228875
An organic compound containing $\mathrm{C}, \mathrm{H}$ and $\mathrm{N}$ gives the following on analysis : $\mathrm{C}=40 \%, \mathrm{H}=$ $13.33 \%$ and $N=46.67 \%$. What would be its empirical formula?
1 $\mathrm{C}_2 \mathrm{H}_7 \mathrm{~N}$
2 $\mathrm{C}_2 \mathrm{H}_7 \mathrm{~N}_2$
3 $\mathrm{CH}_4 \mathrm{~N}$
4 $\mathrm{CH}_3 \mathrm{~N}$
Explanation:
$\mathrm{C}: \mathrm{H}: \mathrm{N}$ $\frac{40}{12}: \frac{13.33}{1}: \frac{46.67}{14}$ (dividing by wt.) $=3.33: 13.33: 3.33$ $=\frac{3.33}{3.33}: \frac{13.33}{3.33}: \frac{3.33}{3.33}$ (dividing by smallest quotient) $=1: 4: 1$ $\therefore$ Empirical formula $=\mathrm{CH}_4 \mathrm{~N}$
CG PET- 2012
Some Basic Concepts of Chemistry
228880
Calculate the amount of $\mathrm{CO}_2$ gas produced, when $32 \mathrm{~g}$ moles of $\mathrm{CH}_4$ is burned with sufficient amount of oxygen. (Given, atomic weights of $C=12, O=16, \mathrm{H}=1$ )
228912
The number of molecules present in $3.5 \mathrm{~g}$ of $\mathrm{CO}$ at $0^{\circ} \mathrm{C}$ and $760 \mathrm{~mm}$ pressure is : 7. Empirical Formula for
1 $6.02 \times 10^{23}$
2 $1.25 \times 6.02 \times 10^{23}$
3 $0.125 \times 6.02 \times 10^{23}$
4 $1.25 \mathrm{~N}_{\mathrm{A}}$
Explanation:
28 gm CO at STP $=6.023 \times 10^{23}$ molecules $\therefore \quad 3.5 \mathrm{gm}$ of CO at STP $=\frac{1}{28} \times 3.5 \times 6.023 \times 10^{23}$ $=0.125 \times 6.02 \times 10^{23}$ Molecular Formula
AP-EAMCET-1992
Some Basic Concepts of Chemistry
228886
$5 \mathrm{~g}$ of $\mathrm{CaCO}_3$ completely reacts with
1 $36.5 \mathrm{~g}$ of $\mathrm{HCl}$
2 $3.65 \mathrm{~g}$ of $\mathrm{HCl}$
3 $0.36 \mathrm{~g}$ of $\mathrm{HCl}$
4 $7.30 \mathrm{~g}$ of $\mathrm{HCl}$
Explanation:
The balanced equation for the reaction is: $\underset{100 \mathrm{~g}}{\mathrm{CaCO}_3}+\underset{2 \times 36.5 \mathrm{~g}}{2 \mathrm{HCl}} \longrightarrow \mathrm{CaCl}_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$ From the above equation, $100 \mathrm{~g}$ of $\mathrm{CaCO}_3$ reacts completely with $73 \mathrm{~g}$ of $\mathrm{HCl}$ So, $5 \mathrm{~g}$ of $\mathrm{CaCO}_3$ will react with $\frac{73}{100} \times 5 \mathrm{~g}$ of $\mathrm{HCl}=3.65 \mathrm{~g}$ of $\mathrm{HCl}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Some Basic Concepts of Chemistry
228875
An organic compound containing $\mathrm{C}, \mathrm{H}$ and $\mathrm{N}$ gives the following on analysis : $\mathrm{C}=40 \%, \mathrm{H}=$ $13.33 \%$ and $N=46.67 \%$. What would be its empirical formula?
1 $\mathrm{C}_2 \mathrm{H}_7 \mathrm{~N}$
2 $\mathrm{C}_2 \mathrm{H}_7 \mathrm{~N}_2$
3 $\mathrm{CH}_4 \mathrm{~N}$
4 $\mathrm{CH}_3 \mathrm{~N}$
Explanation:
$\mathrm{C}: \mathrm{H}: \mathrm{N}$ $\frac{40}{12}: \frac{13.33}{1}: \frac{46.67}{14}$ (dividing by wt.) $=3.33: 13.33: 3.33$ $=\frac{3.33}{3.33}: \frac{13.33}{3.33}: \frac{3.33}{3.33}$ (dividing by smallest quotient) $=1: 4: 1$ $\therefore$ Empirical formula $=\mathrm{CH}_4 \mathrm{~N}$
CG PET- 2012
Some Basic Concepts of Chemistry
228880
Calculate the amount of $\mathrm{CO}_2$ gas produced, when $32 \mathrm{~g}$ moles of $\mathrm{CH}_4$ is burned with sufficient amount of oxygen. (Given, atomic weights of $C=12, O=16, \mathrm{H}=1$ )
228912
The number of molecules present in $3.5 \mathrm{~g}$ of $\mathrm{CO}$ at $0^{\circ} \mathrm{C}$ and $760 \mathrm{~mm}$ pressure is : 7. Empirical Formula for
1 $6.02 \times 10^{23}$
2 $1.25 \times 6.02 \times 10^{23}$
3 $0.125 \times 6.02 \times 10^{23}$
4 $1.25 \mathrm{~N}_{\mathrm{A}}$
Explanation:
28 gm CO at STP $=6.023 \times 10^{23}$ molecules $\therefore \quad 3.5 \mathrm{gm}$ of CO at STP $=\frac{1}{28} \times 3.5 \times 6.023 \times 10^{23}$ $=0.125 \times 6.02 \times 10^{23}$ Molecular Formula
AP-EAMCET-1992
Some Basic Concepts of Chemistry
228886
$5 \mathrm{~g}$ of $\mathrm{CaCO}_3$ completely reacts with
1 $36.5 \mathrm{~g}$ of $\mathrm{HCl}$
2 $3.65 \mathrm{~g}$ of $\mathrm{HCl}$
3 $0.36 \mathrm{~g}$ of $\mathrm{HCl}$
4 $7.30 \mathrm{~g}$ of $\mathrm{HCl}$
Explanation:
The balanced equation for the reaction is: $\underset{100 \mathrm{~g}}{\mathrm{CaCO}_3}+\underset{2 \times 36.5 \mathrm{~g}}{2 \mathrm{HCl}} \longrightarrow \mathrm{CaCl}_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$ From the above equation, $100 \mathrm{~g}$ of $\mathrm{CaCO}_3$ reacts completely with $73 \mathrm{~g}$ of $\mathrm{HCl}$ So, $5 \mathrm{~g}$ of $\mathrm{CaCO}_3$ will react with $\frac{73}{100} \times 5 \mathrm{~g}$ of $\mathrm{HCl}=3.65 \mathrm{~g}$ of $\mathrm{HCl}$
228875
An organic compound containing $\mathrm{C}, \mathrm{H}$ and $\mathrm{N}$ gives the following on analysis : $\mathrm{C}=40 \%, \mathrm{H}=$ $13.33 \%$ and $N=46.67 \%$. What would be its empirical formula?
1 $\mathrm{C}_2 \mathrm{H}_7 \mathrm{~N}$
2 $\mathrm{C}_2 \mathrm{H}_7 \mathrm{~N}_2$
3 $\mathrm{CH}_4 \mathrm{~N}$
4 $\mathrm{CH}_3 \mathrm{~N}$
Explanation:
$\mathrm{C}: \mathrm{H}: \mathrm{N}$ $\frac{40}{12}: \frac{13.33}{1}: \frac{46.67}{14}$ (dividing by wt.) $=3.33: 13.33: 3.33$ $=\frac{3.33}{3.33}: \frac{13.33}{3.33}: \frac{3.33}{3.33}$ (dividing by smallest quotient) $=1: 4: 1$ $\therefore$ Empirical formula $=\mathrm{CH}_4 \mathrm{~N}$
CG PET- 2012
Some Basic Concepts of Chemistry
228880
Calculate the amount of $\mathrm{CO}_2$ gas produced, when $32 \mathrm{~g}$ moles of $\mathrm{CH}_4$ is burned with sufficient amount of oxygen. (Given, atomic weights of $C=12, O=16, \mathrm{H}=1$ )
228912
The number of molecules present in $3.5 \mathrm{~g}$ of $\mathrm{CO}$ at $0^{\circ} \mathrm{C}$ and $760 \mathrm{~mm}$ pressure is : 7. Empirical Formula for
1 $6.02 \times 10^{23}$
2 $1.25 \times 6.02 \times 10^{23}$
3 $0.125 \times 6.02 \times 10^{23}$
4 $1.25 \mathrm{~N}_{\mathrm{A}}$
Explanation:
28 gm CO at STP $=6.023 \times 10^{23}$ molecules $\therefore \quad 3.5 \mathrm{gm}$ of CO at STP $=\frac{1}{28} \times 3.5 \times 6.023 \times 10^{23}$ $=0.125 \times 6.02 \times 10^{23}$ Molecular Formula
AP-EAMCET-1992
Some Basic Concepts of Chemistry
228886
$5 \mathrm{~g}$ of $\mathrm{CaCO}_3$ completely reacts with
1 $36.5 \mathrm{~g}$ of $\mathrm{HCl}$
2 $3.65 \mathrm{~g}$ of $\mathrm{HCl}$
3 $0.36 \mathrm{~g}$ of $\mathrm{HCl}$
4 $7.30 \mathrm{~g}$ of $\mathrm{HCl}$
Explanation:
The balanced equation for the reaction is: $\underset{100 \mathrm{~g}}{\mathrm{CaCO}_3}+\underset{2 \times 36.5 \mathrm{~g}}{2 \mathrm{HCl}} \longrightarrow \mathrm{CaCl}_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$ From the above equation, $100 \mathrm{~g}$ of $\mathrm{CaCO}_3$ reacts completely with $73 \mathrm{~g}$ of $\mathrm{HCl}$ So, $5 \mathrm{~g}$ of $\mathrm{CaCO}_3$ will react with $\frac{73}{100} \times 5 \mathrm{~g}$ of $\mathrm{HCl}=3.65 \mathrm{~g}$ of $\mathrm{HCl}$
228875
An organic compound containing $\mathrm{C}, \mathrm{H}$ and $\mathrm{N}$ gives the following on analysis : $\mathrm{C}=40 \%, \mathrm{H}=$ $13.33 \%$ and $N=46.67 \%$. What would be its empirical formula?
1 $\mathrm{C}_2 \mathrm{H}_7 \mathrm{~N}$
2 $\mathrm{C}_2 \mathrm{H}_7 \mathrm{~N}_2$
3 $\mathrm{CH}_4 \mathrm{~N}$
4 $\mathrm{CH}_3 \mathrm{~N}$
Explanation:
$\mathrm{C}: \mathrm{H}: \mathrm{N}$ $\frac{40}{12}: \frac{13.33}{1}: \frac{46.67}{14}$ (dividing by wt.) $=3.33: 13.33: 3.33$ $=\frac{3.33}{3.33}: \frac{13.33}{3.33}: \frac{3.33}{3.33}$ (dividing by smallest quotient) $=1: 4: 1$ $\therefore$ Empirical formula $=\mathrm{CH}_4 \mathrm{~N}$
CG PET- 2012
Some Basic Concepts of Chemistry
228880
Calculate the amount of $\mathrm{CO}_2$ gas produced, when $32 \mathrm{~g}$ moles of $\mathrm{CH}_4$ is burned with sufficient amount of oxygen. (Given, atomic weights of $C=12, O=16, \mathrm{H}=1$ )
228912
The number of molecules present in $3.5 \mathrm{~g}$ of $\mathrm{CO}$ at $0^{\circ} \mathrm{C}$ and $760 \mathrm{~mm}$ pressure is : 7. Empirical Formula for
1 $6.02 \times 10^{23}$
2 $1.25 \times 6.02 \times 10^{23}$
3 $0.125 \times 6.02 \times 10^{23}$
4 $1.25 \mathrm{~N}_{\mathrm{A}}$
Explanation:
28 gm CO at STP $=6.023 \times 10^{23}$ molecules $\therefore \quad 3.5 \mathrm{gm}$ of CO at STP $=\frac{1}{28} \times 3.5 \times 6.023 \times 10^{23}$ $=0.125 \times 6.02 \times 10^{23}$ Molecular Formula
AP-EAMCET-1992
Some Basic Concepts of Chemistry
228886
$5 \mathrm{~g}$ of $\mathrm{CaCO}_3$ completely reacts with
1 $36.5 \mathrm{~g}$ of $\mathrm{HCl}$
2 $3.65 \mathrm{~g}$ of $\mathrm{HCl}$
3 $0.36 \mathrm{~g}$ of $\mathrm{HCl}$
4 $7.30 \mathrm{~g}$ of $\mathrm{HCl}$
Explanation:
The balanced equation for the reaction is: $\underset{100 \mathrm{~g}}{\mathrm{CaCO}_3}+\underset{2 \times 36.5 \mathrm{~g}}{2 \mathrm{HCl}} \longrightarrow \mathrm{CaCl}_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$ From the above equation, $100 \mathrm{~g}$ of $\mathrm{CaCO}_3$ reacts completely with $73 \mathrm{~g}$ of $\mathrm{HCl}$ So, $5 \mathrm{~g}$ of $\mathrm{CaCO}_3$ will react with $\frac{73}{100} \times 5 \mathrm{~g}$ of $\mathrm{HCl}=3.65 \mathrm{~g}$ of $\mathrm{HCl}$