Laws of chemical combination

The law of multiple proportions applies when the same two elements form more than one compound and the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio. Carbon and oxygen form both \\(CO\\) and \\(CO_2\\), so this is exactly the required situation. For a fixed mass of carbon, the oxygen combined in the two compounds is in the ratio \\(1:2\\), a simple whole-number ratio. Therefore, the correct law is the law of multiple proportions.

Option D is correct. The law of definite proportions applies to one pure compound, whose elements have a fixed mass ratio; it does not compare two different compounds. Conservation of mass concerns total mass in a chemical reaction, while gaseous volumes concerns reacting gas volumes. The key clue is the presence of two compounds made from the same two elements.

The law of multiple proportions applies when the same two elements form more than one compound. In such compounds, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio. This law compares different compounds, not different samples of one compound.

Here, sulfur and oxygen form two compounds, \(SO_2\) and \(SO_3\). For a fixed amount of sulfur, the oxygen amounts are in the ratio \(2:3\), a simple whole-number ratio. Therefore, the comparison is explained by the law of multiple proportions, so option A is correct. The law of definite proportions would concern the constant composition of one pure compound.

The law of multiple proportions compares different compounds made from the same two elements. If the amount of one element is fixed, the different masses of the second element that combine with it show a simple whole-number ratio. This idea is different from the law of definite proportions, which describes the constant composition of one compound.

Water, \(H_2O\), and hydrogen peroxide, \(H_2O_2\), both contain hydrogen and oxygen, but their formulas and compositions differ. For a fixed amount of hydrogen, the oxygen combined in the two compounds is in the ratio \(1:2\). Thus these compounds illustrate the law of multiple proportions, making option B correct.

The law of multiple proportions is used when two elements combine in more than one way to produce different compounds. With a fixed mass of one element, the masses of the other element in the compounds are related by a simple whole-number ratio. It therefore focuses on comparison between compounds with the same constituent elements.

Nitrogen monoxide, \(NO\), and nitrogen dioxide, \(NO_2\), contain nitrogen and oxygen but have different formulas. For the same amount of nitrogen, the oxygen combined in these compounds is in the ratio \(1:2\). This simple ratio shows multiple proportions. Hence option C, the law of multiple proportions, is the correct answer; conservation of mass addresses chemical reactions more generally.

The law of multiple proportions describes the relationship between different compounds formed by the same two elements. When the mass of one element is kept constant, the masses of the other element that combine with it form a simple whole-number ratio. This law helps distinguish several compounds containing the same elements.

Phosphorus and chlorine form \(PCl_3\) and \(PCl_5\). For the same amount of phosphorus, the chlorine amounts are proportional to 3 and 5, giving the simple ratio \(3:5\). Therefore, their comparison is an example of the law of multiple proportions. Option D is correct. The definite-proportions law would describe the fixed composition within one selected compound.

For the fixed 2 g of A, the masses of B are 3 g and 12 g. Thus, their ratio is 3:12. Dividing both terms by 3 gives the simplest ratio, 1:4. Option 4:1 is the reverse order, so it is incorrect. Exam tip: write ratios in the order given and reduce both terms by their highest common factor.

Since the mass of A is fixed at 5 g in both compounds, compare the masses of B: 6 g and 30 g. Their ratio is 6:30, which simplifies to 1:5 on dividing both terms by 6. Therefore, 1:5 is correct. The ratio 5:1 is the reverse order. Exam tip: when one element has a fixed mass, simplify the masses of the other element in the given order.

The mass of A is fixed at 8 g in both compounds, so compare the masses of B: 14:35. Dividing both terms by 7 gives 2:5. The ratio 5:2 is the reverse ratio, so it is not correct. Exam tip: when the mass of one element is fixed, write the masses of the other element in the given order and simplify the ratio.

The mass of A is fixed at 11 g in both compounds, so compare the masses of B: 20:40. Dividing both terms by 20 gives 1:2. The ratio 2:1 is the reversed ratio. Exam tip: identify the element with fixed mass and reduce the masses of the other element to their simplest ratio.

The mass of A is fixed at 6 g in both compounds, so compare the masses of B: 7:28. Dividing both terms by 7 gives 1:4. Option 4:1 is the reverse ratio, so it is incorrect. Exam tip: in multiple-proportion questions, first identify the element whose mass is kept fixed.

Since the mass of A is fixed at 4 g in both compounds, compare the masses of B: 15 g and 25 g. Their ratio is 15:25, which becomes 3:5 after dividing both terms by 5. The ratio 5:3 is the reverse ratio, so it is not correct. Exam tip: reduce a ratio by dividing both terms by their common factor.

The mass of A is fixed at 10 g in both compounds, so compare the masses of B: 18:54. Dividing both terms by 18 gives 1:3. The ratio 3:1 is the reverse ratio, so it is not correct. Exam tip: reduce a ratio to its simplest form by dividing both terms by their highest common factor.

Since the mass of A is fixed at 9 g in both compounds, compare the masses of B: 16:24. Dividing both terms by 8 gives 2:3. The ratio 3:2 is the reverse ratio, so it is not correct. Exam tip: simplify a ratio by dividing both terms by their greatest common factor.

The governing concept is the law of combining volumes: when gases are compared at the same temperature and pressure, their volumes can be represented directly as a ratio. The three given volumes are 1 L, 3 L, and 2 L, so the ratio is 1:3:2. To simplify a ratio, divide all terms by their greatest common divisor. Here the greatest common divisor of 1, 3, and 2 is 1, so no reduction is possible. Therefore option A is correct. Option B rearranges the numbers, option C incorrectly makes all volumes equal, and option D changes the original relative sizes. Since the first volume is already 1, the stated ratio is already in simplest form.

At identical temperature and pressure, gas volumes are compared by writing them in the same order as given. Thus the ratio is 2:5:2. The three numbers have a greatest common divisor of 1 because 5 is not divisible by 2, so the ratio cannot be reduced further. Option B is therefore correct. Option A would incorrectly make the second volume equal to the first and third; option C subtracts or alters the values without justification; option D also does not preserve the original proportions. This is an elementary application of the combining-volume idea: the numerical volume relationship is retained when all measurements use the same unit and conditions.

The relevant concept is comparison and simplification of gas volumes measured under identical conditions. Because temperature and pressure are the same, the volumes can be compared directly; no correction or gas-law calculation is needed. The given ratio is 7:14:7. The greatest common divisor of all three terms is 7, so divide each by 7: 7/7 : 14/7 : 7/7 = 1:2:1. Therefore, option C is correct. Option A is equivalent in proportion but is not in simplest form. Option B incorrectly makes the 14-litre volume equal to 7 litres, while option D changes the relative proportions. The middle volume is twice each outer volume, exactly as represented by 1:2:1.

The governing idea is direct comparison of gas volumes under identical temperature and pressure, as used in the law of combining volumes. Write the data as 4:10:6. The greatest common divisor of 4, 10, and 6 is 2. Dividing every term by 2 gives 4 ÷ 2 : 10 ÷ 2 : 6 ÷ 2 = 2:5:3. Thus option D is correct. Option A repeats the original ratio but does not simplify it. Option B changes the first and third terms, and option C incorrectly states that all three volumes are equal. The reduced ratio still preserves the original order and relative sizes: the second volume is 2.5 times the first, while the third is 1.5 times the first.

For gases measured at the same temperature and pressure, the volume ratio is obtained by reducing the measured volumes by their common factor. The initial ratio is 15:5:10. The greatest common divisor of 15, 5, and 10 is 5. Dividing each term by 5 gives 3:1:2. Therefore option A is correct. Option B is the original, unreduced ratio and does not satisfy the word simplest. Option C changes the order and values, while option D incorrectly claims that all volumes are equal. The result also has a clear physical interpretation: the first gas has three times the volume of the second, and the third has twice the volume of the second.

The relevant concept is the volume relationship of gases under the same temperature and pressure. First write the measured values as 3:12:6. Their greatest common divisor is 3. Dividing all terms by 3 gives 3 ÷ 3 : 12 ÷ 3 : 6 ÷ 3 = 1:4:2. Hence option B is correct. Option A gives the correct unsimplified relationship but not the simplest ratio requested. Option C changes the relative positions of the volumes, and option D falsely makes all three values equal. The reduced ratio preserves the data exactly: for every one volume unit of the first gas, there are four units of the second and two units of the third.

The law of combining volumes allows gas volumes at identical temperature and pressure to be compared directly. The initial ratio is 8:4:12. The greatest common divisor of these three numbers is 4. Dividing each term by 4 gives 8 ÷ 4 : 4 ÷ 4 : 12 ÷ 4 = 2:1:3. Thus option C is correct. Option A is the original ratio and has not been reduced. Option B gives a different relationship, and option D incorrectly treats all three volumes as equal. The reduced form shows that the first gas has twice the volume of the second and the third has three times the volume of the second, exactly matching the original measurements.

The governing principle is direct proportional comparison of gas volumes when temperature and pressure are the same. Begin with 6:6:18. The greatest common divisor of all three terms is 6. Dividing each by 6 gives 6 ÷ 6 : 6 ÷ 6 : 18 ÷ 6 = 1:1:3. Therefore option D is correct. Option A is mathematically equivalent but is not in simplest form. Option B reverses the relative size of the first and third volumes, while option C incorrectly says that 18 L equals 6 L. The final ratio clearly indicates that the first two gases have equal volumes and the third gas has three times either one.

The percentage composition of a pure compound remains fixed. Exam tip: connect percentage composition with definite proportions.

At the same temperature and pressure, the volumes of reacting gases bear a simple whole-number ratio. This is Gay-Lussac’s law of combining volumes, so a 3:2 ratio is explained by this law. The law of definite proportions concerns the mass ratio of elements in a compound, not the volumes of gases. Exam tip: when gas volumes at the same temperature and pressure are compared, use the law of combining volumes.

According to the law of reciprocal proportions, when two elements combine separately with a fixed mass of a third element, the masses of the two elements are in the same ratio, or a simple multiple of the ratio, in which they combine with each other. Hence, the law indicated is the law of reciprocal proportions. The law of definite proportions instead applies to the fixed mass ratio of elements within one compound. Exam tip: the phrase “fixed mass of a third element” is a key clue for reciprocal proportions.

The governing concept is Dalton’s atomic theory and the laws of chemical combination. Matter combines through atoms, and atoms are discrete particles; therefore, compounds are formed when atoms join in definite, simple whole-number ratios. For example, water contains hydrogen and oxygen atoms in a 2:1 atomic ratio, represented by H₂O. The ratio is not related to colour, smell, or the shape of the vessel. Option D is correct because it connects the observed simple ratios with the integral nature of atoms and the formation of definite compounds. The other options describe physical features that do not explain chemical composition.