 Chemistry Quick revision of Basic concepts of chemistry
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### Nature of matter ### Classification of matter ### Properties of matter ### The international system of units (SI) ### Quantity used frequently in chemistry ### Scientific Notation ### Mathematical operation using scientific notation ### Significant figures Precision : It is the closeness of various measurements for the same quantity .

Accuracy : It is the agreement of a particular value to the true value of the result.

### Rules for determining the number of significant figures ### Mathematical operations of significant figures ### Rules for rounding off the numbers ### Dimensional analysis ### Law of conservation of mass Statement : Matter can neither be created nor destroyed.

### Law of definite proportions ### Law of multiple proportions

Statement : If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers. This law was given by Dalton in 1803.

e.g Hydrogen +oxygen → water

\ \ \ \ \ \ \ 2g \ \ \ \ \ \ \ 16g \ \ \ \ \ \ \18g

Hydrogen + Oxygen → Hydrogen peroxide

2g \ \ \ \ \ \ \ \ \ 32 g \ \ \ \ \ \ \ \ \ \ \ \ 34g

Here, the ratio of masses of oxygen (i.e. 16 g and 32 g) which combine with a fixed mass of hydrogen (2 g) is 1 : 2.

### Gay Lussac's law of gaseous volume

Statement : When gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at
same temperature and pressure.

Hydrogen + Oxygen → Water

\ \ \ \ 100 \ mL \ \ \ \ \ \ \ 50 \ mL \ \ \ \ \ \ \100 \ mL

Here, ratio of volumes of hydrogen and oxygen is 2:1.

This law is also known as the law of definite proportions by volume. Statement : Equal volumes of gases at the same temperature and pressure should contain equal number of molecules.

### Dalton's atomic theory ### Mole concept ### Molar mass ### Percentage composition

Percentage composition gives the idea about the purity of a given sample by analyzing the given data.

Mass % of an element = text(mass of that element in the compound)/text(molar mass of the compound)

e.g. For H_2O

Molar mass of water = 18.02g

Mass % of H = (2xx1.008)/18.02 xx100

Mass % of O = 16.00/18.02 xx100= 88.79

### Empirical formula and molecular formula ### Limiting reagents ### Concentration terms  