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Chemistry X Mas > Unit 1 > Flashcards

Unit 1 Flashcards

1
Q

Cos (-x)=

A

Cos X

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2
Q

Sin(-x)

A

-sin x

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3
Q

Avogadro’s law

A

Equal volume of all gases under the same condition of temperature and pressure contains a number of molecules

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4
Q

Avogadro law proposed by

A

Avogadro

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5
Q

Law of multiple proportion

A

If 2 elements combine to form more than one compound the different mass of one of the element that combine with the fixed mass of the Other element are in simple whole number ratio

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6
Q

Daltons atomic theory

A

• matter consists of indivisible atoms
• all the atoms of a given element have identical properties including mass
• atoms of different elements differ in mass
• compounds are found When atoms of different elements combined in a fixed ratio
• atoms can neither be created nor destroyed in a chemical reaction
• Dalton theory could explain the laws of chemical combinations

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7
Q

1 mole is=?

A

6.022*10^23

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8
Q

Unit of amount of substance

A

Mole

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9
Q

1 mole stp =?

A

22.4 L

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10
Q

Definite proportion law proposed by

A

Joseph Proust

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11
Q

No of oxygen present in 5 moles of glucose (C6 H12 O6

A

56.02210^23

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12
Q

Empirical formula

A

Represents simple whole number ratio of various atoms in a compound

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13
Q

What is exact number of different type of atoms in a compound

A

Molecular formula

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14
Q

Limiting reagent

A

The reagent which completely consumed in a chemical reactions

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15
Q

Amount of solute present in given solvent /solution

A

Concentration

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16
Q

Mass %=?

A

Mass of component/
Total mass of solution *100

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17
Q

Volume %=?

A

Volume of component /
Total volume of solution
*100

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18
Q

Mole fraction

A

No of moles of components /
Total no of moles of all components

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19
Q

Sum of mole fraction of all components

A

1

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20
Q

Xa+Xb=?

A

1

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21
Q

Molarity (M) valyamma

A

No of moles of solute / volume of the solution in litre

22
Q

Molality (m) kunjamma

A

Number of moles of solute/ Mass of solvent in kg

23
Q

Electron discovered by

A

Jj Thomson

24
Q

Proton discovered

A

E Goldstein

25
Q

Neutron

A

James Chadwick

26
Q

Mass of electron

A

9.1*10^-31

27
Q

Charge of electron

A

-1.6022*10^-19

28
Q

•The mass of the atom is assumed to be uniformly distributed over the atom

• the total positive charge in an atom is equal to the total negative charge and hence the atom is electrically neutral
Name of model=?

A

Thomson model of atom

29
Q

Scattering experiment done by

A

Rutherford

30
Q

Observations of Rutherford model of atom

A

• most of the a particles pass through the goldfoil without any deviations
• a small fraction of a particles was reflected by small angle
• a very few a particles bounced back that is were deflected by nearly 180 degree

31
Q

Conclusion of Rutherford

A

Most of the space in the atom is empty

The positive charge of atom is concentrated in very small volume and the centre is called nucleus

The volume occupied by the nucleus is negligligibly compared to the total volume of the atom

32
Q

The nuclear model of atom

A

All the positive charge and most of the mass of atom are concentrated in extremely small region called nucleus

Electrons are revolving around the Nucleus with a very high speed in a circular path called orbits

Electrons and nucleus are held Together by electro static force of attraction

33
Q

Rutherford limitations

A

Rutherford model cannot explain the stability of atom

He could not explain electronic structure of atom

34
Q

H=?

A

6.626*10^-34

35
Q

Unit of h

A

Js

36
Q

Highest wavelength and lowest frequency

A

Red

37
Q

Lowest wavelength and highest frequency

A

Violet

38
Q

E=?

A

Hv (v= frequency )

39
Q

Photo electric effect

A

It is a phenomenon of Ejection of electrons by certain metals like K, Rb, Cs , when light of suitable frequency incident on them

It first observed by H Hertz
The electrons are called photo electrons

40
Q

KE=?

A

Hv-hv0

41
Q

W=?

A

Hv0=w

Hv0<hv

42
Q

characteristics of Photoelectric effect

A

There is no time language between the striking of light beam and the ejection of electron between metal surface.

The number of electrons ejected is proportional to the intensity or brightness of light

For each metal there is minimum frequency is the threshold frequency which below which Photoelectric effect is not observed

Kinetic energy of ejected electron is directly proportional to the frequency of incident light

43
Q

Define lbpbp

A

Lymann Balmer paschen bracket pfund

44
Q

The postulates of Bohr model

A

Electron in the hydrogen atom can move around nucleus in circular parts of its fixed radius and energy

The energy of an electron in an Orbit does not change with time

Radius of orbits can be given by equation
Rn=a0n²
a0= 52.9

Energy of electron in an Orbit is given by
E= -Rh/n²
Rh = Rydberg constant=2.18*10^-18

The angular momentum of an electron is an integral multiple of (H/2pi)

45
Q

Limitations of bohr model

A

Do not explain the spectrum of atoms other than hydrogen

It was unable to explain the splitting of spectral lines in the presence of electric field (Stark effect)
And magnetic field (Zeeman effect )

It could not explain ability of atoms to forms molecules by chemical bonds

46
Q

Lamda /wavelength of particle =

A

H/mv

47
Q

De Broglie equation

A

De Broglie propose that radiation,matter also show both particle and wave nature, it is dual behaviour of matter

Electronics should have Momentum as well as wavelength
Lambda equal h/mv

Just like electromagnetic radiation and electron beam also undergoes diffraction

An electron microscope works on the principle of wave nature of electron

48
Q

Heisenberg Uncertainty principle

A

It is impossible to determine simultaneously the exact position and momentum of a moving microscopic particles like electron

49
Q

Dx. Dp ≥h/4pi =

A

Dx.mDv≥h/4pi

=Dx. Dv≥h/4pim

50
Q

Four Quantum numbers

A

Principal quantum number(n)

Azhimuthal Quantum number (l)

Magnetic Quantum number (m)

Spin Quantum number (s)

51
Q

What is Quantum numbers

A

These are numbers used to explain size shape orientation of orbital

52
Q

Bf3 dipole moment zero, but nf3 is not WHY??

A

case of BF3 the dipole moment is zero because it has a regular geometry and no lone pair of electrons is present on B. All the angles are 120 degree. In case of NF3 the dipole moment is not zero because of presence of a lone pair of electrons on Nitrogen atom.

Chemistry X Mas (2 decks)

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