \\ & 1=4, \mathrm{~m}=8, \mathrm{n}=7, \mathrm{x}=8, \mathrm{y}=2 \text { and } \mathrm{z}=8 \\ & \therefore 4\left(\mathrm{FeO} \cdot \mathrm{Cr}_2 \mathrm{O}_3\right)+8 \mathrm{Na}_2 \mathrm{CO}_3+7 \mathrm{O}_2 \longrightarrow \mathrm{Na}_2 \mathrm{CrO}_4+ \\ & 2 \mathrm{Fe}_2 \mathrm{O}_3+8 \mathrm{CO}_2 \end{aligned}$
SRMJEEE - 2011
Some Basic Concepts of Chemistry
228730
$0.14 \mathrm{~g}$ of an element on combustion gives 0.28 gm of its oxide. What is that element?
1 Nitrogen
2 Carbon
3 Fluorine
4 Sulphur
Explanation:
The formula used to calculate the equivalent weight of an element in its oxide : The equivalent weight of element $=\frac{v}{v}$ equivalent wt. of $\mathrm{O}$ atom. The equivalent wt. of $\mathrm{O}$ atom $=\frac{16}{2}=8 \mathrm{~g}$ Equivalent weight of element $=\frac{\text { weight of element }}{\text { weight of oxygen }} \times 8$ $\begin{aligned} & =\frac{0.14}{0.28-0.14} \times 8 \\ & =1 \times 8 \\ & =8 \mathrm{amu} \end{aligned}$ Also, by the balanced reaction: $\mathrm{S}+\mathrm{O}_2 \longrightarrow \mathrm{SO}_2$ Valency of sulphur $=4$ Equivalent weight of sulphur $=\frac{32}{4} \mathrm{amu}=8 \mathrm{amu}$ So, the element is sulphur.
AP- EAMCET(Medical) -2010
Some Basic Concepts of Chemistry
228731
How many ' $\mathrm{mL}$ ' of perhydrol is required to produce sufficient oxygen which can be used to completely convert $2 \mathrm{~L}$ of $\mathrm{SO}_2$ Gas to $\mathrm{SO}_3$ gas?
1 $10 \mathrm{~mL}$
2 $5 \mathrm{~mL}$
3 $20 \mathrm{~mL}$
4 $30 \mathrm{~mL}$
Explanation:
Perhydrol means $30 \%$ solution of $\mathrm{H}_2 \mathrm{O}_2$. $\mathrm{H}_2 \mathrm{O}_2$ decomposes as $2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2$ Volume strength of $30 \% \mathrm{H}_2 \mathrm{O}_2$ solution is 100 that mean $1 \mathrm{~mL}$ of this solution on decomposition gives $100 \mathrm{~mL}$ oxygen. $\begin{array}{lc} \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow & \mathrm{SO}_3 \\ 1 \mathrm{~L} \quad \frac{1}{2} \mathrm{~L} & 1 \mathrm{~L} \\ 2 \mathrm{~L} \quad 1 \mathrm{~L} & 2 \mathrm{~L} \end{array}$ Since, $100 \mathrm{~mL}$ of oxygen is obtained by $1 \mathrm{~mL}$ of $\mathrm{H}_2 \mathrm{O}_2$ $\therefore 1000 \mathrm{~mL}$ of oxygen will be obtained by $\begin{aligned} & =\frac{1}{100} \times 1000 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 \\ & =10 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 \end{aligned}$
AP EAMCET (Engg.)-2009 VITEEE- 2009
Some Basic Concepts of Chemistry
228736
Law of multiple proportions is illustrated by one of the following pairs:
1 $\mathrm{H}_2 \mathrm{~S}$ and $\mathrm{SO}_2$
2 $\mathrm{NH}_3$ and $\mathrm{NO}_2$
3 $\mathrm{Na}_2 \mathrm{~S}$ and $\mathrm{Na}_2 \mathrm{O}$
4 $\mathrm{N}_2 \mathrm{O}$ and $\mathrm{NO}$
Explanation:
According to the law of multiple proportions that if two element combine to form more than one compound, the fixed amount of one element combining with other element have simple whole number ratio. (i) $\mathrm{H}_2 \mathrm{~S}$ and $\mathrm{SO}_2$ are not example of law of multiple proportion because elements are different (ii) $\mathrm{NH}_3$ and $\mathrm{NO}_2$ are not examples of law of multiple proportion because elements are different (iii) $\mathrm{Na}_2 \mathrm{~S}$ and $\mathrm{Na}_2 \mathrm{O}$ are not $\mathrm{NO}$ here 16 parts of oxygen react with 14 and 28 parts of nitrogen. 14:28 is $1: 2$ which is simple whole number ratio. $\therefore$ It follows law of multiple proportion.
228728
$\mathrm{l}\left(\mathrm{FeO} \cdot \mathrm{Cr}_2 \mathrm{O}_3\right)+\mathrm{mNa}_2 \mathrm{CO}_3+\mathrm{nO}_2 \times \mathrm{xNa}_2 \mathrm{CrO}_4+$ $\mathrm{yFe}_2 \mathrm{O}_3+\mathrm{zCO}_2$ $1, m, n, x, y$ and $z$ respectively are
1 $4,8,7,8,2$ and 8
2 $4,8,7,8,8$ and 2
3 $8,8,8,2,4$ and 7
4 2, 4,7,8, 8 and 8
Explanation:
The given reaction is : $\begin{aligned} & \mathrm{l}\left(\mathrm{FeO} \cdot \mathrm{Cr}_2 \mathrm{O}_3\right)+\mathrm{mNa}_2 \mathrm{CO}_3+\mathrm{nO}_2 \longrightarrow \mathrm{xNa}_2 \mathrm{CrO}_4+ \\ & \mathrm{yFe} \mathrm{O}_3+\mathrm{zCO}_2 \\ & \text { after balancing the equation we get - } \\ & 1=4, \mathrm{~m}=8, \mathrm{n}=7, \mathrm{x}=8, \mathrm{y}=2 \text { and } \mathrm{z}=8 \\ & \therefore 4\left(\mathrm{FeO} \cdot \mathrm{Cr}_2 \mathrm{O}_3\right)+8 \mathrm{Na}_2 \mathrm{CO}_3+7 \mathrm{O}_2 \longrightarrow \mathrm{Na}_2 \mathrm{CrO}_4+ \\ & 2 \mathrm{Fe}_2 \mathrm{O}_3+8 \mathrm{CO}_2 \end{aligned}$
SRMJEEE - 2011
Some Basic Concepts of Chemistry
228730
$0.14 \mathrm{~g}$ of an element on combustion gives 0.28 gm of its oxide. What is that element?
1 Nitrogen
2 Carbon
3 Fluorine
4 Sulphur
Explanation:
The formula used to calculate the equivalent weight of an element in its oxide : The equivalent weight of element $=\frac{v}{v}$ equivalent wt. of $\mathrm{O}$ atom. The equivalent wt. of $\mathrm{O}$ atom $=\frac{16}{2}=8 \mathrm{~g}$ Equivalent weight of element $=\frac{\text { weight of element }}{\text { weight of oxygen }} \times 8$ $\begin{aligned} & =\frac{0.14}{0.28-0.14} \times 8 \\ & =1 \times 8 \\ & =8 \mathrm{amu} \end{aligned}$ Also, by the balanced reaction: $\mathrm{S}+\mathrm{O}_2 \longrightarrow \mathrm{SO}_2$ Valency of sulphur $=4$ Equivalent weight of sulphur $=\frac{32}{4} \mathrm{amu}=8 \mathrm{amu}$ So, the element is sulphur.
AP- EAMCET(Medical) -2010
Some Basic Concepts of Chemistry
228731
How many ' $\mathrm{mL}$ ' of perhydrol is required to produce sufficient oxygen which can be used to completely convert $2 \mathrm{~L}$ of $\mathrm{SO}_2$ Gas to $\mathrm{SO}_3$ gas?
1 $10 \mathrm{~mL}$
2 $5 \mathrm{~mL}$
3 $20 \mathrm{~mL}$
4 $30 \mathrm{~mL}$
Explanation:
Perhydrol means $30 \%$ solution of $\mathrm{H}_2 \mathrm{O}_2$. $\mathrm{H}_2 \mathrm{O}_2$ decomposes as $2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2$ Volume strength of $30 \% \mathrm{H}_2 \mathrm{O}_2$ solution is 100 that mean $1 \mathrm{~mL}$ of this solution on decomposition gives $100 \mathrm{~mL}$ oxygen. $\begin{array}{lc} \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow & \mathrm{SO}_3 \\ 1 \mathrm{~L} \quad \frac{1}{2} \mathrm{~L} & 1 \mathrm{~L} \\ 2 \mathrm{~L} \quad 1 \mathrm{~L} & 2 \mathrm{~L} \end{array}$ Since, $100 \mathrm{~mL}$ of oxygen is obtained by $1 \mathrm{~mL}$ of $\mathrm{H}_2 \mathrm{O}_2$ $\therefore 1000 \mathrm{~mL}$ of oxygen will be obtained by $\begin{aligned} & =\frac{1}{100} \times 1000 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 \\ & =10 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 \end{aligned}$
AP EAMCET (Engg.)-2009 VITEEE- 2009
Some Basic Concepts of Chemistry
228736
Law of multiple proportions is illustrated by one of the following pairs:
1 $\mathrm{H}_2 \mathrm{~S}$ and $\mathrm{SO}_2$
2 $\mathrm{NH}_3$ and $\mathrm{NO}_2$
3 $\mathrm{Na}_2 \mathrm{~S}$ and $\mathrm{Na}_2 \mathrm{O}$
4 $\mathrm{N}_2 \mathrm{O}$ and $\mathrm{NO}$
Explanation:
According to the law of multiple proportions that if two element combine to form more than one compound, the fixed amount of one element combining with other element have simple whole number ratio. (i) $\mathrm{H}_2 \mathrm{~S}$ and $\mathrm{SO}_2$ are not example of law of multiple proportion because elements are different (ii) $\mathrm{NH}_3$ and $\mathrm{NO}_2$ are not examples of law of multiple proportion because elements are different (iii) $\mathrm{Na}_2 \mathrm{~S}$ and $\mathrm{Na}_2 \mathrm{O}$ are not $\mathrm{NO}$ here 16 parts of oxygen react with 14 and 28 parts of nitrogen. 14:28 is $1: 2$ which is simple whole number ratio. $\therefore$ It follows law of multiple proportion.
228728
$\mathrm{l}\left(\mathrm{FeO} \cdot \mathrm{Cr}_2 \mathrm{O}_3\right)+\mathrm{mNa}_2 \mathrm{CO}_3+\mathrm{nO}_2 \times \mathrm{xNa}_2 \mathrm{CrO}_4+$ $\mathrm{yFe}_2 \mathrm{O}_3+\mathrm{zCO}_2$ $1, m, n, x, y$ and $z$ respectively are
1 $4,8,7,8,2$ and 8
2 $4,8,7,8,8$ and 2
3 $8,8,8,2,4$ and 7
4 2, 4,7,8, 8 and 8
Explanation:
The given reaction is : $\begin{aligned} & \mathrm{l}\left(\mathrm{FeO} \cdot \mathrm{Cr}_2 \mathrm{O}_3\right)+\mathrm{mNa}_2 \mathrm{CO}_3+\mathrm{nO}_2 \longrightarrow \mathrm{xNa}_2 \mathrm{CrO}_4+ \\ & \mathrm{yFe} \mathrm{O}_3+\mathrm{zCO}_2 \\ & \text { after balancing the equation we get - } \\ & 1=4, \mathrm{~m}=8, \mathrm{n}=7, \mathrm{x}=8, \mathrm{y}=2 \text { and } \mathrm{z}=8 \\ & \therefore 4\left(\mathrm{FeO} \cdot \mathrm{Cr}_2 \mathrm{O}_3\right)+8 \mathrm{Na}_2 \mathrm{CO}_3+7 \mathrm{O}_2 \longrightarrow \mathrm{Na}_2 \mathrm{CrO}_4+ \\ & 2 \mathrm{Fe}_2 \mathrm{O}_3+8 \mathrm{CO}_2 \end{aligned}$
SRMJEEE - 2011
Some Basic Concepts of Chemistry
228730
$0.14 \mathrm{~g}$ of an element on combustion gives 0.28 gm of its oxide. What is that element?
1 Nitrogen
2 Carbon
3 Fluorine
4 Sulphur
Explanation:
The formula used to calculate the equivalent weight of an element in its oxide : The equivalent weight of element $=\frac{v}{v}$ equivalent wt. of $\mathrm{O}$ atom. The equivalent wt. of $\mathrm{O}$ atom $=\frac{16}{2}=8 \mathrm{~g}$ Equivalent weight of element $=\frac{\text { weight of element }}{\text { weight of oxygen }} \times 8$ $\begin{aligned} & =\frac{0.14}{0.28-0.14} \times 8 \\ & =1 \times 8 \\ & =8 \mathrm{amu} \end{aligned}$ Also, by the balanced reaction: $\mathrm{S}+\mathrm{O}_2 \longrightarrow \mathrm{SO}_2$ Valency of sulphur $=4$ Equivalent weight of sulphur $=\frac{32}{4} \mathrm{amu}=8 \mathrm{amu}$ So, the element is sulphur.
AP- EAMCET(Medical) -2010
Some Basic Concepts of Chemistry
228731
How many ' $\mathrm{mL}$ ' of perhydrol is required to produce sufficient oxygen which can be used to completely convert $2 \mathrm{~L}$ of $\mathrm{SO}_2$ Gas to $\mathrm{SO}_3$ gas?
1 $10 \mathrm{~mL}$
2 $5 \mathrm{~mL}$
3 $20 \mathrm{~mL}$
4 $30 \mathrm{~mL}$
Explanation:
Perhydrol means $30 \%$ solution of $\mathrm{H}_2 \mathrm{O}_2$. $\mathrm{H}_2 \mathrm{O}_2$ decomposes as $2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2$ Volume strength of $30 \% \mathrm{H}_2 \mathrm{O}_2$ solution is 100 that mean $1 \mathrm{~mL}$ of this solution on decomposition gives $100 \mathrm{~mL}$ oxygen. $\begin{array}{lc} \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow & \mathrm{SO}_3 \\ 1 \mathrm{~L} \quad \frac{1}{2} \mathrm{~L} & 1 \mathrm{~L} \\ 2 \mathrm{~L} \quad 1 \mathrm{~L} & 2 \mathrm{~L} \end{array}$ Since, $100 \mathrm{~mL}$ of oxygen is obtained by $1 \mathrm{~mL}$ of $\mathrm{H}_2 \mathrm{O}_2$ $\therefore 1000 \mathrm{~mL}$ of oxygen will be obtained by $\begin{aligned} & =\frac{1}{100} \times 1000 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 \\ & =10 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 \end{aligned}$
AP EAMCET (Engg.)-2009 VITEEE- 2009
Some Basic Concepts of Chemistry
228736
Law of multiple proportions is illustrated by one of the following pairs:
1 $\mathrm{H}_2 \mathrm{~S}$ and $\mathrm{SO}_2$
2 $\mathrm{NH}_3$ and $\mathrm{NO}_2$
3 $\mathrm{Na}_2 \mathrm{~S}$ and $\mathrm{Na}_2 \mathrm{O}$
4 $\mathrm{N}_2 \mathrm{O}$ and $\mathrm{NO}$
Explanation:
According to the law of multiple proportions that if two element combine to form more than one compound, the fixed amount of one element combining with other element have simple whole number ratio. (i) $\mathrm{H}_2 \mathrm{~S}$ and $\mathrm{SO}_2$ are not example of law of multiple proportion because elements are different (ii) $\mathrm{NH}_3$ and $\mathrm{NO}_2$ are not examples of law of multiple proportion because elements are different (iii) $\mathrm{Na}_2 \mathrm{~S}$ and $\mathrm{Na}_2 \mathrm{O}$ are not $\mathrm{NO}$ here 16 parts of oxygen react with 14 and 28 parts of nitrogen. 14:28 is $1: 2$ which is simple whole number ratio. $\therefore$ It follows law of multiple proportion.
228728
$\mathrm{l}\left(\mathrm{FeO} \cdot \mathrm{Cr}_2 \mathrm{O}_3\right)+\mathrm{mNa}_2 \mathrm{CO}_3+\mathrm{nO}_2 \times \mathrm{xNa}_2 \mathrm{CrO}_4+$ $\mathrm{yFe}_2 \mathrm{O}_3+\mathrm{zCO}_2$ $1, m, n, x, y$ and $z$ respectively are
1 $4,8,7,8,2$ and 8
2 $4,8,7,8,8$ and 2
3 $8,8,8,2,4$ and 7
4 2, 4,7,8, 8 and 8
Explanation:
The given reaction is : $\begin{aligned} & \mathrm{l}\left(\mathrm{FeO} \cdot \mathrm{Cr}_2 \mathrm{O}_3\right)+\mathrm{mNa}_2 \mathrm{CO}_3+\mathrm{nO}_2 \longrightarrow \mathrm{xNa}_2 \mathrm{CrO}_4+ \\ & \mathrm{yFe} \mathrm{O}_3+\mathrm{zCO}_2 \\ & \text { after balancing the equation we get - } \\ & 1=4, \mathrm{~m}=8, \mathrm{n}=7, \mathrm{x}=8, \mathrm{y}=2 \text { and } \mathrm{z}=8 \\ & \therefore 4\left(\mathrm{FeO} \cdot \mathrm{Cr}_2 \mathrm{O}_3\right)+8 \mathrm{Na}_2 \mathrm{CO}_3+7 \mathrm{O}_2 \longrightarrow \mathrm{Na}_2 \mathrm{CrO}_4+ \\ & 2 \mathrm{Fe}_2 \mathrm{O}_3+8 \mathrm{CO}_2 \end{aligned}$
SRMJEEE - 2011
Some Basic Concepts of Chemistry
228730
$0.14 \mathrm{~g}$ of an element on combustion gives 0.28 gm of its oxide. What is that element?
1 Nitrogen
2 Carbon
3 Fluorine
4 Sulphur
Explanation:
The formula used to calculate the equivalent weight of an element in its oxide : The equivalent weight of element $=\frac{v}{v}$ equivalent wt. of $\mathrm{O}$ atom. The equivalent wt. of $\mathrm{O}$ atom $=\frac{16}{2}=8 \mathrm{~g}$ Equivalent weight of element $=\frac{\text { weight of element }}{\text { weight of oxygen }} \times 8$ $\begin{aligned} & =\frac{0.14}{0.28-0.14} \times 8 \\ & =1 \times 8 \\ & =8 \mathrm{amu} \end{aligned}$ Also, by the balanced reaction: $\mathrm{S}+\mathrm{O}_2 \longrightarrow \mathrm{SO}_2$ Valency of sulphur $=4$ Equivalent weight of sulphur $=\frac{32}{4} \mathrm{amu}=8 \mathrm{amu}$ So, the element is sulphur.
AP- EAMCET(Medical) -2010
Some Basic Concepts of Chemistry
228731
How many ' $\mathrm{mL}$ ' of perhydrol is required to produce sufficient oxygen which can be used to completely convert $2 \mathrm{~L}$ of $\mathrm{SO}_2$ Gas to $\mathrm{SO}_3$ gas?
1 $10 \mathrm{~mL}$
2 $5 \mathrm{~mL}$
3 $20 \mathrm{~mL}$
4 $30 \mathrm{~mL}$
Explanation:
Perhydrol means $30 \%$ solution of $\mathrm{H}_2 \mathrm{O}_2$. $\mathrm{H}_2 \mathrm{O}_2$ decomposes as $2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2$ Volume strength of $30 \% \mathrm{H}_2 \mathrm{O}_2$ solution is 100 that mean $1 \mathrm{~mL}$ of this solution on decomposition gives $100 \mathrm{~mL}$ oxygen. $\begin{array}{lc} \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow & \mathrm{SO}_3 \\ 1 \mathrm{~L} \quad \frac{1}{2} \mathrm{~L} & 1 \mathrm{~L} \\ 2 \mathrm{~L} \quad 1 \mathrm{~L} & 2 \mathrm{~L} \end{array}$ Since, $100 \mathrm{~mL}$ of oxygen is obtained by $1 \mathrm{~mL}$ of $\mathrm{H}_2 \mathrm{O}_2$ $\therefore 1000 \mathrm{~mL}$ of oxygen will be obtained by $\begin{aligned} & =\frac{1}{100} \times 1000 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 \\ & =10 \mathrm{~mL} \text { of } \mathrm{H}_2 \mathrm{O}_2 \end{aligned}$
AP EAMCET (Engg.)-2009 VITEEE- 2009
Some Basic Concepts of Chemistry
228736
Law of multiple proportions is illustrated by one of the following pairs:
1 $\mathrm{H}_2 \mathrm{~S}$ and $\mathrm{SO}_2$
2 $\mathrm{NH}_3$ and $\mathrm{NO}_2$
3 $\mathrm{Na}_2 \mathrm{~S}$ and $\mathrm{Na}_2 \mathrm{O}$
4 $\mathrm{N}_2 \mathrm{O}$ and $\mathrm{NO}$
Explanation:
According to the law of multiple proportions that if two element combine to form more than one compound, the fixed amount of one element combining with other element have simple whole number ratio. (i) $\mathrm{H}_2 \mathrm{~S}$ and $\mathrm{SO}_2$ are not example of law of multiple proportion because elements are different (ii) $\mathrm{NH}_3$ and $\mathrm{NO}_2$ are not examples of law of multiple proportion because elements are different (iii) $\mathrm{Na}_2 \mathrm{~S}$ and $\mathrm{Na}_2 \mathrm{O}$ are not $\mathrm{NO}$ here 16 parts of oxygen react with 14 and 28 parts of nitrogen. 14:28 is $1: 2$ which is simple whole number ratio. $\therefore$ It follows law of multiple proportion.
