Useful information

Useful information



1. *PAN* - permanent account number.
2. *PDF* - portable document format.
3. *SIM* - Subscriber Identity Module.
4. *ATM* - Automated Teller machine.
7. *Wi-Fi* - Wireless fidelity.
8. *GOOGLE* - Global Organization Of Oriented Group Language Of Earth.
9. *YAHOO* - Yet Another Hierarchical Officious Oracle.
10. *WINDOW* - Wide Interactive Network Development for Office work Solution.
11. *COMPUTER* - Common Oriented Machine. Particularly United and used under Technical and Educational Research.
12. *VIRUS* - Vital Information Resources Under Siege.
13. *UMTS* - Universal Mobile Telecommunicati ons System.
14. *AMOLED* - Active-matrix organic light-emitting diode.
15. *OLED* - Organic light-emitting diode.
16. *IMEI* - International Mobile Equipment Identity.
17. *ESN* - Electronic Serial Number.
18. *UPS* - Uninterruptible power supply.
19. *HDMI* - High-Definition Multimedia Interface.
20. *VPN* - Virtual private network.
21. *APN* - Access Point Name.
22. *LED* - Light emitting diode.
23. *DLNA* - Digital Living Network Alliance.
24. *RAM* - Random access memory.
25. *ROM* - Read only memory.
26. *VGA* - Video Graphics Array.
27. *QVGA* - Quarter Video Graphics Array.
28. *WVGA* - Wide video graphics array.
29. *WXGA* - Widescreen Extended Graphics Array.
30. *USB* - Universal serial Bus.
31. *WLAN* - Wireless Local Area Network.
32. *PPI* - Pixels Per Inch.
33. *LCD* - Liquid Crystal Display.
34. *HSDPA* - High speed down-link packet access.
35. *HSUPA* - High-Speed Uplink Packet Access.
36. *HSPA* - High Speed Packet Access.
37. *GPRS* - General Packet Radio Service.
38. *EDGE* - Enhanced Data Rates for Globa Evolution.
39. *NFC* - Near field communication.
40. *OTG* - On-the-go.
41. *S-LCD* - Super Liquid Crystal Display.
42. *O.S* - Operating system.
43. *SNS* - Social network service.
44. *H.S* - HOTSPOT.
45. *P.O.I* - Point of interest.
46. *GPS* - Global Positioning System.
47. *DVD* - Digital Video Disk.
48. *DTP* - Desk top publishing.
49. *DNSE* - Digital natural sound engine.
50. *OVI* - Ohio Video Intranet.
51. *CDMA* - Code Division Multiple Access.
52. *WCDMA* - Wide-band Code Division Multiple Access.
53. *GSM* - Global System for Mobile Communications.
54. *DIVX* - Digital internet video access.
55. *APK* - Authenticated public key.
56. *J2ME* - Java 2 micro edition.
57. *SIS* - Installation source.
58. *DELL* - Digital electronic link library.
59. *ACER* - Acquisition Collaboration Experimentation Reflection.
60. *RSS* - Really simple syndication.
61. *TFT* - Thin film transistor.
62. *AMR*- Adaptive Multi-Rate.
63. *MPEG* - moving pictures experts group.
64. *IVRS* - Interactive Voice Response System.
65. *HP* - Hewlett Packard.
*Do we know actual full form of some words???*
66. *News paper =*
_North East West South past and present events report._
67. *Chess =*
_Chariot, Horse, Elephant, Soldiers._
68. *Cold =*
_Chronic Obstructive Lung Disease._
69. *Joke =*
_Joy of Kids Entertainment._
70. *Aim =*
_Ambition in Mind._
71. *Date =*
_Day and Time Evolution._
72. *Eat =*
_Energy and Taste._
73. *Tea =*
_Taste and Energy Admitted._
74. *Pen =*
_Power Enriched in Nib._
75. *Smile =*
_Sweet Memories in Lips Expression._
76. *etc. =*
_End of Thinking Capacity_
77. *OK =*
_Objection Killed_
78. *Or =*
_Orl Korec (Greek Word)_
79. *Bye =*
_Be with you Everytime

Law of combining volumes (Gay-Lussac, 1808) Avogadro law (1811)

Law of combining volumes

(Gay-Lussac, 1808)

"When gases react, the volumes consumed and produced, measured at the same temperature and pressure, are in ratios of small whole numbers".
Consequence. Stoichiometric coefficients in equations of chemical reactions for molecules of gaseous substances shows in which volume combinations gaseous substances are got or react.

examples
a)    
2CO + O2 ® 2CO2

In the time of oxidizing of two volumes of carbon (II) oxide by one volume of oxygen forms 2 volume of carbon (IV) oxide, i.e. the volume of source reaction mixture decrease on 1 volume.
b)     In the time of synthesis of ammonia from elements:

N2 + 3H2 ® 2NH3

One volume of nitrogen is reacting with three volumes of hydrogen: 2 volumes of ammonia are forming – the volume of source reaction mixture will be decreased in 2 times.


Avogadro law
(1811)

Equal volumes of all gases at the same conditions (temperature, pressure) contain the same number of molecules.
This law truth for gaseous substances only.
Consequences:
1.      On mole of any substance in the gaseous state occupies the same volume at the same temperature and pressure.
2.      One mole of any gas in standard conditions (0°C = 273°K, 1 atm = 101.3 kPa) occupies a volume of 22.4 litres.

example 1
What volume of hydrogen at s.t.p. would be evolved at dissolution 4.8 g magnesium in excess of hydrochloric acid?

solution
Mg + 2HCl ® MgCl2 + H2­

At dissolution of 24 g (1 g-mol) magnesium in HCl –– 22.4 l (1 g-mol) of hydrogen is evaluated; at dissolution of 4.8 g of magnesium –– X l of hydrogen.

X =
4.8 * 22.4
————
24
= 4.48 l of hydrogen

example 2
3.17 g of chlorine is borrowing volume, which equal 1 l (at s.t.p.). Calculate the molecular mass of chlorine.

solution
Find mass of 22.4 l chlorine.
1 l
––
3.17 g of hydrogen
22.4 l
––
X g of hydrogen

X = 3.17 * 22.4 = 71 g

Law of constant composition

Law of constant composition:

For the first time has formulated by G. Prust (1808)

"All individual chemical substances have constant quality and quantity composition and definite chemical structure and does not depend on how this substance was prepared."
From the law of constant composition follows that at complex substance formation the elements combine with each other in definite mass proportions.

example
CuS- copper sulphide. m (Cu) : m (S) = Ar (Cu) : Ar (S) = 64 : 32 = 2 : 1
To get copper sulphide (CuS) it is necessary to mix up the powders of copper and sulphur in mass relations 2:1.
If taken amounts of source substances do not correspond their correlation in the chemical formula of compound one of them stay in the excess.
For instance, if take 3 g. copper and 1 g. sulphur than after the reaction 1 g. copper, which did not enter in the chemical reaction will stay.
Substances with non-molecular structure do not possess strictly constant composition. Their composition depends on conditions of preparation.
The mass share of element w(e) shows what part forms the mass of given element from the whole mass of substance: where n – a number of atoms; Ar(e) – relative atomic mass of element; Mr – relative molecular mass of substance.



n * Ar(e)     w(e)=———Mr

Knowing the quantitative element composition of substance its simplest molecular formula is possible to determine:
1.      Mark formula of compounds Ax By Cz.
2.      Calculate an attitude X: Y: Z through the mass shares of elements:



х * Ar(А)


y * Ar(B)


z * Ar(C)
w(A)
=
—————
w(B)
=
—————
w(C)
=
—————


Mr(AxByCz)


Mr(AxByCz)


Mr(AxByCz)

X =
w(A) * Mr
—————
Ar(А)
Y =
w(B)  * Mr
—————
Ar(B)
Z =
w(C)  * Mr
—————
Ar(C)


x : y : z
=
w(A)
——
Ar(А)
:
w(B)
——
Ar(B)
:
w(C)
——
Ar(C)

3.      Obtained numerals divide on the least for getting total numbers.
4.      Write a formula of compound.

Law of multiple proportions
(D. Dalton, 1803)

If two elements form several chemical compounds with each other, then the masses of one of the elements corresponding to the same mass of the other element in these compounds are in a simple integral proportion.

N2O                 N2O3               NO2(N2O4)     

             N2O5

A number of oxygen atoms in molecules of such compounds corresponding to the two nitrogen atoms are in a proportion 1:3:4:5.

Calculations on chemical equations and its solution with examples

 Calculations on chemical equations

Calculations on chemical equations (stoichiometric calculations) are based on the mass conservation law of substances. In real chemical processes because of incomplete running chemical reactions and losses a products mass usually less theoretically calculated.
Yield of reaction (h) is a ratio of real product mass (mr) to theoretically possible (mt), expressed in shares of units or percent.
If in conditions of problems a yield of reactions products does not specified, its take in calculations as 100% should be minded that;
 (quantitative output).

h=mr

––* 100%mt


example 

Calculated the mass of copper formed when reduction 8 g copper oxide by hydrogen. The yield of reaction – 82% from theoretical.

solution
CuO + H2 ® Cu + H2O


example :
Calculate the yield of reaction of tungsten preparation by aluminizing, if from 33.14 g of ore concentrate, containing WO3 and non-reduce admixtures (mass share of admixtures 0.3), were obtained 12.72 g of metal.

solution
1.      Define the WO3 mass (g) in 33.14 g of ore concentrate.

w(WO3) = 1.0 – 0.3 = 0.7
m(WO3) = w(WO3) * more = 0.7 * 33.14 = 23.2 g

2.      Define the theoretical yield of tungsten as a result of reduction 23.2 g by aluminum powder.

WO3 + 2Al ® Al2 + O3 + W

When reducing 232 g (1g-mol) WO3 form 187 g (1g-mol) W, but at 23.2 g WO3 – X g W

X =
23.2 * 187
—————
232
= 18.7 g W

3.      Calculate the practical yield of tungsten

18.7 g W
––
100%
12.72 g W
––
Y%

Y =
12.72 * 100
–––––––––
18.7
= 68 %

example;
How much grams of barium sulphate will be formed at pouring together the solutions containing 20.8 g of barium chloride and 18.0 g of sodium sulphate?

solution
BaCl2 + Na2SO4 ® BaSO4Ї + 2NaCl

Calculation of product amount carries on the source substance, which is taken in the deficit.
a)     Beforehand define which substance stand in the deficit.
Mark the amount of Na2SO4 – X
208 g (1 g-mol) BaCl2 react with 132 g (1 g-mol) Na2SO4
20.8 g with X g

X =20.8 * 132—————208= 13.2 g Na2SO4

We found, that on the reaction with 20.8 g BaCl2 will be spend 13.2 g Na2SO4, but we have 18.0 g. Thus, sodium sulphate in the reaction is taken in the excess and further calculation is necessary to carry on the BaCl2, which is taken in deficit.


b)     Define quantity of BaSO4 precipitate. 208 g (1 g-mol) BaCl2 form 233 g BaSO4.
20.8 g –– Y g.



233 * 20.8


Y
=
—————
=
23.3 g


208

Arranging of chemical equations

Arranging of chemical equations:


Include three stages:

1.      Record formulas of substances: entered in the reaction (on the left) and products of reaction (on the right), having connected them on the sense by signs «+», «®»:
eg;

HgO ® Hg + O2

2.      Selection the coefficients for each substance so that amount of atoms of each element in left and right part of equation will be equally:
eg;

2HgO ® 2Hg + O2

3.      Checking a number of atoms of each element in left and right parts of equation.

Punjab Phaemacy Technician Practical DateSheet

M.Sc. Zoology Syllabus Syllabus of Zoology as prescribed by various Universities and Colleges.

M.Sc. Zoology Syllabus
Syllabus of Zoology as prescribed by various Universities and Colleges.

Sem. I

Sr. No.

Subjects of Study

1

Structure and Functions of Non-chordates

2

Ecological theories and applications

3

Genetics and Genetic Engineering                            

4

Tissue structure, Function and Chemistry              

5

Parasites and Immunity                                            

6

Laboratory course

Sem. II

1

Structure and Functions of Chordates                      

2

Comparative Animal Physiology

3

Endocrinology and Neuroscience

4

Cell and Receptor Biology

5

Biochemistry and Molecular biology

6

Laboratory course

Sem. III

1

Taxonomy and Biodiversity

2

Evolution and Animal Behaviour

3

Development and Differentiation

4

Laboratory course

5

Elective I Theory

6

Elective I Practical

Sem. IV

1

Conservation Biology and Wild life

2

Biostatistics, Bioinformatics and Instrumentation

3

Biotechnology and Applications

4

Laboratory course (1 module)

5

Elective II Theory

6

Elective II Practical

7

Seminar and Dissertation or Review

1

Ecology

2

Endocrinology

3

Entomology

4

Environmental Biology and Toxicology




5

Fisheries and Aquaculture

6

Genetics and Molecular Biology

7

Immunology

8

Wildlife biology

9

Stem cells and developmental biology