This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 28801. |
The correct decreasing order for acid strength is(1) NO2 CH2 COOH > NCCH2 COOH > FCH2 COOH > ClCH2COOH (2) FCH2COOH > NCCH2 COOH > NO2 CHCOOH > ClCH2 COOH (3) NO2CH2 COOH > FCH2 COOH > NCCH2 COOH > ClCH2 COOH (4) NCCH2COOH > O2NCH2 COOH > FCH2 COOH > ClCH2COOH |
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Answer» Correct option (1) NO2 CH2 COOH > NCCH2 COOH > FCH2 COOH > ClCH2COOH Explanation: EWG increase acidic strength NO2 CH2 COOH > NCCH2 COOH > FCH2 COOH > ClCH2COOH |
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| 28802. |
The decreasing order of electrical conductivity of the following aqueous solutions is : 0.1 M Formic acid (A), 0.1 M Acetic acid (B), 0.1 M Benzoic acid (C) (1) A > B > C (2) A > C > B (3) C > B > A (4) C > A > B |
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Answer» Correct option (2) A > C > B Explanation: Order of acidity : A > C > B Order of electrical conductivity : A > C > B |
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| 28803. |
Suggest two materials other than hydrogen that can be used as fuels in fuel cells |
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Answer» Methane and Ethane |
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| 28804. |
Suggest two materials other than hydrogen that can be used as fuels in fuel cells. |
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Answer» Methane and methanol. |
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| 28805. |
Why does the conductivity of a solution decrease with dilution ? |
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Answer» Conductivity of a solution is dependent on the number of ions per unit volume. On dilution, the number of ions per unit volume decreases, hence the conductivity decreases. |
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| 28806. |
Depict the galvanic cell in which the reaction Zn (s) + 2Ag+ -->Zn2+ + 2Ag (s)takes place. Further show : (a) Which of the electrode is negatively charged? (b) The carriers of the current in the cell. (c) Individual reaction at each electrode. |
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Answer» Zn (s)|Zn2+ (aq) || Ag+ (aq)|Ag (s) (a) Zn electrode (anode) (b) Ions are carriers of the current in the cell. (c) At anode : Zn (s) --> Zn2+ + 2e- At cathode : Ag+ + e- --> Ag (s) |
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| 28807. |
What are sacred groves? Where are they found? |
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Answer» Sacred Groves:- Sacred groves are the secret patches around places of worship. They are of a great religious value among tribal communities. In such cases nature is protected by prevailling religious and cultural traditions. Here tracts of forests are set a side and all plants and animals are venerated and provided with complete protection. Example of sacred groves are Kashi and Jaintia hills in Meghalaya, Aravali Hills in Rajasthan, westen Ghat regions of Karnataka, Maharastra and Sarguja (MP), Chanda and Batsar areas of M.P., Kaimul in Bihar, Rajgir and Brabar Hills in Bihar regions Hazaribagh in Jharkhand. In Sikkim, Khecheo palri lake is decleared sacred by peope, thus protecting the aquatic flora and fauna. Role in conservatino:- Many rare and threatened plants have bun protected in sacredgroves of Meghalaya. Such areas have been found to be most undistrubed and they are usually surrounded bymost degraded lands capes. |
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| 28808. |
Which one of the following is not the Characteristics of the Social Class ?(a) Class Concioussness(b) Achieved membership(c) Based on birth(d) Open System. |
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Answer» Based on birth is not the Characteristics of the Social Class. |
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| 28809. |
Which Statement is not Correct amongst the following ?(a) Restriction on food are found in the caste system(b) Caste system is based on birth(c) Hierarchy is not found in the Caste system(d) Hierarchy is found in the Caste system. |
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Answer» Hierarchy is not found in the Caste system is not Correct . |
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| 28810. |
Bullet train in india was run with the help of which country?1. Japan2. China3. Russia4. USA |
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Answer» Correct Answer - Option 1 : Japan Concept:
Explanation:
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| 28811. |
In which state is Sriharikota located?1. Andhra Pradesh2. Tamilnadu3. Orissa4. Gujarat |
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Answer» Correct Answer - Option 1 : Andhra Pradesh The correct answer is Andhra Pradesh.
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| 28812. |
What name is given to the path of a planet around the sun?1. Route2. Milky Way3. Cosmic highway4. Orbit |
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Answer» Correct Answer - Option 4 : Orbit The correct answer is Orbit.
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| 28813. |
Most southern state of India?1. Karnataka2. Kerla3. Tamil Nadu4. Andhra Pradesh |
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Answer» Correct Answer - Option 3 : Tamil Nadu Concept:
Explanation:
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| 28814. |
Rafale jet come from which country?1. France2. Japan3. China4. Russia |
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Answer» Correct Answer - Option 1 : France Explanation:
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| 28815. |
What is responsible for the chemical transmission of messages between cells?1. Synapse2. Neurotransmitters3. Axons4. Dendrites |
Answer» Correct Answer - Option 1 : Synapse
Let us discuss other options as well:
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| 28816. |
Electronic City of India?1. Delhi2. Chandigarh3. Bangalore4. Indore |
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Answer» Correct Answer - Option 3 : Bangalore Explanation:
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| 28817. |
Neither my friend ______ I like playing soccer. A) or B) and C) but D) nor |
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Answer» Correct option is D) nor |
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| 28818. |
Who wrote 'Origin of Species'?1.Einstein2. Gibbs3. Newton4.Charles Darwin |
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Answer» Correct Answer - Option 4 : Charles Darwin Origin of Species by Natural Selection:
Charles Darwin:
Thus, Charles Darwin wrote 'Origin of Species'.
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| 28819. |
In which of the following years Kabir Purushkar award started?1. 19872. 19883. 19894. 1986 |
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Answer» Correct Answer - Option 4 : 1986 The correct answer is 1986.
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| 28820. |
Out of NaCl and AgCl, which one shows Frenkel defect and why? |
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Answer» AgCl shows Frenkel defect due to large difference in size between the positive (Ag+) and negative (Cl−) ion |
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| 28821. |
what gives covalent bond its strength? |
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Answer» The strength of a covalent bond is measured by its bond dissociation energy, that is, the amount of energy required to break that particular bond in a mole of molecules. Multiple bonds are stronger than single bonds between the same atoms. |
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| 28822. |
The coagulation of `100mL` of a colloidal solution of gold is completely prevented by adding `0.25g` of starch to it before adding `10mL` of `10%NaCl` solution. The gold number of starch isA. `24`B. `240`C. `120`D. `360` |
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Answer» Correct Answer - A `0.024g` or `24mg` |
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| 28823. |
In which of the following, first memeber is more stable than second?A. `(C_(6)H_(5))_(2)overset(**)(C)CH_(3),(C_(6)H_(5))_(3)overset(**)C`B. `:CH_(2),*CH_(2)*`C. `(C_(6)H_(5))_(3)overset(+)C,overset(+)CH_(3)`D. `(C_(2)H_(5))_(3),overset(**)C,(CH_(3))_(3)overset(**)C` |
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Answer» Correct Answer - C `(C_(6)H_(5))_(3)overset(-)C` is more stable than `overset(-)CH_(3)` because three phenyl groups not only decrease the electron density but also stabilise it by the phenomena of resonance. |
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| 28824. |
Ammonia changes the moist red litmus paper into-(A) Blue(B) Green(C) Black(D) White |
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Answer» Answer is (A) Blue |
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| 28825. |
How will you distinguish between: (i). Propan-2-ol and propan-1-ol (ii). Phenol and ethanol |
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Answer» (i) propan-2-ol when treated with iodine and alkali gives yellow precipitate of iodoform whereas propan -1- ol does not gives iodoform test. (ii) Only phenol gives violet colour with FeCl3 solution. 3C6H5OH + FeCl3 (C6H3)3Fe + 3HCl |
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| 28826. |
(I) Mention one important application of Henry's law. (II) Calculate the osmotic pressure of a solution obtained by mixing 100 cm3 of 0.25m and urea 100 cm3 of 0.1 m solution of cane sugar [R= 0.0821 L atm/mol/K] |
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Answer» (i) soft drink and soda water packing under high pressure (ii) Number of mole of cane sugar = \(\frac{0.1\times100}{1000}\) = 0.01 mol Total volume, V = (100 + 100) cm3 = 200 cm3 = 0.2L For osmotic pressure, \(\pi=\frac{(n_1+n_2)}{v}RT\) = \(\frac{(0.025+0.01)}{0.2}\) x 0.082 x 293 = \(\frac{0.035}{2}\) × 0.082 × 293 = 4.2 atm |
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| 28827. |
When NaCl solution is added to Fe(OH)3 colloidal solution then-(a) [Fe(OH3)]Fe3+ is formed(b) [Fe(OH3)]Cl-(c) [Fe(OH3)]Na+ is formed(d) Fe(OH)3 is coagulated |
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Answer» Answer is (a) [Fe(OH3)]Fe3+ is formed |
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| 28828. |
What will be the van't hoff factor for a solution of K4[FeCN6] if its dissociation is 50% |
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Answer» Degree of dissociation, x=50%=0.5 = (1+4x)/1 i=1+4(0.5)=1+2=3 |
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| 28829. |
Strips of zinc, iron, magnesium and copper are taken in the test tubes \( A, B, C \) and \( D \) respectively. Same quantity of ferrous sulphate solution is added to these test tubes. In which test tubes chemical reaction will occur? Why? Write the chemical equations for the reactions taking place here. |
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Answer» According to the metal reactivity series- zinc, magnesium metals are highly reactive metals and iron metal is also move reactive than copper metal. Highly reactive metal can displace les reactive metals from their solutions. Order of reactivity- Mg > Zn > Fe > Cu Therefore- In test tube 'A & B' chemical reaction occurs- In test tube -A Zn(5) + FeSO4(aq) → Zn SO4(aq) + Fe(s) In test tube - B Mg(s) + FeSO4 (aq) → MgSO4(aq) + Fe(s) In test tube - C & D → No chemical reactions. |
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| 28830. |
On the basis of crystal field theory the write the electronic configuration of d4 in terms of t2g and eg in an octahedral field , when (I) Δ0 > P (II) Δ0 < P |
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Answer» If ∆o > P, it becomes more energetically favourable for the fourth electron to occupy a t2g orbital with configuration t2g4 eg0 . Ligands which produce this effect are known as strong field ligands and form low spin complexes. If ∆o < P. Ligands which produce this effect are known as strong field ligands and form low spin complexes. In this case the configuration is t2g3 eg1 |
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| 28831. |
(i). why is adsorption always exothermic? (ii). Physisorption is multi layered while chemisorption is mono layered. |
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Answer» (i) The particles are attached to the surface of another medium for the reaction to take place. During this process the residual forces on the surface of adsorbent decreases which in turn decreases the surface energy. This energy appears in the form of heat, hence adsorption is exothermic in nature. (ii) Chemisorption takes place as a result of the reaction between adsorbent and adsorbate. When the surface of the adsorbent is covered with one layer, no further reaction can take place. Physisorption is simply by Vander Wall’s forces. So any number of layers may be formed one over the other on the surface of the adsorbent. |
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| 28832. |
(i). copper iron is not stable in aqueous solution(ii). Many of the transition metals shows maximum paramagnetic character amongst the bivalent ions of 1st transition series. |
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Answer» (i) Cu+ ion in an aqueous solution is unstable. It disproportionate to give Cu2+ and Cu. (ii) Many of the transition metals shows maximum paramagnetic character amongst the bivalent ions of 1st transition series due to maximum number of unpaired electrons. |
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| 28833. |
Explain the following (i). With the same d orbital configuration (d4) before Cr2+ is a reducing agent while Mn3+ is an oxidizing agent (ii). Many of the transition metals are known to form interstitial compounds |
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Answer» (i) Cr2+ has the configuration 3d4 . It can lose electron to form Cr3+ which has stable 3d3 configuration. Hence it is reducing agent. On the other hand, Mn3+ also has 3d4 configuration but it can gain electron to form Mn2+ which has stable 3d5 configuration (as it is exactly half - filled). Hence it is oxidising agent. (ii) Many of the transition elements are known to form interstitial compounds due to presence of unpaired electrons in the d-orbital. |
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| 28834. |
Give reasons:a). thermal stability decreases from H2O to H2Te b). fluoride ion has higher hydration enthalpy than chloride ion |
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Answer» (a) Thermal stability decreases from H2O to H2Te. The thermal stability of the hydrides of group 16 elements decreases down the group. This is because M-H bond dissociation energy decreases down the group with the increase in the size of a central atom. (b) The size of halogen decreases down the group, therefore hydration enthalpy decrease. |
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| 28835. |
Corbonyl compound (p) `underset((2). H_(2)O)overset((1). "grignard reagent")toPh-CH_(2)-underset(OH)underset(|)overset(H)overset(|)C-CH_(3)` Carbonyl compound can be:A. `Ph-CH_(2)-underset(O)underset(||)C-CH_(3)`B. `Ph-CH_(2)-CH=O`C. `CH_(3)-CH_(2)-CH=O`D. `CH_(3)-underset(O)underset(||)C-CH_(3)` |
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Answer» Correct Answer - B `Ph-CH_(2)-CH-=Ounderset((2). H_(2)O)overset((1). CH_(3)MgBr)toPh-CH_(2)-underset(OH)underset(|)overset(H)overset(|)C-CH_(3)` |
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| 28836. |
In which of the following reaction incorrect product is given?A. `PhMgBroverset(O_(2))tooverset(H^(+))toPh-OH`B. `CH_(3)-CH_(2)-MgBroverset(CO_(2))tooverset(H^(+))toCH_(3)-CH_(2)-COOH`C. `CH_(3)-CH=Ooverset(PhMgBr)tooverset(H^(+))toCH_(3)-underset(OH)underset(|)CH-Ph`D. `CH_(3)-C-=Noverset(PhMgBr)tooverset(H_(2)O)toCH_(3)-underset(OH)underset(|)CH-Ph` |
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Answer» Correct Answer - D `CH_(3)-C-=Noverset(PhMgBr)tooverset(H_(2)O)toCH_(3)-underset(O)underset(||)C-Ph` |
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| 28837. |
The metal extracted by leaching with cyanide isA. MgB. AgC. CuD. Na |
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Answer» Correct Answer - B Silver ore forms a soluble complex when leached with NaCN solution and from which silver is precipitated using scrap zinc. `Ag_(2)S("conc. Ore")+2NaCNoverset(Air)iff2AgCN+Na_(2)S` `4Na_(2)S+5O_(2)+2H_(2)Oto2Na_(2)SO_(4)+4NaOH+2S` `Na_(2)S` is converted in to `Na_(2)SO_(4)` Hence equilibrium shifts towards right side. `AgCN+NaCNtoNa[Ag(CN)_(2)]` (soluble complex) |
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| 28838. |
Reagent for the above reaction will be:A. hot acidic `KMnO_(4)`B. `CrO_(3),H^(+)`C. `CrO_(3)`, Pyridine, `CH_(2)Cl_(2)`D. dil alkaline `KMnO_(4)` |
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Answer» Correct Answer - C Mild oxidizing agent like PCC can oxidize only alcohols into carbonyl compounds. |
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| 28839. |
`NH_3` is heated initially at 15 atm from `27^@C` to `127^@C` at constant volume.At `127^@C` equilibrium is established.The new pressure at equilibrium at `127^@C` becomes 30 atm for the reaction `2NH_3(g)hArrN_2(g)+3H_2(g)` Then find the % of moles of `NH_3` actually decomposed. |
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Answer» Correct Answer - 50 `{:(2NH_3hArr,N_2+,3H_2),(a,0,0),(a-2x,x,3x):}` `P_1/T_1=P_2/T_2implies15/300=P_2/400implies P_2=20` atm Now `(a+2x)/a=30/20` 2a+4x=3a `implies x=1/4a " " :.` % of `NH_3` decomposed =`(2x)/axx100=50%` |
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| 28840. |
Match the following : `{:("Column I(Assume only reactant were present initially)","Column II"),((A)"For the equilibrium" NH_4I(s)hArr NH_3(g)+HI(g)",""If pressure is increased at equilibrium",(p)"Forward shift"),((B)"For the equilibrium" N_2(g)+3H_2(g)hArr 2NH_3(g)"volume is increased at equilibrium",(q)"No shift in equilibrium"),((C )"For the equilibrium" H_2O(g)+CO(g)hArr H_2(g)+CO_2(g)"inert gas is added at constant pressure at equilibrium",(r)"Backward shift"),((D)"For the equilibrium" PCl_5hArr PCl_3+Cl_2 Cl_2"is removed at equilibrium",(s)"Final pressure is more than initial pressure "):}` |
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Answer» Correct Answer - A-r ; B-r ; C-q ; D-p (A)`Deltan_(g)` is +ve so as P is increased, backward shifting will take place.Total pressure even after shifting will remain same. (B)`Deltan_g` is -ve so as V is increased, backward shifting will take place. But `P_("final")ltP_("Initial")` ( C)No change but `P_("final")ltP_("Initial")` as volume has increased. (D)Forward shifting will take place and `P_("final")ltP_("Initial")` |
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| 28841. |
For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula.The observed densities decreases towards a limit as the temperature is raised.This is due to the splitting of the molecular into simpler ones.The process is reversible and is called thermal dissociation. Example : `NH_4Cl hArr NH_3+HCl I_2hArr 2I` `N_2O_4 hArr 2NO_2 PCl_5hArr PCl_3+Cl_2` With increase in the number of molecules, the volume increases (pressure remaining constant) and in consequence, the density decreases.As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occurs the density reaches its lowest limit. The extent of dissociation, i.e., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation.Gas density measurements can be used to determine the degree of dissociation.Let us take by general case where one molecule of a substance A splits up into n molecule of A on heating , i.e., `A_n(g)hArr nA(g)` `t=0 " " a" " 0` `t=t_(eq) a-x n.x alpha=x/a rArr x=a alpha.` `a-a alpha naalpha` Total no. of moles =`a-a alpha+n a alpha` =`[1+(n-1)alpha]a` Observed molecular weight or molar mass of the mixture `M_("mixture")=M_(A_(n))/([1+(n-1)alpha]), M_(A_(n))`=Molar mass of gas `A_n` The `K_P` for the reaction `N_2O_4 hArr 2NO_2` is 640 mm at 775 K.The percentage dissociation of `N_2O_4` at equilibrium pressure of 160 mm is :A. `80%`B. `30%`C. `50%`D. `70%` |
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Answer» Correct Answer - D `{:(,N_2O_4" "hArr,2NO_2),("Mole before equilibrium",1,),("Mole at equilibrium",(1-x),2x):}` `K_p=(4x^2)/((1-x))xx[P/(sumn)]^(Deltan)implies 640=(4x^2)/((1-x))160/((1+x))` `4=(4x^2)/((1-x))` or `1-x^2=x^2` or `2x^2=1` `:. " " x^2=1//2 or x=0.707=70.7%` |
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| 28842. |
For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula.The observed densities decreases towards a limit as the temperature is raised.This is due to the splitting of the molecular into simpler ones.The process is reversible and is called thermal dissociation. Example : `NH_4Cl hArr NH_3+HCl I_2hArr 2I` `N_2O_4 hArr 2NO_2 PCl_5hArr PCl_3+Cl_2` With increase in the number of molecules, the volume increases (pressure remaining constant) and in consequence, the density decreases.As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occurs the density reaches its lowest limit. The extent of dissociation, i.e., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation.Gas density measurements can be used to determine the degree of dissociation.Let us take by general case where one molecule of a substance A splits up into n molecule of A on heating , i.e., `A_n(g)hArr nA(g)` `t=0" " a" " 0` `t=t_(eq) a-x n.x alpha=x/a rArr x=a alpha.` `a-a alpha naalpha` Total no. of moles =`a-a alpha+n a alpha` =`[1+(n-1)alpha]a` Observed molecular weight or molar mass of the mixture `M_("mixture")=M_(A_(n))/([1+(n-1)alpha]), M_(A_(n))`=Molar mass of gas `A_n` The equation `alpha=(D-d)/((n-1)d)` is correctly matched for :A. `A hArr nB//2+nC//3`B. `A hArr nB//3+(2n//3)C`C. `A to(n//2)B+(n//4)C`D. `A hArr (n//2)B+C` |
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Answer» Correct Answer - B `alpha=(D-d)/((n-1)d) implies (n-1)alpha=D/d-1` `implies D/d=1+(n-1)alpha` Hence one mole of reactant should produce total n moles of product |
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| 28843. |
For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula.The observed densities decreases towards a limit as the temperature is raised.This is due to the splitting of the molecular into simpler ones.The process is reversible and is called thermal dissociation. Example : `NH_4Cl hArr NH_3+HCl I_2hArr 2I` `N_2O_4 hArr 2NO_2 PCl_5hArr PCl_3+Cl_2` With increase in the number of molecules, the volume increases (pressure remaining constant) and in consequence, the density decreases.As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occurs the density reaches its lowest limit. The extent of dissociation, i.e., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation.Gas density measurements can be used to determine the degree of dissociation.Let us take by general case where one molecule of a substance A splits up into n molecule of A on heating , i.e., `A_n(g)hArr nA(g)` `t=0 " "a" " 0` `t=t_(eq) a-x n.x alpha=x/a rArr x=a alpha.` `a-a alpha naalpha` Total no. of moles =`a-a alpha+n a alpha` =`[1+(n-1)alpha]a` Observed molecular weight or molar mass of the mixture `M_("mixture")=M_(A_(n))/([1+(n-1)alpha]), M_(A_(n))`=Molar mass of gas `A_n` x(degree of dissociation) varies with `D/d` in the above reaction according to :A. B. C. D. |
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Answer» Correct Answer - B `D/d=1+(n-1)x` `implies D/d=1+x " " [n=2]` which is a line with positive slope and X intercept =1 |
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| 28844. |
The following reaction attains equlibrium at high temperature `N_(2)(g)+2H_(2)O(g)+heathArr2NO(g)+2H_(2)(g)` The concentration of NO(g)is affected byA. increasing the nitrogen concentrationB. decreasing the nitrogen concentrationC. Compressing the reaction mixtureD. None of these |
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Answer» Correct Answer - A,B,C As concentration of reaction (gaseous) are increased at equilibrium reaction will go in the forward direction.Also above reaction is endothermic therefore increase in temperature will favour it. |
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| 28845. |
For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula.The observed densities decreases towards a limit as the temperature is raised.This is due to the splitting of the molecular into simpler ones.The process is reversible and is called thermal dissociation. Example : `NH_4Cl hArr NH_3+HCl I_2hArr 2I` `N_2O_4 hArr 2NO_2 PCl_5hArr PCl_3+Cl_2` With increase in the number of molecules, the volume increases (pressure remaining constant) and in consequence, the density decreases.As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occurs the density reaches its lowest limit. The extent of dissociation, i.e., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation.Gas density measurements can be used to determine the degree of dissociation.Let us take by general case where one molecule of a substance A splits up into n molecule of A on heating , i.e., `A_n(g)hArr nA(g)` `t=0 " " a " " 0` `t=t_(eq) a-x n.x alpha=x/a rArr x=a alpha.` `a-a alpha naalpha` Total no. of moles =`a-a alpha+n a alpha` =`[1+(n-1)alpha]a` Observed molecular weight or molar mass of the mixture `M_("mixture")=M_(A_(n))/([1+(n-1)alpha]), M_(A_(n))`=Molar mass of gas `A_n` A sample of mixture of A(g),B(g) and C(g) under equilibrium has a mean molecular weight (observed ) is 80. The equilibrium is `underset((mol.wt.=100))(A(g))hArrunderset((mol.wt.=60))(B(g))+underset((mol.wt.=40))(C(g))` Find the degree of dissociation `alpha` for A(g)A. `0.25`B. `0.5`C. `0.75`D. `0.8` |
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Answer» Correct Answer - A `A(g) hArr B(g)+C(g)` `M_(Ob)=M_(th)/(M+(n-1)alpha)implies 80=100/(1+alpha)implies =1/4=0.25` |
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| 28846. |
Statement-1 :Graphite can be converted into diamond by application of very high pressure and temperature. Using a suitable catalyst. Statement-2: Graphite is thermodynamically more stable and less dense than diamond.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - A `C_(("graphite"))hArrC_(("diamond"))`(at high T and high P) (It is a fact) |
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| 28847. |
Name the elements present in anode mud during refining of copper. Why does it contain such elements ? |
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Answer» Au and Ag. They are not oxidized at anode. They are less electropositive than copper. |
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| 28848. |
What is the effect of adding catalyst on the free energy of a reaction ? |
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Answer» No change in AG. |
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| 28849. |
What are collectors in froth floatation process ? Give one example. |
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Answer» The correct answer " Pine oil. " |
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| 28850. |
Give the names and formulae of three ores which are concentrated by froth fioatation process. |
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Answer» Galena (PbS), Zinc blende (ZnS), Cinnabar (HgS). |
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