NCERT Solutions for Class 12 Chemistry Chapter 1 The Solid State are part of NCERT Solutions for Class 12 Chemistry. Here we have given NCERT Solutions for Class 12 Chemistry Chapter 1 The Solid State.
|Chapter Name||The Solid State|
|Number of Questions Solved||50|
NCERT Solutions for Class 12 Chemistry Chapter 1 The Solid State
NCERT IN-TEXT QUESTIONS
Why are solids rigid ?
Solids are rigid because the constituent particles are very closely packed. They donot have any translatory movement
and can only oscillate about their mean positions.
Why do solids have definite volume ?
Solids keep their volume because of rigidity in their structure. The interparticle forces are very strong. Moreover, the interparticle spaces are very few and small as well. As a result, their volumes cannot change by applying pressure.
Classify the following as amorphous and crystalline solids ; polyurethane, naphthalene, benzoic acid, teflon,
potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.
Amorphous solids : Polyurethane, naphthalene, teflon, cellophane, polyvinyl chloride, fibre glass. Crystalline solids : Benzoic acid, potassium nitrate, copper.
Why is glass considered as super cooled liquid ? (C.B.S.E. Delhi 2013)
Glass is considered to be super cooled liquid because it shows some of the characteristics of liquids, though it is an amorphous solid. For example, it is slightly thicker at the bottom. This can be possible only if it has flown like liquid, though very slowly.
Refractive index of a solid is observed to have the same value along all the directions. Comment on the nature of the solid. Would it show cleavage property ?
As the solid has the same refractive index along all the directions, it is isotropic in nature and is therefore, an amorphous solid. It is not expected to show a clean cleavage when cut with a special type of knife. It will break into pieces with irregular surfaces.
Classify the following solids in different categories based on the nature of the inter molecular forces : sodium sulphate, copper, benzene, urea, ammonia, water, zinc sulphide, diamond, rubedium, argon, silicon
Ionic, metallic, molecular, molecular, molecular (hydrogen bonded), molecular (hydrogen bonded), ionic, covalent, metallic, molecular, covalent (net work).
A solid substance ‘A’ is very hard and electrical insulator both in the solid state as well as in molten state. It has also very high melting point. Is the solid metal like silver or network solid like silicon carbide (SiC) ?
Since the solid behaves as an insulator even in the molten state, it cannot be a metal like silver. Therefore, it is a covalent or network solid like SiC.
Why are ionic solids conducting in the molten state and not in the solid state ?
In the ionic solids, the electrical conductivity is due to the movement of the ions. Since the ionic mobility is negligible in the solid state, these are non-conducting in this state. Upon melting, the ions present acquire some mobility. Therefore, the ionic solids become conducting
What types of solids are electrical conductors, malleable and ductile ? (C.B.S.E. Outside Delhi 2013)
Metallic solids exhibit these characteristics. Their atoms are linked to one another by metallic bonds.
Give the significance of a lattice point.
The lattice point denotes the position of a particular constituent in the crystal lattice. It may be atom, ion or a molecule. The arrangement of the lattice points in space is responsible for the shape of a particular crystalline solid.
Name the parameters which characterise a unit cell.
A unit cell is characterised by two types of parameters. These are : edges (a, b, c) which may or may not be mutually perpendicular and angles between the edges (α, β and γ).
(i) Edges or edge lengths. The edges a, b and c represent the dimensions of the unit cell in psace along the three axes. The edges may or may not be mutually perpendicular.
(ii) Angles between the edges. There are three angles between the edges. These are denoted as α (between b and c), β (between a and c) and γ (between a and b). Thus, a unit call may be characterised by six parameters as a shown in the Fig. 1.12. The various types of crystal systems differ with respect to edge lengths as a well as angles between the edges.
Distinguish between :
(i) Hexagonal and monoclinic unit cells
(ii) Face-centred and end-centred unit cells.
(i) In a hexagonal unit cell :
a = b # c; α = β = 90° and γ = 120°
In a monoclinic unit cell :
a # b # c and α = γ = 90° and β # 90°
(ii) In a face centred unit cell, constituent particles are located at all the corners as well as at the centres of all the faces.
In end-centred unit cell, constituent particles are located at all the corners as well as at the centres of two opposite faces. (C.B.S.E Foreign 2015)
Explain how much portion of an atom located at the
(i) corner and
(ii) body centre of a cubic unit cell is a part of the neighbouring unit cell.
(i) An atom located at the corner is shared by eight unit cells. Therefore, its contribution to a particular unit cell is 1/8.
(ii) An atom located at the body of the unit cell is not shared by any unit cell. It belongs to one particular unit cell only.
What is the two dimensional co-ordination number of a molecule in a square close packed layer ?
In the two dimensional square close packed layer, a particular molecule is in contact with four molecules. Hence, the co-ordination number of the molecule is four.
A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it ? How
many of these are tetrahedral voids ?
No. of atoms in 0.5 mole of the compound = 0.5 x N0 = 0.5 x 6.022 x 1023 = 3.011 x 1023
No. of octahedral voids = No. of atoms = 3.011 x 1023
No. of tetrahedral voids = 2 x 3.011 x 1023 = 6.022 x 1023
Total no. of voids = (3.011 + 6.022) x 10223 = 9.033 x 1023
A compound is formed by two elements M and N. The element N forms ccp and atoms of the element M occupy 1/3 of the tetrahedral voids. What is the formula of the compound ? (C.B.S.E. Foreign 2015)
Let us suppose that,
the no. of atoms of N present in ccp = x
Since 1/3rd of the tetrahedral voids are occupied by the atoms of M, therefore,
the no. of tetrahedral voids occupied = 2x/3
The ratio of atoms of N and M in the compound = x : 2x/3 or 3 : 2
∴ The formula of the compound = N3M2 or M2N3
Which of the following lattices has the highest packing efficiency :
(i) simple cubic
(ii) body-centred cubic and
(iii) hexagonal close packed lattice ?
The packing efficiency of the different types of arrangement is :
(i) Simple cubic = 52.4%
(ii) Body-centred cubic = 68%
(iii) Hexagonal close packed = 74%
his means that hexagonal close packed arrangement has the maximum packing efficiency (74%).
An element with molar mass 2:7 x 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2:7 x 103 kg m-3, what is the nature of the cubic unit cell ? (C.B.S.E. Delhi 2015)
Since there are four atoms per unit cell, the cubic unit cell must be face centred (fcc) or cubic close packed (ccp).
What type of defect can arise when a solid is heated ? Which physical property is affected by it and in what
When a solid is heated, some atoms or ions may leave the crystal lattice. As a result, vacancies are created and this leads to vacancy defects in the crystalline solid. Since the number of atoms/ions per unit volume decreases, the vacancy defects lead to decrease in the density.
What types of stoichiometric defects are shown by (C.B.S.E. Delhi 2013)
(i) ZnS crystals may show Frenkel defects since the cationic size is smaller as compared to anionic size.
(ii) AgBr crystals may show both Frenkel and Schottky defects.
Explain how vacancies are introduced in the ionic solid when a cation of higher valence is added as impurity to it.
Let us consider an ionic solid sodium chloride (Na+Cl–) to which a small amount of strontium chloride (SrCl2) has been added to act as impurity. Since the crystal as a whole is to remain electrically neutral, two Na+ ions have to leave their sites to create two vacancies. Out of these, one will be occupied by Sr2+ ion while the other will be vacant. Thus, vacancies will be created in the ionic solid. When a cation of higher valency is added as an impurity in the ionic solid, some of the sites of the original cations are occupied by the cations of higher valency. Each cation of higher valency replaces two or more original cations and occupies the site of one original cation and the other site(s) remains vacant.
Ionic solids which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Let us illustrate by sodium chloride (Na+Cl–) crystals. Upon heating in the atmosphere of sodium (Na) vapours, sodium atoms get deposited on the surface of the crystals. The Cl– ions from the crystal lattice leave their sites and diffuse into the surface. They tend to combine with sodium atoms present in the vapours which in turn get ionised to form Na+ ions by releasing electrons. The latter are trapped by the anionic vacancies created by Cl–ions in order to maintain the crystals electrically neutral. Now, the electrons absorb radiations corresponding to a certain colour from white light and start vibrating. They emit radiations corresponding to yellow colour. That is how, the crystals of sodium chloride develop yellow colour. These electrons are called F-centres because these are responsible for colours (In German, F = Farbe meaning colour).
A group 14 element is to be converted into n-type semi-conductor by doping it with a suitable impurity. To which group should the impurity element belong ?
n-type semiconductors are conducting due to the presence of excess of negatively charged electrons. In order to convert group 14 elements (e.g. Si, Ge) into n-type semi-conductors, doping is done with some elements of group 15 (e.g. P, As)
What type of substances would make better permanent magnets ; ferromagnetic or ferrimagnetic ? Justify your answer. (C.B.S.E. Outside Delhi 2013)
Ferromagnetic substances make better permanent magnets than ferrimagnetic substances. The metal ions of a ferromagnetic substance are grouped into small regions known as domains and these are randomly oriented. Under the influence of the applied magnetic field, all domains are oriented in the direction of the magnetic field and as a result, a strong magnetic field is produced. The ferromagnetic substance behaves as a magnet. This characteristic of the domains persists even when the external magnetic field is removed. This imparts permanent magnetic character to these substances. However, this property is lacking in ferrimagnetic substances. Therefore ferromegnetic substances are better magnets.
Define the term ‘amorphous’. Give a few examples of amorphous solids.
Amorphous or amorphous solids imply those solids in which the constituent particles have short range order. These have irregular shapes and are isotropic in nature. Apart from that they do not have sharp melting points. A few examples of amorphous solids are : glass, rubber, plastic, celluose etc.
What makes glass different from a solid such as quartz ? Under what conditions could quartz be converted into glass?
Glass is a super cooled liquid and an amorphous substance. Quartz is the crystalline form of silica (SiO2) in which tetrahedral units SiO4 are linked with each other in such a way that the oxygen atom of one tetrahedron is shared with another Si atom. Quartz can be converted into glass by melting it and cooling the melt very rapidly. In glass, Si04 tetrahedra are joined in a random manner.
Classify each of the following solids as ionic, metallic, molecular, net work (covalent) or amorphous :
(a) Tetra phosphorus decoxide (P4O10)
(d) Ammonium phosphate (NH4)3PO4
(a) Molecular solid
(b) Covalent (Net-work) solid
(c) Metallic solid
(d) Ionic solid
(e) Covalent solid (Network)
(f) Metallic solid
(g) Molecular solid
(h) lonic solid
(i) Molecular solid
(j) Covalent solid
(k) Amorphous solid.
(a) What is meant by the term coordination number ?
(b) What is the co-ordination number of atoms
(i) in a cubic close packed structure
(ii) in a body centred cubic structure ?
(a) The co-ordination number of a constituent particle (atom, ion or molecule) is the number of nearest neighbours in its contact in the crystal lattice.
How can you determine the atomic mass of an unknown metal if you know its density and dimensions of its unit cell ? Explain your answer. (C.B.S.E. Outside Delhi 2011)
(a) Stability of a crystal is reflected in the magnitude of the melting point. Comment.
(b) Collect the melting point of
- ethyl alcohol
- diethyl ether
- methane from a data book. What can you say about intermolecular forces between the molecules ?
(a) The stability of a crystal depends upon the magnitude of force of interaction in the constituting particles. Greater the force of attraction present, more will be the stability of the crystal. For example, ionic solids such as Naci, KCl etc. have very high melting and boiling points while the molecular crystals such as naphthalene, iodine etc. have low values of melting and boiling points.
(b) The melting points of different substances are :
- Ice = 273 K
- Ethyl alcohol = 155.7 K
- Diethyl ether = 156.8 K
- Methane = 90-5 K
The intermolecular forces in molecules of ice and ethyl alcohol are mainly hydrogen bonding. The magnitude is more in ice than in alcohol as is evident from the values of melting points.
The forces in the molecules of diethyl ether are dipolar forces while in methane, these are mainly the van der Waal’s forces which is quite evident from its melting point (least among the compounds listed).
How will you distinguish between the following pairs of terms :
(a) Hexagonal close packing and cubic close packing
(b) Crystal lattice and unit cell
(c) Tetrahedral void and octahedral void.
(a) In hexagonal close packing (hcp), the spheres of the third layer are vertically above the spheres of the first layer
(ABABAB……. type). On the other hand, in cubic close packing (ccp), the spheres of the fourth layer are present above the spheres of the first layer (ABCABC…..type).
(b) Crystal lattice: It deplicts the actual shape as well as size of the constituent particles in the crystal. It is therefore, called space lattice or crystal lattice.
Unit cell: Each bricks represents the unit cell while the block is similar to the space or crystal lattice. Thus, a unit cell is the fundamental building block of the space lattice.
(c) Tetrahedral void: A tetrahedral void is formed when triangular void made by three spheres of a particular layer and touching each other.
Octahedral void: An octahedral void or site is formed when three spheres arranged at the corners of an equilateral triangle are placed over anothet set of spheres.
How many lattice points are there in one unit cell of each of the following lattices
(a) face centred cubic
(b) face centred tetragonal
(c) body centred cubic ?
(a) In face centred cubic arrangement,
Lattice points located at the corners of the cube = 8
Lattice points located at the centre of each face = 6
Total no. of lattice points = 8 + 6 = 14
(b) In face centred tetragonal, the number of lattice points is also the same i.e., 8 + 6 = 14.
(In both the cases, particles per unit cell = 8 xX 1/8 + 6 x 1/2 = 4)
(c) In body centred cubic arrangement,
Lattice points located at the corners of the cube = 8
Lattice points located in the centre of the body = 1
Total no. of lattice points = 8 + 1 = 9
(the total number of particles per unit cell = 8 x 1/8 + 1 = 2)
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
(i) Basis of similarities. The basis of similarities between the metallic and ionic crystals are the presence of strong electrostatic forces of attraction. These are present among the ions in the ionic crystals and among the kernels and
valence electrons in the metallic crystals. That is why both metals and ionic compounds are good conductors of electricity and have high melting points.
Basis of differences. The basis of differences is the absence of mobility of ions in the ionic crystals while the same is present in the valence electrons and kernels in case of metallic crystals. As a consequence, the ionic compounds conduct electricity only in the molten state while the metals can do so even in the solid state.
(ii) The ionic solids are hard and brittle because of strong electrostatic forces of attraction which are present in the oppositely charged ions.
The ionic solids are hard because of the presence of strong inter ionic forces of attraction in the oppositely charged ions. These ions are arranged in three dimensional space. The ionic solids are brittle because the ionic bond
Calculate the efficiency of packing in case of metal crystal for :
(i) Simple cubic
(ii) Body centred cubic
(iii) Face centred cubic (with the assumption that the atoms are touching each other). (C.B.S.E. Outside Delhi 2011) Answer:
(i) Simple cubic: We know that in a simple cubic unit cell, there is one atom (or one sphere) per unit cell. If r is the radius of the sphere, volume occupied by one sphere present in the unit cell = 4/3πr3.
(ii) Body-centred cubic: We know that a body- centered cubic unit cell has 2 spheres (atoms) per unit cell. If r is the radius of the sphere Volume of one sphere = 4/3πr,sup>3
(iii) Face centred cubic: We know that a face centered cubic unit cell (fcc) contains four spheres (or atoms) per unit cell.
Silver crystallises in a face centred cubic lattice with all the atoms at the lattice points. The length of the edge of
the unit cell as determined by X-ray diffraction studies is found to be 4.077 x 10-8 cm. The density of silver is 10.5 g cm-3. Calculate the atomic mass of silver. (C.B.S.E. Sample Paper 2012)(Uttarakhand Board 2015)
A cubic solid is made of two elements P and Q. Atoms Q are at the corners of the cube and P at the body centre. What is the formula of the compound ? What is the co-ordination number of P and Q?
Contribution by atoms Q present at the eight corners of the cube = = x 8 = 1
Contribution by atom P present at the body centre = 1
Thus, P and Q are present in the ratio 1:1.
∴ Formula of the compound is PQ.
Co-ordination number of atoms P and Q = 8.
Niobium crystallises in body centred cubic structure. If the density is 8.55 g cm-?, calculate atomic radius of niobium given that atomic mass of niobium is 93 g mol-1. (C.B.S.E. Delhi 2008)
Step I. Calculation of edge length of unit cell.
No. of particles in b.c.c. type unit cell (Z) = 2
Atomic mass of the element (M) = 93 g mol-1
Step II. Calculation of radius of unit cell
If the radius of octahedral void is r and the radius of the atom in close packing is R, derive the relation between r and R. (C.B.S.E. Sample Paper 2017)
Let length of each side of the square is a and the radii of the void and the sphere are r and R respectively. Consider the right angled triangle ABC.
∴ Radius of octahedral void is 0.414R(or 41.4% as compared to that of the sphere).
Copper crystallises into a foc lattice with edge length 3•61 x 10-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm. (C.B.S.E. Delhi 2009 Comptt.)
The calculated value of the density is nearly the same as the measured value.
Analysis shows that nickel oxide has formula Nin.os 01.00. What fraction of nickel exists as Ni2+ and as Ni3+ ions ?
The ratio of Ni and O atoms in pure nickel oxide (NiO) = 1:1
Let x be the no. of Ni (II) atoms replaced by Ni (III) atoms in the oxide.
∴ No. of Ni (II) atoms present = (0.98 – x)
Since the oxide is neutral in nature,
Charge on Ni atoms = Charge on oxygen atoms
2(0.98 – x) + 3x = 2
1.96 – 2x + 3x = 2
x = 2 – 1.96 = 0.04
% of Ni (II) atoms in nickel oxide = 100 – 4:01 = 95.99%
What are semi-conductors ? Describe the two main types of semiconductors and contrast their conduction mechanisms.
Semi-conductors are the substances whose conductivity lies in between those of conductors and insulators. The two
main types of semi-conductors are n-type and p-type.
(i) n-type semiconductor: When a silicon or germanium crystal is doped with group 15 element like P or As, the dopant atom forms four covalent bonds like Si or Ge atom but the fifth electron, not used in bonding, becomes delocalised and continues its share towards electrical conduction. Thus silicon or germanium doped with P or As is called H-type semiconductor, a-indicative of negative since it is the electron that conducts electricity.
(ii) p-type semiconductor: When a silicon or germanium is doped with group 13 element like B or Al, the dopant is present only with three valence electrons. An electron vacancy or a hole is created at the place of missing fourth electron. Here, this hole moves throughout the crystal like a positive charge giving rise to electrical conductivity. Thus Si or Ge doped with B or Al is called p-type semiconductor, p stands for positive hole, since it is the positive hole that is responsible for conduction.
Non-stoichiometric cuprous oxide (Cu2O) can be prepared in the laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semi-conductor ?
The ratio less than 2:1 in Cu2O shows that some cuprous (Cu2+) ions have been replaced by cupric (Cu2+) ions. In order to maintain the electrical neutrality, every two Cut ions will be replaced by one Cu2+ ion which results in creating cation vacancies leading to positive holes. Since the conduction is due to positive holes, it is a p-type semiconductor.
Ferric oxide crystallises in a hexagonal close packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of ferric oxide.
There is one octahedral hole for each atom in hexagonal close packed arrangement.
If the number of oxide ions (O2-) per unit cell is 1, then the number of Fe3+ ions = 2/3 x octahedral holes = 2/3 x 1 = 2/3.
Thus, the formula of the compound = Fe2/3O1, or Fe2O3.
Classify each of the following as being either a p-type or an n-type semi-conductor.
(i) Ge doped with In
(ii) B doped with Si.
(a) p-type semi conductor
(b) n-type semi conductor.
(i) Ge is group 14 element and In is group 13 element. Hence, an electron deficient hole is created and therefore, it is a p – type semiconductor.
(ii) B is group 13 element and Si is group 14 element, there will be a free electron, So, it is an n-type semiconductor.
Gold (atomic radius = 0.144 nm) crystallises in a face centred unit cell. What is the length of the side of the unit cell ?
For a face centred cubic unit cell (fcc)
Edge length (a) = = 2 x 1.4142 x 0.144 mm = 0.407 nm
In terms of Band Theory, what is the difference
(i) between a conductor and an insulator
(ii) between a conductor and semi-conductor ?
The variation in the electrical conductivity of the solids can be explained with the help of the band theory.
(i) In insulators, the energy gaps are very large and the no electron jump is feasible from the valence band to the conduction band. The energy gaps also called as forbidden zones. The insulators therefore, do not conduct electricity.
(ii) In semi-conductors, there is small energy gap between valence band and conduction band. However, some electrons may jump to the conduction band and these semi-conductors can exhibit a little electrical conductivity.
Explain the following terms with suitable examples.
(i) Schottky defect,
(ii) Frenkel defect,
(i) Schottky defect: This rises because certain ions are missing from the crystal lattice and vacancies or holes are created at their repective positions. Since a crystal is electrically neutral, the number of such missing cations (A+) and anions (B–) must be the same. e.g., KCl, NaCl, KBr, etc.
(ii) Frenkel defect: It results when certain ions leave their normal sites and occupy positions elsewhere in the crystal lattice. Holes are created at their respective positions. Since cations are smaller in size as compared to anions normally these are involved in Frenkel defect. e.g., AgBr, ZnS, etc.
(iii) Interstitials: This defect is noticed when constituent particles (atoms or molecules) occupy the interstutal sites in the crystal lattice. As a result, the number of particles per unit volume increases and so the density of the solid.
(iv) F-centres: These are the anionic sites occupied by unpaired electrons. F-centres impart colour to crystals. The colour results by the excitation of electrons when they absorb energy from the visible light falling on the crystal.
Aluminium crystallises in a cubic close packed structure. Its metallic radius is 125 pm.
(a) What is the length of the side of the unit cell ?
(b) How many unit cells are there in 1.00 cm? of aluminium ? (C.B.S.E. Outside Delhi 2013)
Step I. Calculation of length of side of the unit cell
For f.c.c. unit cell, a = = (125pm) = 2 x 1.4142 x (125 pm) = 354 pm.
Step II. Calculation of no. of unit cells in 1:00 cm3 of aluminium.
Volume of one unit cell = (354 pm)3 = (354 x 10-10 cm)3 = 44174155 x 10-30 cc.
If NaCl is doped with 10-3 mol % of SrCl2, what is the concentration of cation vacancies?
Explain the following with suitable examples.
(b) Piezoelectric effect
(e) Antifluoride structure
(f) 12 – 16 and 13 – 15 compounds.
(a) Ferromagnetism: A few solids like iron. cobalt, nickel, gadolinium and CeO,sub.2 are attracted very strongly by magnetic fields. These are known as ferromagnetic solids. Apart from that, they can be even permanently magnetised or become permanent magnet. e.g., Fe, Ni, Co and CrO2
(b) Piezoelectric effect: A dielectic crystal which has a resultant dipole moment can produce electricity or show electrical property when external pressure is applies. Such a crystal is known as piezoelectric crystal and this property is called piezoelectricity or pressure electricity. e.g., PbZrO2, Nh4H2PO4 etc.
(c) Paramagnetism: These are the solids attracted by a magnet. Actually, the atoms of the elements present have certain unpaired electrons. Their spins or magnetic moments may lead to magnetc character. Many transition metals such as Co, Ni, Fe, Cu, etc. and their ions are paramagnetic. e.g., O2, Cu2+, Fe3+, etc
(d) Ferrimagnetism: They have certain resultant magnetic moment or magnetic character which is of permanent nature. However, ferrimanetic solids are less magnetic than ferromagnetic solids. For example, magnetic oxide of iron (Fe,sub.3O4) and ferrites with general formula MFe2O4. e.g., Fe3O4
(e) Antifluoride structure: In this structure, the positions of the cations and anions as compared to fluorite structure get reversed i.e. the smaller cations occupy the position of fluoride ions while the anions with bigger size occupy the positions of calcium ions. e.g., Li2O, K2O, Rb2O and Rb2S.
(f) 12 – 16 and 13 – 15 compounds: A large variety of solid state materials have been prepared by the combination of elements belonging to group 13 and 15 or group 12 and 16. A few examples of compounds 13-15 combinations are: InSb, Alp and GaAs. Similarly, compounds resultinf from 12 – 16 combination are AdS, CdSe, HgTe.
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