1 Mark Questions2

1. Which of the following statements are cor-rect?X: Kerosene is optically denser than water.Y: The mass density of kerosene is less

than water.A: Both X, Y are correct.2. Refractive index of glass relative to water

9is . What is the refractive index of water

8relative to glass?

A: Refractive index of water relative to glass

8is .

93. The refractive index of transparent material

3is . Then find the speed of light in that

2

medium?C

A: Refractive index (n) = V

3 3 108 = 2 VSpeed of light in the medium (V)

2= 3 108

3

= 2 108 m/s.4. What happens to the light when the angle

of incidence is greater than critical angle?A: When the angle of the incidence is greater

than critical angle, the light ray reflects intodenser medium at interface.

5. A ray of light passes from denser mediumto rarer medium. How does the refractedray travel in rarer medium?

A: The refracted ray bonds away from thenormal.

6. Which phenomenon is involved in the for-mation of a mirage?

A: Total internal reflection

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Refraction of light at plain surfaces

Refraction of light at curved surfaces

1. Why is it difficult to shoot a fish swimmingin water?

A: It is very difficult to shoot a fish swimming inwater. Because its position appears to beshifted up from its original position due torefraction.

2. A ray of light enters from air to medium 'X'.The speed of light in the medium is 1.5 108 m/s and speed of light in air 3 108 m/s.Find the refractive index of the medium X.

A: Refractive index of medium.velocity of light in vacuum or air

= velocity of light in the medium

3 108 m/s= = 2

1.5 108 m/s 3. Mention two properties of image formed by

a concave lens.A: i) Image produced is virtual.

ii) Image produced is erect and diminished.4. The critical angle of a material substance is

145. What is its refractive index?(sin 45=)

2

A: Critical angle of water (C) = 45The 1

refractive index of the material (n) = sin C

1n = = 2 =1.414

sin 455. Pose a question to understand the difference

between concave lens and convex lens.A: F Are all lenses have same surfaces?

F Are the lenses used by dentists and ophthalmologist same?

6. Write any two questions about the 'forma-

tion of mirages'?A: i) How do the mirages form?

ii) What is the science behind formation ofmirages?

iii) At what situations do mirages form?7. Lens used: Convex

i) What is the nature of the image?ii) What is magnification?

A: i) The image is real and inverted.ii) It's magnification is 1.

8. Find the focal length of a convex lens, if it canform image at 25 cm of an object at 10 cm.

A. u = 10 cm, v = +25 cmuv 10(25)

f = = u – v 10 – 25

250= = 7.14

35 focal length (f) = 7.14 cm

D.V. SubbanaiduSubject Expert

Writer

u v f

10 cm - 10 cm 5 cm

1. How do you find the focal length of aconvex lens by u - v method?

A: Aim: To determine the focal length of aconvex lens.Apparatus: Convex lens, Meter scale,V-stand, Screen.Procedure: H Take a V - stand andplace it on a long table at the middle.H Place a convex lens on the V - stand.H Light the candle and place it at a dis-

tance of 60 cm from the lens on theprinciple axis.

H Adjust the screen which is on theother side of the lens to get animage on it.

H Measure the distance between thecandle and the stand of the lens, thisvalue noted as the object distance (u).

H Measure the distance of the imagefrom the stand of the lens, this valueis noted as image distance (v).

H Repeat the experiment for variousobject distances (u) like 50 cm, 40cm, 30 cm and measure the dis-tance of image (v) in all cases andnoted in the following table.

H From the above table f = uv/u - vvalue is constant.

H This average constant value givesthe focal length of the lens.

2. Draw the ray diagrams for the given situations.i) Object placed between focal point

and optic centre of a convex lens.A:

ii) Object placed between focal pointand optic centre of concave lens.

1. The Focal length of a converging lens is 20cm. where will the image be formed, if anobject is placed at 60 cm from the lens?Write characteristics of the image.

A: Focal length of the lens (f) = 20 cmObject distance (u) = 60 cmImage distance (v) = ?

1 1 1 Lens formula =

f v u 1 1 1 = + v f u 1 1 1 = + v 20 60 1 3 1 2 1 = = = v 60 60 30 v = 30 cm

2. Ravi wants to make a lens. Which formulahe has to follow? Write the formula andexplain the terms in it .

A: Lens maker's formula1 1 1 = (n 1) ( )f R1 R2

R1, R2 - Radii of curvaturen - Refractive index

f - Focal length3. Suppose you are inside the water in a

swimming pool. Your friend is standing onthe edge. Do you find your friend taller orshorter than this actual height? Why?

A: My friend appears to be taller than thisactual height for me. Because when I sawhim from water, the light travels fromdenser medium (water) to rarer medium(air). So it bends away from the normal(refraction of light) so he appears to betaller.

4. Distinguish between convex lens and con-cave lens?

5. A double convex lens has two surfaces of

equal radii 'R' and refractive index n = 1.5.Find the focal length?

A: Radii of curvature of both surfaces areequal R1 = R2 = R

Refractive index of the material (n) = 1.5

focal length of the lens (f) = ?

Lens maker's formula

1 1 1 = (n 1) ( )f R1 R2For double convex lens R1 is +ve, R2 is -ve

1 1 1 = (n 1) ( + )f R1 R2

1 2 = (n 1) ( ..

. R1 R2)f R

R Rf = =

2(n 1) 2(1.5 1) R

= = R1

f = R6. Why do stars appear twinkling?

A: Stars appear twinkling due to multiplerefractions at different layer of air of differ-ent densities, the light undergoes all theway to reach our eye.

A:- Convex lensi) A convex lens is

thick at middleand thin at edges.

ii) It is also known asa converging lens.

iii) This lens is usedto treat hypermetropia.

Concave lens i) A concave lens is

thin at middle andthick at edges.

ii) It is also known asa diverging lens.

iii) This lens is usedto treat myopia.

2 Marks Questions

Why do stars appear twinkling?

F1 2F1

2F2 F2

Obj

ect

Image

F2 2F2

2F1 F1Image

Object

Target-2020

TenthPhysical Science

100100

4 Marks Questions

1 Mark Questions

S. Object Image Focal No. distance (u) distance (v) length

uvf = u v1 50 cm

2 40 cm

3 30 cm

4 Marks1. An electric lamp and a screen are

placed on the table, in a line at a dis-tance of 1 m. In what positions of con-vex lens of focal length of f = 21cm willthe image of lamp be sharp?

2. Derive a relation between thickness ofglass slab, vertical shift and refractiveindex of glass slab by an experiment.

2 Marks1. Why does a diamond shine more than

a glass piece cut to the same shape?2. Write the role of lenses in our daily

life?3. What happens to the light ray, when it

strikes the interface normally?

Additional Questions

2 Marks Questions

1. Sigma () bond is stronger than pi()bond, predict the reasons.

A: If two atoms form multiple bonds betweenthem the first bond is due to the overlapof orbitals along the inter nuclear axis giv-ing a stronger sigma () bond. After for-mation of () bond the other bonds areformed due to the overlap of orbitals sidewise or laterally giving weaker bonds.The ‘’ bond is stronger because the

electron pair shared is concentrated morebetween the two nuclei due to end - end orhead on overlap and attracted to both thenuclei. The bond overlap gives a weakerbond due to the lateral overlap of ‘p’ orbitalswhich is not to greater extent.

2. What type of bonds are present in the fol-lowing molecules? Draw their electronicstructure. i) O2 ii) CH4

3. The electronic configurations of two ele-ments X and Y are given below:X - 2, 6Y - 2, 8, 1i) What type of chemical bond is formed

between the two atoms of X?ii) What type of chemical bond will be

formed between the atoms of X and Y?A: i) Covalent bond

ii) Ionic bond4. Draw the Lewis dot structures of B, C, N

and O atoms.. . . . . . .

A: :B :C :N . :O:5. Draw the dot diagrams of

i) Methane ii) Water

6. Draw the dot diagrams of formation of themolecules i) Ammonia ii) Water

A: i) Ammonia (NH3) molecule:

ii) Water (H2O) Molecule:

7. Write down the characteristics of the ele-ments having atomic number 12.i) period number ii) Group number iii) Valency iv) Element family

A: i) Period number - 3 ii) Group number - IIA (2)iii) Valency - 2 iv) Element family - Alkali earth metal

family.

email: help@eenadupratibha.netøŒ‰vÚÛî¦ô¢Ù 27 è…šúÙñô¢ª 2019

K. Gagan Kumar

Subject Expert

Writer

Target-2020

TenthPhysical Science

100100

Classification of Elements

Chemical Bonding

A:

ii) Covalent bond

H. . . . . ..N: + H + H + H H : N: or. . .H

N

HH

H

. .

. . . . . . . .:O . + H + H :O: H or :O | H. . .

H H

One Mark Questions

1. Ca, Sr, Ba are Dobereiner’s triads;atomic weights of Ca and Ba are 40.0and 137.0, then what is the atomicweight of strontium?

40.0 + 137.0A: Atomic weight of Sr =

2177

= = 88.52

2. What are the new elements discoveredin place of Eka-boran, Eka-aluminiumand Eka-silicon?

A: Scandium, Gallium and Germanium arediscovered in place of Eka-boran, Eka-aluminium and Eka-silicon respectively.

3. Alkali earth metals belongs to whichgroup?

A: 2 (IIA) - Group4. Which group elements of periodic table

has highest electronegativity values?A: Halogens - VII A Group5. What is the valency of Carbon Family?A: 46. How many elements are present in 6th

period of long form of periodic table?A: 327. Which of the block in a periodic table

have non metals, metals and metal-loids?

A: P - block8. In periodic table, which of the group has

lanthanides and ActinidesA: 3, (IIIB)9. Among Na+, Mg2+ and Al3+ write

increasing order of ionic radii.A: Al3+ < Mg2+ < Na+

10. In the modern periodic table, which arethe metals among the first ten ele-ments?

A: Lithium, Beryllium

1. Draw the figures of BeCl2, BF3, H2O, HClmolecules based on hybridization.

2. Write the main points of valence shellElectron pair repulsion theory (VSEPRT)?

A: i) VSEPRT considers electrons in thevalence shells which are in covalentbonds and in lone pairs as chargeclouds that repel one another and stayas far apart as possible. This is the rea-

son why molecules get specific shapes.ii) If we know the total number of electron

pairs in the valence shell as covalentbonds and lone pairs in the Centralatom, it will help us to predict thearrangement of those pairs around thenucleus of the central atom and fromthat the shape of the molecule.

iii) Lone pairs occupy more space aroundthe central atom than bond pairs. Lonepair means unshared electron pair ornon-binding electron pair. These areattracted to only one nucleus where asthe bond pair is shared by two nuclei.

Thus, the presence of lone pairs on thecentral atom causes slight distortion ofthe bond angles from the expected regu-lar shape. If the angle between lone pairand bond pair increases at the centralatom due to more repulsion, the actualbond angles between atoms must bedecreased.iv) If two bond pairs are present in two

covalent bonds around the nucleus ofthe central atom without any lone pairsin the valence shell, they must be sep-arated by 180 to have minimum repul-sion between them. Thus, the mole-cule would belinear.

e.g.: Berylliumchloride

v) If three bond pairs are there in threecovalent bonds around the nucleus of

the central atom, without any lone pairsthey get separated by 120 along threecorners of a triangle. Therefore, theshape of themolecule is trigo-nal - planar.

e.g.: Boran trifluo-ride

vi) If there are fourbond pairs in the valence shell of thecentral atom, the four bond pairs willorient along the four corners of atetrahedron and the bond angleexpected is109 28' .

e.g.: Methanevii) If there are

three bondpairs and onelone pair i.e.,unsharedelectron pair,then the lone pair occupies morespace around the nucleus of the cen-tral atom. The remaining three bondpairs come relatively closer as in NH3molecule.

e.g.: Ammoniaviii) If there are two bond

pairs and two lonepairs of electronsaround the nucleus of the central atom inits valence shell, lone pair - lone pairrepulsion is greater than lone pair - bond

pair repulsion. Therefore, the anglebetween bond pairs further decreases.

e.g.: water ix) Valence shell

Electron pairrepulsion.Theory(VSEPRT)mainly fails inexplaining the strengths of the bonds.

Four Marks Questions

A:

180

Cl Be Cl

F| 120B

F F

i) Covalent bond

A:i) Methane

(CH4) molecule:

ii) Water (H2O) Molecule:

i) BeCl2:

ii) BF3:

iii) H2O:

iv) HCl:

NH

HH

12

Mark Questions

1. Identify the Dobereiner triad among thefollowing.A) F, Cl, Br B) Ne, Ar, KrC) Li, Na, K D) S, Sr, Ba

2. Number of elements in period 4 of thelong form of periodic table?A) 2 B) 8 C) 18 D) 32

3. The order of the electron affinity of halo-gens is A) Cl > F > Br > I B) F > Cl > Br > IC) I > Br > Cl > F D) Br > F > Cl > I

4. Which element has lowest electronega-tivity A) Fr B) Li C) F D) Cl

5. Anomalous pairs in Mendeleev’s peri-odic table

A) Mn, Cl B) Te, I C) Na, K D) Cu, Zn

ANSWERS1-C 2-C 3-A 4-A 5-B

10748'

H H10431'

10928'

Valency of Carbon Family is?

p-sp

p-sp2

p-sp2

sp2-p

Becl cl

sp-p

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4 ÷«ô¢ª\õ ví£øŒoõª

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2.- SF6 Ôô¢p-è[-åÙö˺ S áJ›í ú£ÙÚÛ-K-ÚÛ-ô¢-é°Eo N÷-JÙ-àŸÙè….-

á:- S = 1s2 2s2 2p6

3s1 3px1 3py1 3pz1 3d1 3d1

S CyBóŸª ÑvCÚÛh þ§–ô³ö˺ÖÚÛ 3s, ÷´è[ª 3p, ·ôÙè[ª3d ÎJ(-æ°üŒ‰x ÖÚÛ-ë¯-EêÁÖÚÛæ¨ ÚÛLú‡ Ö¸Ú ô¢«í£Ù ÚÛL_ìÎô¢ª sp3d2 ú£ÙÚÛô¢ ÎJ(-æ°-üŒxìª Ôô¢p-ô¢ª-þ§hô³.- ÏN ÷ªSx6F í£ô¢-÷«-éª-õÚÛª àμÙCì pÎJ(-æ°-üŒxêÁ ÍA-ð§êŸÙ àμÙC 90 ñÙëÅ]-ÚÁ-é°õª ÚÛL-Tì ÎÚ¥d-šï°-vè[öËÀ ÎÚÛ”-A-E Ï-þ§hô³.-% s = 16.66%, % p = 50%

% d = 33.34%

3.- PCl5 Ôô¢p-è[-åÙö˺ P áJ›í ú£ÙÚÛ-K-ÚÛ-ô¢-é°Eo N÷-JÙ-àŸÙè…

á:- P = 1s2 2s2 2p6

3s1 3px1 3py1 3pz1 3d1

ðƧú£pÄ-ô¢úà 1÷ ÑvCÚÛhþ§–ô³-ö˺E ÖÚÛ 3s,÷´è[ª 3p, ÖÚÛ 3dÎJ(-æ°üŒ‰x ÖÚÛ-ë¯-EêÁ ÖÚÛæ¨ ÚÛLú‡Ö¸Ú-ô¢«í£Ù ÚÛL_ìÍô³ë]ª sp3dú£ÙÚÛô¢ ÎJ(-æ°-üŒxìª Ôô¢p-ô¢ª-þ§hô³.- ÏN ÷ªSx 5Clí£ô¢-÷«-éª-÷±-õÚÛª àμÙCì p ÎJ(-æ°-üŒxêÁ ÍA-ð§êŸÙàμÙC 90, 120 ñÙëÅ]-ÚÁ-é°õª ÚÛL-Tì wçËμjÞÁ-ìöËÀ òËμjí‡ô¢-Nª-è[öËÀ ÎÚÛ”AE Ôô¢p-ô¢ª-þ§hô³.-% s = 20% % p = 60% % d = 20%

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á:- Ö¸Ú Íéª÷± öË¶ë¯ î�¶¸ôyô¢ª Íéª-÷±-õÚÛª àμÙCì H+,X (X = F, O öË¶ë¯ N) õ ÷ªëÅ]u ñõ-ì ú‡–ô¢-N-ë]ªuê ÎÚÛ-ô¢{é ÷´õÙÞ¥ Ôô¢pè˶ ñÙëÅ] šïj°vèÁ-áû ñÙëÅ]Ù.-ÍÙêŸ-ô¢-éªÚÛ šïj°vèÁ-áû ñÙëÅ]Ù:- Ö¸Ú öË¶ë¯ î�¶¸ôyô¢ªú£î¶ªt-üŒ-û¦-õÚÛª àμÙCì ·ôÙè[ª öË¶ë¯ ÍÙêŸ-ÚÛÙç˶ÓÚÛª\÷ Íéª-÷±õ ÷ªëÅ]u Ôô¢pè˶ šïj°vèÁ-áû ñÙëÅ]Ù.-Ñë¯:- HF, H2O

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á:- H î¦óŸ³ Íéª-÷±õ ú£Þœåª ÞœA-øŒÚ¨h T.-H î¦óŸ³ Íéª-÷±õ ÷ªëÅ]u ÎÚÛ-ô¢{-é- ñ-ö°õª öË¶ë¯ NÚÛ-

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ëůEo êμL-óŸª-â˶›ú ú£Oª-ÚÛ-ô¢éÙ 1

V sò°ô³öËÀ EóŸª÷ªÙz.-.-.-.-.-.-.-.-(1) P

V T sàŦ·ôxúà EóŸª÷ªÙz.-.-.-.-.-.-.-.-.-(2)

V n sÍ÷-Þ¥vèÁ EóŸª÷ªÙz.-.-.-.-.-.-.-.-.-(3)

nT nTR1, 2, 3õ ìªÙ# V =

P P

PV = nTR

7.- ÖÚÛ ÚÛô¢(ì ú£î¶ªt-üŒ-ìÙö˺ 12.8% C, 2.1% H,85.1% Br Ñû¦oô³.- Î ú£î¶ªt-üŒìÙ Íéª-òÅ°ô¢Ù187.9 Íô³ê¶ ë¯E Íéª ðƧ-ô¢ªt-ö°ìª öËμÚ¨\Ù-àŸÙè….-

á:- C H Br 12.8 2.1 85.1 12 1 80

1.06 2.1 1.06 1.06 1.06 1.06

1 2 1

Íìª-òÅ°-NÚÛ ðƧô¢ªtö° = CH2Br

Íìª-òÅ°-NÚÛ ðƧô¢ªtö° òÅ°ô¢Ù =- 94

Í-éª-ðƧ-ô¢ªtö° òÅ°ô¢Ù 187.9n = = = 2

-Íìª-òÅ°-NÚÛ ðƧô¢ªtö° òÅ°ô¢Ù 94

Íéª-ðƧ-ô¢ªtö° = (CH2Br)2 = C2H4Br28.- šï°úà ú‡–ôÁù£g EóŸª-÷«Eo Eô¢y-#Ù# N÷-JÙ-àŸÙè….-á:- ÖÚÛ ô¢þ§-óŸªì àŸô¢uö˺ ÚÛL¸Þ ÓÙëǯLpÄ îμ³êŸhÙ

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1C + O2 CO H = 110.5 k.J

21

CO + O2 CO2 H = 283 k.J2

C + O2 CO2 H = 393.5 k.J

9.- 2SO2 sî¦z `- O2 sî¦z 2SO3 sî¦z àŸô¢uÚÛªKc, Kp õ ÷ªëÅ]u ú£ÙñÙ-ëůEo Ñê¦p-CÙ-àŸÙè….-

[SO3]2á:- Kc = [SO2]2 [O2]

P2SO3

Kp = P2

SO2.O2

PSO3= [SO3]RT ; PSO2 = [SO2] RT ;

PO2 = [O2] RT[SO3]2 [RT]2

Kp = [SO2]2 [RT]2 [O2] [RT]

= Kc [RT]2 (2 + 1)

Kp = Kc. [RT]1

10. M ÿ§-æ¨x-óŸªô ú£«vê¦Eo Eô¢y-#Ù#, NH3 êŸ-óŸ«-Kö˺íˆè[ìÙ, Ñÿ¼g-vÞœêŸ ví£òÅ°-î¦ö˺x Ð ú£«vêŸÙ Ñí£-óμ«-Þ¥Eo êμõ-í£Ùè….-

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Ú¥JÈ-ìu-êŸìª Óö° ê•õ-T-þ§hô¢ª?á:- ú£ñª(êÁ ê¶LÞ¥_ ìªô¢-Þœ-E-÷yE áö°Eo ÚÛJÈì áõÙ

ÍÙæ°ô¢ª.- Ca+2, Mg+2, Cl, SO4-2, HCO3

õªFæ¨ Ú¥JÈ-ìu-êŸìª ÚÛL-T-þ§hô³.-Ú¥-ö°_û í£ë]ÌÄA:- þ¼-è…óŸªÙ šï°Ú¥q îμªæ° ðƧ›úpÄæÀ(NaPO3)6 ìª Ú¥ö°_û ÍÙæ°ô¢ª.- (Cal =Calcium, gon = gone). ÚÛ-JÈì áö°-EÚ¨ Ú¥ö°_-ûÂEÚÛLí‡ Ca+2, Mg+2 Í-óŸ«-ìxìª ê•õ-T-þ§hô¢ª.-Na2[Na4(PO3)6] + 2 Ca+2

Na2[Ca2(PO3)6] + 4Na+

13.- ò˺ô¦ÚÂq ÍÙç˶ ÔNªæ¨? Ñë¯--ô¢-éêÁ ò˺ô¦ÚÂqí£²ú£ í£K¤Ûìª N÷-JÙ-àŸÙè….-

á:- Na2B4O7.- 10H2O E ò˺ô¦ÚÂq ÍÙæ°ô¢ª.- DEoî�¶è… ඛúh Þ¥V- ö°Ùæ¨ í£²ú£ìª Ôô¢p-ô¢ª-ú£ªhÙC.- NNëÅ]í£J-÷-ô¢hì ÷´õ-Ú¥õª GÅìo ô¢ÙÞœª-õª ÑÙè˶ ú£ÙñÙ-CÅêŸ îμªæ° -ò˺-̧ô-æÀ-õìª Ôô¢p-ô¢ª-þ§hô³.- ò˺ô¦-ÚÂqECoOêÁ î�¶è… ඛúh FL ô¢ÙÞœª í£²ú£ Ôô¢p-è[ª-꟪ÙC.-

Na2B4O7.10H2O Na2B4O710H2O

B2O3 + 2NaBO2

B2O3 + CoO Co(BO2)2 sFL ô¢ÙÞœªz

ÔÓûÂ-Óúà øŒÙÚÛ-ô¢-ô¦÷±Nù£óŸª Eí£±éªõª

ô¢àŸô³êŸ

WEóŸªô ÏÙåôÂ

ô¢þ§-óŸªì ø‹ú£YÙ100100

æ°·ô_æÀ 2020

Ñë¯:- H O : + H+ [ H O H]+

H H

: :

S

FF

F

FF

F

P

ClCl

ClCl

Cl120

90

1. H2O2 óμ³ÚÛ\ ·ôÙè[ª ÎÚ©q-ÚÛ-ô¢é, ¤ÛóŸª-ÚÛ-ô¢é ëÅ]ô¦t-õìª êμõ-í£Ùè….-

á:- Î-Ú©q-ÚÛ-ô¢é ëÅ]ô¦tõª:- H H2O2 ì-õxE PbS ìª êμõxE PbSO4Þ¥ ÎÚ©q-ÚÛ-

ô¢éÙ à¶ú£ªhÙC.-PbS + 4H2O2 PbSO4 + 4H2 O

H H2O2, Fe+2E Fe+3 Þ¥ ÎÚ©q-ÚÛ-ô¢éÙ à¶ú£ªhÙC.-2Fe+2 + H2O2 2Fe+3 + 2OH

¤Û-óŸª-ÚÛ-ô¢é ëÅ]ô¦tõª:-H H2O2, I2E I Þ¥ ¤ÛóŸª-ÚÛ-ô¢éÙ à¶ú£ªhÙC.-

I2 + H2O2 + 2 OH 2I + 2H2O + O2

H H2O2, HOClE ClÞ¥ ¤ÛóŸª-ÚÛ-ô¢éÙ à¶ú£ªhÙC.-HOCl + H2O2 H3O+ + Cl + O2

2. i) Ó-õ-vÚ¥dû ڕô¢êŸ Þœõ šïj°wèËμj-èÂõª ii) Ó-õ-vÚ¥d-ûÂõªÍCÅ-ÚÛÙÞ¥ Þœõ šïj°wèËμj -èÂõª iii) Ó-õ-vÚ¥d-ûÂõª ÚÛ#a-êŸÙÞ¥ Ñìo šïj°wèËμj-èÂõª iv) Í-óŸ«-EÚÛ šïj°wèËμj-èÂõª ÍÙç˶ ÔNªæ¨?

á:- i) õ«ô´ Eô¦t-é°õª ô¦óŸª-è¯-EÚ¨ Ú¥î¦-Lqì Óõ-vÚ¥d-ûÂõ ÚÛÙç˶ êŸÚÛª\÷ Óõ-vÚ¥d-ûÂõª Þœõ 13÷vÞœ«í£± ÷´õ-Ú¥õ šïj°wèËμj-èÂõª. Ñë¯:- B2H6

ii) õ«ô´ Eô¦t-é°õª ô¦óŸª-è¯-EÚ¨ Ú¥î¦-Lqì Óõ-vÚ¥d-ûÂõ ÚÛÙç˶ ÓÚÛª\÷ Óõ-vÚ¥d-ûÂõª Þœõ 15,16, 17÷ vÞœ«í£± ÷´õ-Ú¥õ šïj°wèËμj-èÂõª.-Ñë¯: H2O, NH3

iii) õ«ô´ Eô¦t-é°õª ô¦óŸª-è¯-EÚ¨ Ú¥î¦-LqìÓõ-vÚ¥d-ûÂõª Þœõ 14÷ vÞœ«í£± ÷´õ-Ú¥õšïj°wèËμj-èÂõª.- Ñë¯:- CH4

iv) ÍCÅÚÛ ëÅ]ì Në]ªu-ë¯-êŸt-ÚÛêŸ Þœõ s ò°xÚÛª÷´õ-Ú¥õª Ôô¢p-Jචšïj°wèËμj-èÂõª.- Ñë¯:- NaH

3.- pH E Eô¢y-#Ù#, 0.05 M H2SO4 óμ³ÚÛ\ pHE öËμÚ¨\Ù-àŸÙè….-

á:- H+ Í-óŸ«û ޥèÅ[êŸ óμ³ÚÛ\ ô¢ªé ú£Ù÷-ô¢_-÷«-û¦EopH ÍÙ-æ°ô¢ª.-N = [H+] = 0.05 2 = 0.1 = 101

pH = - log [101] = (1) = +1

÷ªJÚ•Eo..

Ðúˆšï°àÂÓúÃö˺ Ý°Sõª šïj°ë]ô¦ò°ëÂö˺E ÓÚÂqú£KyúÃîμªûËÂÚÛÙvæ¨ñ«uæK šï°öËÀh úˆ\îªsÐúˆšï°àÂÓúÃz Ú¨ÙC Ý°SöËòÅ¡KhÚ¨ ë]ô¢Ý°ú£ªhõª ÚÁô¢ªêÁÙC. H îμ³êŸhÙ Ý°Sõª: 114ð¼ú£ªdõª: îμªè…ÚÛöËÀ šúpù£LúÃd, îμªè…ÚÛöËÀ Îíƈú£ôÂ, èËμÙæöËÀÎíƈú£ôÂ, ìJqÙÞ êŸCêŸô¦õª. Íô¢|êŸ: ð¼ú£ªdEÍìªú£JÙ# ú£ÙñÙCÅêŸ ú£òËμbÚÛªdö˺x è…ð»x÷«, HÓúˆq,Hè†ÓúÃ, HðƧô¢túˆ, ÓÙHHÓúÃ, ÓÙè†/ ÓÙÓúÃ/ è†ÓûËÂHÑBhô¢gêŸ, ÍìªòÅ¡÷Ù.ÓÙí‡ÚÛ NëůìÙ: ÏÙæô¢«yuÎëůô¢ÙÞ¥. ÏÙæô¢«yu ê¶Dõª: 2020 áì÷J 28,íƇvñ÷J 05. ë]ô¢Ý°ú£ªh NëůìÙ: ÎíÆÃöËμjûËÂ. #÷Jê¶C: áì÷J 8, 2020. #ô¢ªû¦÷«:Ðúˆšï°àÂÓúà šúöËÀ, êμöËÙÞ¥é ÍÙè ÎÙvëů ú£òËÀÔJóŸ«, ò˹ö°xô¢Ù ð¼úÃd, ú‡Ú¨Ùvë¯ò°ëÂ.îËμòËÀšújæËÀ: https://tasaechs.telangana.gov.in

ÓûËÂÕæ©ÕÐ n 2020 ÷³Ùñô³s÷ªï£„ô¦ù£Zzö˺E û¶ù£ìöËÀÏûËÂú‡då«uæËÀ ÎíÆà ÏÙè[vú‡dóŸªöËÀÏÙ>FJÙÞ sÓûËÂÕæ©ÕÐz n 2020 Në¯uú£Ù÷êŸqô¦EÚ¨ Ú¨ÙC vð¼vÞ¥÷³ö˺x ví£î˶ø‹öËÚÛªë]ô¢Ý°ú£ªhõª ÚÁô¢ªêÁÙC.H ÓûËÂÕæ©ÕÐ íˆ@è† vð¼vÞ¥÷³õªn2020ÚÁô¢ªqõª: ð¼úÃd vÞ¥è[ªuó¶ªæËÀ è…ð»x÷« sÏÙè[vú‡dóŸªöËÀÏÙ>FJÙÞÂ, ÏÙè[vú‡dóŸªöËÀ ›úíƈd ÍÙè ÓEyô¦ûËÂîμªÙæËÀû¶âËËÀîμªÙæËÀ, ÏÙè[vú‡dóŸªöËÀ û¶âËËÀîμªÙæËÀ, ÷«ìªuðƧÚÛaJÙÞ û¶âËËÀîμªÙæËÀ, vð§âËμÚÂd û¶âËËÀîμªÙæËÀz.Íô¢|êŸ: ñ°u#öËôÂq è…vU, ÏÙ>FJÙÞ è…vU ÑBhô¢gêŸ.ë]ô¢Ý°ú£ªh NëůìÙ: ÎûËÂöËμjûËÂ.ÓÙí‡ÚÛ NëůìÙ: Ú¥uæËÀ, ¸ÞæËÀ þ¼\ôÂ, vÞœ«í£± è…ú£\ù£ûËÂ,í£ô¢qìöËÀ ÏÙæô¢«yu Îëůô¢ÙÞ¥. #÷Jê¶C: íƇvñ÷J 3, 2020.

îËμòËÀšújæËÀ: https://www.nitie.edu/

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