Laws of chemical combination

The greatest common divisor of 15 and 20 is 5. Dividing both terms by 5 gives \(15:20 = 3:4\). The ratio 4:3 reverses the terms, so it is not equivalent. Exam tip: simplify a ratio by dividing both terms by their greatest common divisor.

The greatest common divisor of 18 and 27 is 9. Dividing both terms by 9 gives \(18:27 = 2:3\). The ratio \(3:2\) reverses the terms, so it is not equivalent. Exam tip: simplify a ratio by dividing both terms by their greatest common divisor.

The greatest common factor of 14 and 35 is 7. Dividing both terms by 7 gives \(14:35 = 2:5\), so option A is correct. \(5:2\) reverses the terms, while \(1:5\) and \(2:7\) are not equivalent to 14:35. Exam tip: simplify a ratio by dividing both terms by their greatest common factor.

The greatest common factor of 24 and 36 is 12. Dividing both terms by 12 gives \(24:36 = 2:3\). The ratio \(3:2\) is the reverse ratio, so it is incorrect. Exam tip: simplify a ratio by dividing both terms by their greatest common factor.

The governing concept is the law of conservation of mass. When 4 g of A combines with 12 g of B and no material escapes or enters, the mass of the compound equals the sum of the masses of its reactants. Calculation: total mass = mass of A + mass of B = 4 g + 12 g = 16 g. Therefore, option C is correct. Option A is not the sum, option B counts only B and ignores A, while option D results from multiplication rather than mass addition. The conclusion applies to a closed system in which both reactants are completely included in the product.

The total mass of the compound is the sum of the masses of A and B. Therefore, mass of B = 54 g − 18 g = 36 g. The value 18 g is the mass of A, not B. Exam tip: subtract the known component’s mass from the total mass to find the unknown component.

The total mass of a compound equals the sum of the masses of components A and B. Therefore, mass of B = 63 g − 21 g = 42 g. Here, 21 g is the mass of A only, while 63 g is the mass of the complete compound. Exam tip: To find an unknown component, subtract the known component’s mass from the total mass.

The greatest common factor of 9 and 6 is 3. Dividing both terms by 3 gives 9:6 = 3:2, so 3:2 is correct. In 2:3, the order of the terms is reversed, so it does not represent A:B. Exam tip: simplify both terms by the same number and do not change their order.

Since the mass of A is fixed at 12 g in both compounds, compare the masses of B: 18:24. Dividing both terms by 6 gives 3:4. The ratio 4:3 is the reverse order, so it is incorrect. Exam tip: always reduce a mass ratio to its simplest whole-number form.

Since the mass of A is fixed, compare the masses of B in the given order: 20:50. Dividing both terms by 10 gives the simplest ratio, 2:5. The ratio 5:2 is the reverse order, so it is not correct here. Exam tip: write ratios in the same order as the quantities stated in the question.

The greatest common divisor of 14 and 7 is 7. Dividing both terms of 14:7 by 7 gives the simplest ratio, 2:1. The ratio 1:2 is the reverse ratio, so it is not correct for the given order 14 litres : 7 litres. Exam tip: simplify a ratio by dividing both terms by their greatest common divisor.

The greatest common factor of 9 and 12 is 3. Dividing both terms by 3 gives 9:12 = 3:4, so option A is correct. The ratio 4:3 reverses the order, whereas the volumes are given as 9 litres : 12 litres. Exam tip: simplify a ratio by dividing both terms by their greatest common factor.

18 divided by 72 multiplied by 100 gives 25 percent. Exam tip: apply percentage calculation carefully.

20 divided by 80 multiplied by 100 is 25 percent. Exam tip: keep total mass as denominator while finding percent.

A gas is matter: it occupies space and has mass. If a gas escapes from an open container, the measured mass of the container may appear to decrease, but the total mass of the complete closed system, including the gas, is conserved. Option C wrongly treats this open-system observation as destruction of mass. Exam tip: in questions involving gases, check whether the system is open or closed.

The governing concept is conservation of mass in a closed system. The law states that the total mass before a reaction equals the total mass after it when every reactant and product is included. In an open bottle, some gaseous product escapes, so the measured mass after reaction appears smaller because that gas is no longer on the balance. The law has not failed; the observation is incomplete. Thus option B is correct. Option A wrongly rejects the law, option C falsely claims that gas has no mass, and option D confuses loss from measurement with failure to form products.

In a closed vessel matter does not escape so total mass is correct. Exam tip: closed vessel is the safe answer.

Total mass before and after is equal. Exam tip: choose conservation from before-after equality.

The governing concept is the law of conservation of mass, which applies to the total mass of a complete reaction system, not to each individual substance. A chemical reaction rearranges atoms, so reactants are converted into products with different identities and often different individual masses. What remains equal is the sum of all reactant masses and the sum of all product masses, provided the system is closed: total reactant mass = total product mass. Therefore, option A is correct. Options B and C make false generalisations, while D incorrectly suggests that products cannot be measured.

According to the law of definite proportions, the constituent elements of a pure compound are always present in a fixed ratio by mass. In pure NaCl, sodium and chlorine have a fixed mass ratio, so option A is correct. Tea, air, and soil are mixtures, and the amounts of their components can vary. Exam tip: a fixed composition indicates a compound, whereas a variable composition indicates a mixture.

According to the law of definite proportions, a pure compound contains its constituent elements in a fixed mass ratio, regardless of the sample source. If the composition changes from one sample to another, the material may be a mixture because the proportions of components in a mixture can vary. Options B and C describe a pure compound, so they do not explain the stated observation. Exam tip: variable composition indicates a mixture; fixed composition indicates a pure compound.

Same percentage composition proves constant composition. Exam tip: connect percentage composition with definite proportion.

Sea water can contain different salts in different amounts. Exam tip: treat natural water samples as mixtures.

In a pure compound the ratio of elements remains fixed. Exam tip: look for same ratio everywhere clue.

Nitrogen and oxygen form several compounds so comparison is possible. Exam tip: look for same elements different compounds.