306684
The law of multiple proportions was proposed by
1 Lavoisier
2 Dalton
3 Proust
4 Gay- Lussac
Explanation:
Law of multiple proportions was proposed by Dalton.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306685
\({\text{4}}{\text{.4 g}}\) of an oxide of nitrogen gives \(\mathrm{2.24 \mathrm{~L}}\) of nitrogen and \(\mathrm{60 \mathrm{~g}}\) of another oxide of nitrogen gives \(\mathrm{22.4 \mathrm{~L}}\) of nitrogen at S.T.P. The data illustrates:
1 Law of conservation of mass
2 Law of constant proportions
3 Law of multiple proportions
4 Law of reciprocal proportions
Explanation:
In first oxide mass of \(\mathrm{2.24 \mathrm{~L}}\) of nitrogen at \(\mathrm{\mathrm{STP}=2.8 \mathrm{~g}}\) \(\mathrm{\therefore}\) mass of oxygen \(\mathrm{=4.4-2.8=1.6 \mathrm{~g}}\) In second oxide, mass of \(\mathrm{22.4 \mathrm{~L}}\) of nitrogen at \(\mathrm{\mathrm{STP}=28 \mathrm{~g}}\) \(\mathrm{\therefore}\) Mass of oxygen \(\mathrm{=60-28=32 \mathrm{~g}}\) (or) in second oxide \(\mathrm{2.8 \mathrm{~g}}\) of nitrogen combines with \(\mathrm{3.2 \mathrm{~g}}\) of oxygen. The masses of oxygen which combines with 2.8 \(\mathrm{\mathrm{g}}\) of nitrogen in the oxides are \(\mathrm{1.6 \mathrm{~g}: 3.2 \mathrm{~g}}\) or 1 \(\mathrm{: 2}\). This is a simple whole number ratio. \(\mathrm{\mathrm{T} \mathrm{h}}\) is illustrates the law of multiple proportions.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306686
The combining ratios of hydrogen and oxygen in water and hydrogen peroxide are 1 : 8 and 1 : 16. Which law is illustrated in this example?
1 Law of definite proportions
2 Law of conservation of mass
3 Gay Lussac’s law of combining volumes of gases
4 Law of multiple proportions
Explanation:
The combining ratios of hydrogen and oxygen in water and hydrogen peroxide are 1: 8 and 1: 16. It is an example of law of multiple proportions. According to this law, if two elements combine together to form several compounds then weight of one of these elements, which combines with a fixed weight of the other, are in ratio of simple whole numbers.
MHTCET - 2019
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306687
\({{\rm{H}}_{\rm{2}}}{\rm{S}}\) contains 5.88% hydrogen, \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) contains 11.11% hydrogen while \({\rm{S}}{{\rm{O}}_{\rm{2}}}\) contains 50% sulphur. These figures illustrate the
1 Conservation of mass
2 Constant proportions
3 Multiple proportions
4 Reciprocal proportions
Explanation:
In \({{\rm{H}}_{\rm{2}}}{\rm{S}}\), Mass of hydrogen = 5.88 g Mass of sulphur = 100 – 5.88 = 94.12 g In \({{\rm{H}}_{\rm{2}}}{\rm{O}}\), Mass of hydrogen = 11.11 g Mass of oxygen = 100 – 11.11 = 88.89 g 11.11 g of hydrogen combines with 88.89 g of oxygen 5.88 g of hydrogen combines with \(\frac{{{\rm{88}}{\rm{.89 \times 5}}{\rm{.88}}}}{{{\rm{11}}{\rm{.11}}}}{\rm{ = 47}}{\rm{.04}}\) g of oxygen. The ratio of masses of sulphur and oxygen which combine with fixed amount of hydrogen is 94.12 : 47.04 = 2 : In \({\rm{S}}{{\rm{O}}_{\rm{2}}}\), Mass of sulphur = 50 g Mass of oxygen = 100 – 50 = 50 g The ration of masses of sulphur and oxygen = 50 : 50 = 1 : From eq. (1) and (2), The ratio of sulphur and oxygen = 2 : Hence, the given data follows Law of reciprocal proportion.
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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306684
The law of multiple proportions was proposed by
1 Lavoisier
2 Dalton
3 Proust
4 Gay- Lussac
Explanation:
Law of multiple proportions was proposed by Dalton.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306685
\({\text{4}}{\text{.4 g}}\) of an oxide of nitrogen gives \(\mathrm{2.24 \mathrm{~L}}\) of nitrogen and \(\mathrm{60 \mathrm{~g}}\) of another oxide of nitrogen gives \(\mathrm{22.4 \mathrm{~L}}\) of nitrogen at S.T.P. The data illustrates:
1 Law of conservation of mass
2 Law of constant proportions
3 Law of multiple proportions
4 Law of reciprocal proportions
Explanation:
In first oxide mass of \(\mathrm{2.24 \mathrm{~L}}\) of nitrogen at \(\mathrm{\mathrm{STP}=2.8 \mathrm{~g}}\) \(\mathrm{\therefore}\) mass of oxygen \(\mathrm{=4.4-2.8=1.6 \mathrm{~g}}\) In second oxide, mass of \(\mathrm{22.4 \mathrm{~L}}\) of nitrogen at \(\mathrm{\mathrm{STP}=28 \mathrm{~g}}\) \(\mathrm{\therefore}\) Mass of oxygen \(\mathrm{=60-28=32 \mathrm{~g}}\) (or) in second oxide \(\mathrm{2.8 \mathrm{~g}}\) of nitrogen combines with \(\mathrm{3.2 \mathrm{~g}}\) of oxygen. The masses of oxygen which combines with 2.8 \(\mathrm{\mathrm{g}}\) of nitrogen in the oxides are \(\mathrm{1.6 \mathrm{~g}: 3.2 \mathrm{~g}}\) or 1 \(\mathrm{: 2}\). This is a simple whole number ratio. \(\mathrm{\mathrm{T} \mathrm{h}}\) is illustrates the law of multiple proportions.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306686
The combining ratios of hydrogen and oxygen in water and hydrogen peroxide are 1 : 8 and 1 : 16. Which law is illustrated in this example?
1 Law of definite proportions
2 Law of conservation of mass
3 Gay Lussac’s law of combining volumes of gases
4 Law of multiple proportions
Explanation:
The combining ratios of hydrogen and oxygen in water and hydrogen peroxide are 1: 8 and 1: 16. It is an example of law of multiple proportions. According to this law, if two elements combine together to form several compounds then weight of one of these elements, which combines with a fixed weight of the other, are in ratio of simple whole numbers.
MHTCET - 2019
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306687
\({{\rm{H}}_{\rm{2}}}{\rm{S}}\) contains 5.88% hydrogen, \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) contains 11.11% hydrogen while \({\rm{S}}{{\rm{O}}_{\rm{2}}}\) contains 50% sulphur. These figures illustrate the
1 Conservation of mass
2 Constant proportions
3 Multiple proportions
4 Reciprocal proportions
Explanation:
In \({{\rm{H}}_{\rm{2}}}{\rm{S}}\), Mass of hydrogen = 5.88 g Mass of sulphur = 100 – 5.88 = 94.12 g In \({{\rm{H}}_{\rm{2}}}{\rm{O}}\), Mass of hydrogen = 11.11 g Mass of oxygen = 100 – 11.11 = 88.89 g 11.11 g of hydrogen combines with 88.89 g of oxygen 5.88 g of hydrogen combines with \(\frac{{{\rm{88}}{\rm{.89 \times 5}}{\rm{.88}}}}{{{\rm{11}}{\rm{.11}}}}{\rm{ = 47}}{\rm{.04}}\) g of oxygen. The ratio of masses of sulphur and oxygen which combine with fixed amount of hydrogen is 94.12 : 47.04 = 2 : In \({\rm{S}}{{\rm{O}}_{\rm{2}}}\), Mass of sulphur = 50 g Mass of oxygen = 100 – 50 = 50 g The ration of masses of sulphur and oxygen = 50 : 50 = 1 : From eq. (1) and (2), The ratio of sulphur and oxygen = 2 : Hence, the given data follows Law of reciprocal proportion.
306684
The law of multiple proportions was proposed by
1 Lavoisier
2 Dalton
3 Proust
4 Gay- Lussac
Explanation:
Law of multiple proportions was proposed by Dalton.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306685
\({\text{4}}{\text{.4 g}}\) of an oxide of nitrogen gives \(\mathrm{2.24 \mathrm{~L}}\) of nitrogen and \(\mathrm{60 \mathrm{~g}}\) of another oxide of nitrogen gives \(\mathrm{22.4 \mathrm{~L}}\) of nitrogen at S.T.P. The data illustrates:
1 Law of conservation of mass
2 Law of constant proportions
3 Law of multiple proportions
4 Law of reciprocal proportions
Explanation:
In first oxide mass of \(\mathrm{2.24 \mathrm{~L}}\) of nitrogen at \(\mathrm{\mathrm{STP}=2.8 \mathrm{~g}}\) \(\mathrm{\therefore}\) mass of oxygen \(\mathrm{=4.4-2.8=1.6 \mathrm{~g}}\) In second oxide, mass of \(\mathrm{22.4 \mathrm{~L}}\) of nitrogen at \(\mathrm{\mathrm{STP}=28 \mathrm{~g}}\) \(\mathrm{\therefore}\) Mass of oxygen \(\mathrm{=60-28=32 \mathrm{~g}}\) (or) in second oxide \(\mathrm{2.8 \mathrm{~g}}\) of nitrogen combines with \(\mathrm{3.2 \mathrm{~g}}\) of oxygen. The masses of oxygen which combines with 2.8 \(\mathrm{\mathrm{g}}\) of nitrogen in the oxides are \(\mathrm{1.6 \mathrm{~g}: 3.2 \mathrm{~g}}\) or 1 \(\mathrm{: 2}\). This is a simple whole number ratio. \(\mathrm{\mathrm{T} \mathrm{h}}\) is illustrates the law of multiple proportions.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306686
The combining ratios of hydrogen and oxygen in water and hydrogen peroxide are 1 : 8 and 1 : 16. Which law is illustrated in this example?
1 Law of definite proportions
2 Law of conservation of mass
3 Gay Lussac’s law of combining volumes of gases
4 Law of multiple proportions
Explanation:
The combining ratios of hydrogen and oxygen in water and hydrogen peroxide are 1: 8 and 1: 16. It is an example of law of multiple proportions. According to this law, if two elements combine together to form several compounds then weight of one of these elements, which combines with a fixed weight of the other, are in ratio of simple whole numbers.
MHTCET - 2019
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306687
\({{\rm{H}}_{\rm{2}}}{\rm{S}}\) contains 5.88% hydrogen, \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) contains 11.11% hydrogen while \({\rm{S}}{{\rm{O}}_{\rm{2}}}\) contains 50% sulphur. These figures illustrate the
1 Conservation of mass
2 Constant proportions
3 Multiple proportions
4 Reciprocal proportions
Explanation:
In \({{\rm{H}}_{\rm{2}}}{\rm{S}}\), Mass of hydrogen = 5.88 g Mass of sulphur = 100 – 5.88 = 94.12 g In \({{\rm{H}}_{\rm{2}}}{\rm{O}}\), Mass of hydrogen = 11.11 g Mass of oxygen = 100 – 11.11 = 88.89 g 11.11 g of hydrogen combines with 88.89 g of oxygen 5.88 g of hydrogen combines with \(\frac{{{\rm{88}}{\rm{.89 \times 5}}{\rm{.88}}}}{{{\rm{11}}{\rm{.11}}}}{\rm{ = 47}}{\rm{.04}}\) g of oxygen. The ratio of masses of sulphur and oxygen which combine with fixed amount of hydrogen is 94.12 : 47.04 = 2 : In \({\rm{S}}{{\rm{O}}_{\rm{2}}}\), Mass of sulphur = 50 g Mass of oxygen = 100 – 50 = 50 g The ration of masses of sulphur and oxygen = 50 : 50 = 1 : From eq. (1) and (2), The ratio of sulphur and oxygen = 2 : Hence, the given data follows Law of reciprocal proportion.
306684
The law of multiple proportions was proposed by
1 Lavoisier
2 Dalton
3 Proust
4 Gay- Lussac
Explanation:
Law of multiple proportions was proposed by Dalton.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306685
\({\text{4}}{\text{.4 g}}\) of an oxide of nitrogen gives \(\mathrm{2.24 \mathrm{~L}}\) of nitrogen and \(\mathrm{60 \mathrm{~g}}\) of another oxide of nitrogen gives \(\mathrm{22.4 \mathrm{~L}}\) of nitrogen at S.T.P. The data illustrates:
1 Law of conservation of mass
2 Law of constant proportions
3 Law of multiple proportions
4 Law of reciprocal proportions
Explanation:
In first oxide mass of \(\mathrm{2.24 \mathrm{~L}}\) of nitrogen at \(\mathrm{\mathrm{STP}=2.8 \mathrm{~g}}\) \(\mathrm{\therefore}\) mass of oxygen \(\mathrm{=4.4-2.8=1.6 \mathrm{~g}}\) In second oxide, mass of \(\mathrm{22.4 \mathrm{~L}}\) of nitrogen at \(\mathrm{\mathrm{STP}=28 \mathrm{~g}}\) \(\mathrm{\therefore}\) Mass of oxygen \(\mathrm{=60-28=32 \mathrm{~g}}\) (or) in second oxide \(\mathrm{2.8 \mathrm{~g}}\) of nitrogen combines with \(\mathrm{3.2 \mathrm{~g}}\) of oxygen. The masses of oxygen which combines with 2.8 \(\mathrm{\mathrm{g}}\) of nitrogen in the oxides are \(\mathrm{1.6 \mathrm{~g}: 3.2 \mathrm{~g}}\) or 1 \(\mathrm{: 2}\). This is a simple whole number ratio. \(\mathrm{\mathrm{T} \mathrm{h}}\) is illustrates the law of multiple proportions.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306686
The combining ratios of hydrogen and oxygen in water and hydrogen peroxide are 1 : 8 and 1 : 16. Which law is illustrated in this example?
1 Law of definite proportions
2 Law of conservation of mass
3 Gay Lussac’s law of combining volumes of gases
4 Law of multiple proportions
Explanation:
The combining ratios of hydrogen and oxygen in water and hydrogen peroxide are 1: 8 and 1: 16. It is an example of law of multiple proportions. According to this law, if two elements combine together to form several compounds then weight of one of these elements, which combines with a fixed weight of the other, are in ratio of simple whole numbers.
MHTCET - 2019
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306687
\({{\rm{H}}_{\rm{2}}}{\rm{S}}\) contains 5.88% hydrogen, \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) contains 11.11% hydrogen while \({\rm{S}}{{\rm{O}}_{\rm{2}}}\) contains 50% sulphur. These figures illustrate the
1 Conservation of mass
2 Constant proportions
3 Multiple proportions
4 Reciprocal proportions
Explanation:
In \({{\rm{H}}_{\rm{2}}}{\rm{S}}\), Mass of hydrogen = 5.88 g Mass of sulphur = 100 – 5.88 = 94.12 g In \({{\rm{H}}_{\rm{2}}}{\rm{O}}\), Mass of hydrogen = 11.11 g Mass of oxygen = 100 – 11.11 = 88.89 g 11.11 g of hydrogen combines with 88.89 g of oxygen 5.88 g of hydrogen combines with \(\frac{{{\rm{88}}{\rm{.89 \times 5}}{\rm{.88}}}}{{{\rm{11}}{\rm{.11}}}}{\rm{ = 47}}{\rm{.04}}\) g of oxygen. The ratio of masses of sulphur and oxygen which combine with fixed amount of hydrogen is 94.12 : 47.04 = 2 : In \({\rm{S}}{{\rm{O}}_{\rm{2}}}\), Mass of sulphur = 50 g Mass of oxygen = 100 – 50 = 50 g The ration of masses of sulphur and oxygen = 50 : 50 = 1 : From eq. (1) and (2), The ratio of sulphur and oxygen = 2 : Hence, the given data follows Law of reciprocal proportion.