Laws of chemical combination

The correct answer is D, 24 g. The mass ratio is B:A = 12:4 = 3:1. Therefore, for 8 g of A, the required mass of B is 8 × 3 = 24 g. This fixed mass ratio illustrates the law of definite proportions.

The direct answer is 12 grams, so option C is correct. The compound always combines its elements in a fixed mass ratio. Here, 3 grams of A combine with 2 grams of B. The new amount of B is 8 grams. First compare B: 8 ÷ 2 = 4, so the amount of B has become four times larger. The amount of A must also become four times larger to keep the same compound ratio. Therefore, A = 3 × 4 = 12 grams. Option A, 6 grams, is wrong because it is only twice 3 grams, while B has become four times 2 grams. Option B, 10 grams, does not follow the fixed ratio. Option C, 12 grams, correctly preserves the ratio 3:2, because 12:8 simplifies to 3:2. Option D, 16 grams, is too large and would give a different ratio. The important idea is the law of definite proportions: a pure compound has its elements in a fixed mass ratio. A useful exam cue is to find the multiplier of the known mass first, then multiply the other mass by the same number.

For the fixed mass of 1 g hydrogen, the masses of oxygen are 8 g and 16 g. Their ratio is 8:16, which simplifies to 1:2. Option 2:1 is the reverse ratio, so it is incorrect. Exam tip: always reduce a mass ratio to its simplest form.

The mass of element A is fixed at 2 g in both compounds. Therefore, compare the masses of element B: 3 g and 6 g. Thus, 3:6 = 1:2. Hence, the correct ratio is 1:2. The ratio 2:1 is the reverse ratio, so it is incorrect. Exam tip: when one element has a fixed mass, simplify the ratio of the masses of the other element.

The direct answer is 1:3, so option B is correct. The question compares the masses of B that combine with the same mass of A, namely 10 grams. In the first compound, 10 grams of A combine with 5 grams of B. In the second, the same 10 grams of A combine with 15 grams of B. Therefore the mass ratio of B in the two compounds is 5:15. Divide both terms by their common factor 5: 5 ÷ 5 = 1 and 15 ÷ 5 = 3. Thus the simplest whole-number ratio is 1:3. Option A, 1:2, would represent 5:10, not 5:15. Option B, 1:3, is correct because it is the simplified form of 5:15. Option C, 2:3, does not simplify the given numbers. Option D, 3:5, reverses or changes the required comparison and is not equal to 5:15. This illustrates the law of multiple proportions: when the same masses of one element combine with different masses of another, those different masses are in a small whole-number ratio. The exam method is to write the two masses in order and divide both by their greatest common divisor.

The volume ratio is 6:3. Dividing both terms by their greatest common factor, 3, gives 2:1. The ratio 1:2 represents the reverse order, so it is not correct. Exam tip: simplify a ratio by dividing both terms by their greatest common factor.

The volume ratio of gas A to gas B is 3:9. Dividing both terms by 3 gives the simplest ratio, 1:3. At the same temperature and pressure, gas volumes are proportional to their number of moles, so this is the reaction volume ratio. 3:1 is the reversed ratio. Exam tip: simplify a ratio by dividing both terms by their highest common factor.

The volumes of gas A and the gaseous product are 2 L : 2 L. Dividing both terms by 2 gives the simplest whole-number ratio, 1:1. A ratio of 2:1 would mean that the volume of the first gas is twice the volume of the product. Exam tip: always reduce gas-volume ratios to the smallest whole numbers.

At the same temperature and pressure, the ratio of the given gas volumes is 5:10. Dividing both terms by 5 gives the simplest whole-number ratio, 1:2. The ratio 2:1 reverses the given order, so it is incorrect. Exam tip: write volume ratios in the stated order and then simplify them.

The direct answer is option A, conservation of mass. In a closed system, matter does not disappear and new matter is not created during a chemical reaction. The total mass before the reaction must equal the total mass after it. Here, the reactants have mass 50 g. The products have masses 35 g and 15 g, so product mass is 35 + 15 = 50 g. Therefore, the mass is conserved. Option A is correct because it states exactly this rule. Option B, definite proportions, concerns the fixed mass ratio of elements in one pure compound, which is not being tested here. Option C, multiple proportions, compares different compounds made from the same elements in small whole-number ratios. Option D, gaseous volumes, concerns simple volume ratios of reacting gases. Memory cue: in a closed system, add all product masses and compare them with the reactant mass.

According to the law of conservation of mass, the total mass remains constant in a chemical reaction. Here, the 4 g gas is also a product, so total product mass = 10 + 4 = 14 g. The 10 g value is only the mass of the main product, not of all products. Exam tip: always include the mass of a gas when calculating total product mass.

The law states that these masses are in a simple whole-number ratio. The ratio 1:2 is a whole-number ratio, so option A fits the law.

The given ratio is A : B = 1 : 4. If 1 part of A is 5 g, then 4 parts of B = 4 × 5 = 20 g. Therefore, the correct answer is 20 g. Taking 25 g of B would make the ratio 1:5, not the given 1:4. Exam tip: Find the value of one ratio part first, then multiply it by the required number of parts.

The 5 parts for B equal 25 g, so 1 part is 25 ÷ 5 = 5 g. A is 4 parts, so its mass is 4 × 5 = 20 g.

The mass ratio A:B is 11:44. Dividing both terms by their greatest common factor, 11, gives 1:4. The ratio 4:1 reverses the order, so it is not correct for A:B. Exam tip: write ratios in the stated order and simplify both terms by the same common factor.

The mass of element A is fixed at 1 gram in both compounds. Therefore, the ratio of the masses of B is 2:6, which simplifies to 1:3 on dividing both terms by 2. The ratio 2:1 is obtained by reversing the order, so it is not correct. Exam tip: when the mass of one element is fixed, reduce the masses of the other element to the simplest whole-number ratio.

Water contains only hydrogen and oxygen, so the oxygen mass is the total mass minus the hydrogen mass: 18 g − 2 g = 16 g.

The governing concept is conservation of mass in a compound. The total mass of water is made up of the masses of hydrogen and oxygen, so the mass of oxygen equals the total mass minus the mass of hydrogen. Calculation: oxygen mass = 27 g − 3 g = 24 g. Therefore, option C is correct. A quick check is 3 g hydrogen + 24 g oxygen = 27 g water. Option A and option B do not provide the complementary mass required to reach 27 g, while option D simply repeats the total mass and ignores the hydrogen already included. The calculation assumes the stated sample contains only these two components, as water does.

Since the compound contains only A and B, their masses add to 9 g. Thus, the mass of B is 9 g − 3 g = 6 g.

Sugar dissolved in water is a mixture, so the relative amounts of sugar and water can vary. The law of definite proportions applies to pure compounds, not mixtures.

The direct answer is option A, 2:3:2. For a gaseous reaction, write the volumes in the same order as the substances are named: gas A, gas B, and gas C. Their given volumes are 2 L, 3 L, and 2 L. Thus the ratio is 2:3:2. It is already a simple whole-number ratio because 2, 3, and 2 have no common factor greater than 1. Option A is correct. Option B, 1:3:2, incorrectly changes the volume of A from 2 L to 1 L without dividing every term by the same number. Option C, 2:1:2, changes B incorrectly. Option D, 3:2:1, reverses and rearranges the values, so it does not follow the stated order. The numbers here describe gaseous-volume proportions, not necessarily the balanced chemical equation unless temperature and pressure conditions are understood to be the same. Memory cue: copy the three volumes in question order before forming the ratio.

Divide every term of 4:8:4 by their highest common factor, 4, to get 1:2:1. Hence, this is the simplest whole-number ratio. Although 2:4:2 is equivalent, it can still be divided by 2, so it is not in simplest form. Exam tip: divide all terms by their HCF.

In a closed system, the total product mass equals the total reactant mass: 9 g + 11 g = 20 g. The other product therefore has a mass of 20 g − 13 g = 7 g.

By conservation of mass, the products together must have a mass of 16 g. The gas mass is therefore 16 g − 10 g = 6 g.

The ratio has 2 + 5 = 7 parts, so each part is 35 g ÷ 7 = 5 g. Component B has 5 parts, giving 5 × 5 g = 25 g.