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Hybridisation of N in HNO_3 is

$$\mathrm{Hybridisation}\:\mathrm{of}\:\mathrm{N}\:\mathrm{in}\:\mathrm{HNO}_{\mathrm{3}} \:\mathrm{is} \\ $$

Question Number 23394 Answers: 1 Comments: 0

Which is correct statement? (1) The entropy of the universe increases and tends towards the maximum value (2) All natural processes are irreversible (3) For reversible isolated processes, change of entropy is zero (4) For irreversible expansion of isolated processes, entropy change < 0

$$\mathrm{Which}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{statement}? \\ $$$$\left(\mathrm{1}\right)\:\mathrm{The}\:\mathrm{entropy}\:\mathrm{of}\:\mathrm{the}\:\mathrm{universe}\:\mathrm{increases} \\ $$$$\mathrm{and}\:\mathrm{tends}\:\mathrm{towards}\:\mathrm{the}\:\mathrm{maximum} \\ $$$$\mathrm{value} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{All}\:\mathrm{natural}\:\mathrm{processes}\:\mathrm{are}\:\mathrm{irreversible} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{For}\:\mathrm{reversible}\:\mathrm{isolated}\:\mathrm{processes}, \\ $$$$\mathrm{change}\:\mathrm{of}\:\mathrm{entropy}\:\mathrm{is}\:\mathrm{zero} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{For}\:\mathrm{irreversible}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\mathrm{isolated}\:\mathrm{processes},\:\mathrm{entropy}\:\mathrm{change}\:<\:\mathrm{0} \\ $$

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Two blocks of masses 2 kg and 3 kg are kept on a smooth inclined plane. A constant force of magnitude 20 N is applied on 2 kg block parallel to the inclined. The contact force between the two blocks is

$$\mathrm{Two}\:\mathrm{blocks}\:\mathrm{of}\:\mathrm{masses}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{and}\:\mathrm{3}\:\mathrm{kg} \\ $$$$\mathrm{are}\:\mathrm{kept}\:\mathrm{on}\:\mathrm{a}\:\mathrm{smooth}\:\mathrm{inclined}\:\mathrm{plane}. \\ $$$$\mathrm{A}\:\mathrm{constant}\:\mathrm{force}\:\mathrm{of}\:\mathrm{magnitude}\:\mathrm{20}\:\mathrm{N}\:\mathrm{is} \\ $$$$\mathrm{applied}\:\mathrm{on}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{block}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{inclined}.\:\mathrm{The}\:\mathrm{contact}\:\mathrm{force}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{blocks}\:\mathrm{is} \\ $$

Question Number 23381 Answers: 1 Comments: 0

A women swimming upstream is not moving with respect to the ground. Is she doing any work, if she stops swimming and merely floats is work done on her?

$$\mathrm{A}\:\mathrm{women}\:\mathrm{swimming}\:\mathrm{upstream}\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{moving}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{Is} \\ $$$$\mathrm{she}\:\mathrm{doing}\:\mathrm{any}\:\mathrm{work},\:\mathrm{if}\:\mathrm{she}\:\mathrm{stops} \\ $$$$\mathrm{swimming}\:\mathrm{and}\:\mathrm{merely}\:\mathrm{floats}\:\mathrm{is}\:\mathrm{work} \\ $$$$\mathrm{done}\:\mathrm{on}\:\mathrm{her}? \\ $$

Question Number 23369 Answers: 1 Comments: 0

Which of the following pair would correct inequality for standard molar entropy? (1) NO(g) < NO_2 (g) (2) C_2 H_2 (g) > C_2 H_6 (g) (3) CH_3 COOH (l) < HCOOH (l) (4) CO_2 (g) < CO(g)

$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{pair}\:\mathrm{would} \\ $$$$\mathrm{correct}\:\mathrm{inequality}\:\mathrm{for}\:\mathrm{standard}\:\mathrm{molar} \\ $$$$\mathrm{entropy}? \\ $$$$\left(\mathrm{1}\right)\:\mathrm{NO}\left(\mathrm{g}\right)\:<\:\mathrm{NO}_{\mathrm{2}} \left(\mathrm{g}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{C}_{\mathrm{2}} \mathrm{H}_{\mathrm{2}} \left(\mathrm{g}\right)\:>\:\mathrm{C}_{\mathrm{2}} \mathrm{H}_{\mathrm{6}} \left(\mathrm{g}\right) \\ $$$$\left(\mathrm{3}\right)\:\mathrm{CH}_{\mathrm{3}} \mathrm{COOH}\:\left(\mathrm{l}\right)\:<\:\mathrm{HCOOH}\:\left(\mathrm{l}\right) \\ $$$$\left(\mathrm{4}\right)\:\mathrm{CO}_{\mathrm{2}} \left(\mathrm{g}\right)\:<\:\mathrm{CO}\left(\mathrm{g}\right) \\ $$

Question Number 23368 Answers: 0 Comments: 0

Assertion: The first ionization energy of Be is greater than that of B. Reason: 2p-orbital is lower in energy than 2s-orbital.

$$\boldsymbol{\mathrm{Assertion}}:\:\mathrm{The}\:\mathrm{first}\:\mathrm{ionization}\:\mathrm{energy} \\ $$$$\mathrm{of}\:\mathrm{Be}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{that}\:\mathrm{of}\:\mathrm{B}. \\ $$$$\boldsymbol{\mathrm{Reason}}:\:\mathrm{2}{p}-\mathrm{orbital}\:\mathrm{is}\:\mathrm{lower}\:\mathrm{in}\:\mathrm{energy} \\ $$$$\mathrm{than}\:\mathrm{2}{s}-\mathrm{orbital}. \\ $$

Question Number 23361 Answers: 1 Comments: 0

A 50 g lead bullet, sp. heat 0.02 is initially at 30°C. It is fired vertically upwards with a speed of 840 m/s and on returning to the starting level strikes a cake of ice at 0°C. How much ice is melted? Assume that all energy is spent in melting only (L = 80 cal/g).

$$\mathrm{A}\:\mathrm{50}\:\mathrm{g}\:\mathrm{lead}\:\mathrm{bullet},\:\mathrm{sp}.\:\mathrm{heat}\:\mathrm{0}.\mathrm{02}\:\mathrm{is} \\ $$$$\mathrm{initially}\:\mathrm{at}\:\mathrm{30}°\mathrm{C}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{fired}\:\mathrm{vertically} \\ $$$$\mathrm{upwards}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{840}\:\mathrm{m}/\mathrm{s}\:\mathrm{and} \\ $$$$\mathrm{on}\:\mathrm{returning}\:\mathrm{to}\:\mathrm{the}\:\mathrm{starting}\:\mathrm{level}\:\mathrm{strikes} \\ $$$$\mathrm{a}\:\mathrm{cake}\:\mathrm{of}\:\mathrm{ice}\:\mathrm{at}\:\mathrm{0}°\mathrm{C}.\:\mathrm{How}\:\mathrm{much}\:\mathrm{ice}\:\mathrm{is} \\ $$$$\mathrm{melted}?\:\mathrm{Assume}\:\mathrm{that}\:\mathrm{all}\:\mathrm{energy}\:\mathrm{is} \\ $$$$\mathrm{spent}\:\mathrm{in}\:\mathrm{melting}\:\mathrm{only}\:\left({L}\:=\:\mathrm{80}\:\mathrm{cal}/\mathrm{g}\right). \\ $$

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A uniform chain of mass M and length L is hanging from the table. The chain is in limiting equilibrium when l length of chain over hangs. It is slightly disturbed from this position. Find the speed of the chain just after it completely comes off the table.

$$\mathrm{A}\:\mathrm{uniform}\:\mathrm{chain}\:\mathrm{of}\:\mathrm{mass}\:{M}\:\mathrm{and}\:\mathrm{length} \\ $$$${L}\:\mathrm{is}\:\mathrm{hanging}\:\mathrm{from}\:\mathrm{the}\:\mathrm{table}.\:\mathrm{The}\:\mathrm{chain} \\ $$$$\mathrm{is}\:\mathrm{in}\:\mathrm{limiting}\:\mathrm{equilibrium}\:\mathrm{when}\:{l}\:\mathrm{length} \\ $$$$\mathrm{of}\:\mathrm{chain}\:\mathrm{over}\:\mathrm{hangs}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{slightly} \\ $$$$\mathrm{disturbed}\:\mathrm{from}\:\mathrm{this}\:\mathrm{position}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{chain}\:\mathrm{just}\:\mathrm{after}\:\mathrm{it}\:\mathrm{completely} \\ $$$$\mathrm{comes}\:\mathrm{off}\:\mathrm{the}\:\mathrm{table}. \\ $$

Question Number 23274 Answers: 0 Comments: 2

Assertion: Both N_2 and NO^+ are diamagnetic substances. Reason: NO^+ is isoelectronic with N_2 .

$$\boldsymbol{\mathrm{Assertion}}:\:\mathrm{Both}\:\mathrm{N}_{\mathrm{2}} \:\mathrm{and}\:\mathrm{NO}^{+} \:\mathrm{are} \\ $$$$\mathrm{diamagnetic}\:\mathrm{substances}. \\ $$$$\boldsymbol{\mathrm{Reason}}:\:\mathrm{NO}^{+} \:\mathrm{is}\:\mathrm{isoelectronic}\:\mathrm{with}\:\mathrm{N}_{\mathrm{2}} . \\ $$

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The reaction CH_4 (g) + Cl_2 (g) → CH_3 Cl(g) + HCl(g) has ΔH = −25 kcal and bond dissociation energy of C − Cl, H − Cl, C − H and Cl − Cl is given as 84 kcal/mol, 103 kcal/mol, x kcal/mol and y kcal/mol respectively. Given x : y = 9 : 5, then bond energy of Cl − Cl bond in kcal/mol is

$$\mathrm{The}\:\mathrm{reaction}\:\mathrm{CH}_{\mathrm{4}} \left(\mathrm{g}\right)\:+\:\mathrm{Cl}_{\mathrm{2}} \left(\mathrm{g}\right)\:\rightarrow\:\mathrm{CH}_{\mathrm{3}} \mathrm{Cl}\left(\mathrm{g}\right) \\ $$$$+\:\mathrm{HCl}\left(\mathrm{g}\right)\:\mathrm{has}\:\Delta\mathrm{H}\:=\:−\mathrm{25}\:\mathrm{kcal}\:\mathrm{and}\:\mathrm{bond} \\ $$$$\mathrm{dissociation}\:\mathrm{energy}\:\mathrm{of}\:\mathrm{C}\:−\:\mathrm{Cl},\:\mathrm{H}\:−\:\mathrm{Cl}, \\ $$$$\mathrm{C}\:−\:\mathrm{H}\:\mathrm{and}\:\mathrm{Cl}\:−\:\mathrm{Cl}\:\mathrm{is}\:\mathrm{given}\:\mathrm{as}\:\mathrm{84}\:\mathrm{kcal}/\mathrm{mol}, \\ $$$$\mathrm{103}\:\mathrm{kcal}/\mathrm{mol},\:\mathrm{x}\:\mathrm{kcal}/\mathrm{mol}\:\mathrm{and}\:\mathrm{y}\:\mathrm{kcal}/\mathrm{mol} \\ $$$$\mathrm{respectively}.\:\mathrm{Given}\:\mathrm{x}\::\:\mathrm{y}\:=\:\mathrm{9}\::\:\mathrm{5},\:\mathrm{then} \\ $$$$\mathrm{bond}\:\mathrm{energy}\:\mathrm{of}\:\mathrm{Cl}\:−\:\mathrm{Cl}\:\mathrm{bond}\:\mathrm{in}\:\mathrm{kcal}/\mathrm{mol} \\ $$$$\mathrm{is} \\ $$

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Consider the system shown in the figure. Initially the system was in rest. (i) Find the acceleration of block if man climbs the rod with acceleration a (w.r.t. rod) (ii) If the man climb to the top of the rod then find the distance moved by the block.

$$\mathrm{Consider}\:\mathrm{the}\:\mathrm{system}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{figure}.\:\mathrm{Initially}\:\mathrm{the}\:\mathrm{system}\:\mathrm{was}\:\mathrm{in}\:\mathrm{rest}. \\ $$$$\left(\mathrm{i}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{block}\:\mathrm{if}\:\mathrm{man} \\ $$$$\mathrm{climbs}\:\mathrm{the}\:\mathrm{rod}\:\mathrm{with}\:\mathrm{acceleration}\:{a}\:\left(\mathrm{w}.\mathrm{r}.\mathrm{t}.\right. \\ $$$$\left.\mathrm{rod}\right) \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{If}\:\mathrm{the}\:\mathrm{man}\:\mathrm{climb}\:\mathrm{to}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{rod}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{moved}\:\mathrm{by}\:\mathrm{the} \\ $$$$\mathrm{block}. \\ $$

Question Number 23205 Answers: 0 Comments: 0

thank u.....

$${thank}\:{u}..... \\ $$

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plz anyone answer the question 23181...plz plz plz...

$${plz}\:\:\:{anyone}\:{answer}\:{the}\:{question}\: \\ $$$$\mathrm{23181}...{plz}\:{plz}\:{plz}... \\ $$

Question Number 23181 Answers: 0 Comments: 2

show that the curve with parametric equcations x=t^2 −3t+5, y=t^3 +t^2 −10t+9 intersect at the point (3,1).

$${show}\:{that}\:{the}\:{curve}\:{with}\:{parametric}\: \\ $$$${equcations}\:\:{x}={t}^{\mathrm{2}} \:−\mathrm{3}{t}+\mathrm{5}, \\ $$$${y}={t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} \:−\mathrm{10}{t}+\mathrm{9}\:{intersect}\:{at}\:{the}\: \\ $$$${point}\:\left(\mathrm{3},\mathrm{1}\right). \\ $$

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Assertion: The element with electronic configuration [Xe]^(54) 4f^1 5d^1 6s^2 is a d- block element. Reason: The last electron enters the d- orbital.

$$\boldsymbol{\mathrm{Assertion}}:\:\mathrm{The}\:\mathrm{element}\:\mathrm{with}\:\mathrm{electronic} \\ $$$$\mathrm{configuration}\:\left[\mathrm{Xe}\right]^{\mathrm{54}} \:\mathrm{4}{f}^{\mathrm{1}} \:\mathrm{5}{d}^{\mathrm{1}} \:\mathrm{6}{s}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{a}\:{d}- \\ $$$$\mathrm{block}\:\mathrm{element}. \\ $$$$\boldsymbol{\mathrm{Reason}}:\:\mathrm{The}\:\mathrm{last}\:\mathrm{electron}\:\mathrm{enters}\:\mathrm{the}\:{d}- \\ $$$$\mathrm{orbital}. \\ $$

Question Number 23146 Answers: 0 Comments: 5

Square planar complex is formed by hybridisation of which atomic orbitals? (1) s, p_x , p_y , p_z (2) s, p_x , p_y , d_z^2 (3) s, p_x , p_y , d_(x^2 −y^2 ) (4) s, p_x , p_y , d_z^3

$$\mathrm{Square}\:\mathrm{planar}\:\mathrm{complex}\:\mathrm{is}\:\mathrm{formed}\:\mathrm{by} \\ $$$$\mathrm{hybridisation}\:\mathrm{of}\:\mathrm{which}\:\mathrm{atomic}\:\mathrm{orbitals}? \\ $$$$\left(\mathrm{1}\right)\:{s},\:{p}_{{x}} ,\:{p}_{{y}} ,\:{p}_{{z}} \\ $$$$\left(\mathrm{2}\right)\:{s},\:{p}_{{x}} ,\:{p}_{{y}} ,\:{d}_{{z}^{\mathrm{2}} } \\ $$$$\left(\mathrm{3}\right)\:{s},\:{p}_{{x}} ,\:{p}_{{y}} ,\:{d}_{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:{s},\:{p}_{{x}} ,\:{p}_{{y}} ,\:{d}_{{z}^{\mathrm{3}} } \\ $$

Question Number 23135 Answers: 1 Comments: 0

The standard enthalpy of formation of gaseous H_2 O at 298 K is −241.82 kJ mol^(−1) . Estimate its value of 100°C given the following values of the molar heat capacities at constant pressure: H_2 O(g) : 35.58 JK^(−1) mol^(−1) , H_2 (g) : 28.84 J mol^(−1) K^(−1) and O_2 (g) : 29.37 J mol^(−1) K^(−1) . Assume heat capacity to be independent of temperature.

$$\mathrm{The}\:\mathrm{standard}\:\mathrm{enthalpy}\:\mathrm{of}\:\mathrm{formation}\:\mathrm{of} \\ $$$$\mathrm{gaseous}\:\mathrm{H}_{\mathrm{2}} \mathrm{O}\:\mathrm{at}\:\mathrm{298}\:\mathrm{K}\:\mathrm{is}\:−\mathrm{241}.\mathrm{82}\:\mathrm{kJ} \\ $$$$\mathrm{mol}^{−\mathrm{1}} .\:\mathrm{Estimate}\:\mathrm{its}\:\mathrm{value}\:\mathrm{of}\:\mathrm{100}°\mathrm{C}\:\mathrm{given} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{values}\:\mathrm{of}\:\mathrm{the}\:\mathrm{molar}\:\mathrm{heat} \\ $$$$\mathrm{capacities}\:\mathrm{at}\:\mathrm{constant}\:\mathrm{pressure}: \\ $$$$\mathrm{H}_{\mathrm{2}} \mathrm{O}\left(\mathrm{g}\right)\::\:\mathrm{35}.\mathrm{58}\:\mathrm{JK}^{−\mathrm{1}} \:\mathrm{mol}^{−\mathrm{1}} ,\:\mathrm{H}_{\mathrm{2}} \left(\mathrm{g}\right)\:: \\ $$$$\mathrm{28}.\mathrm{84}\:\mathrm{J}\:\mathrm{mol}^{−\mathrm{1}} \:\mathrm{K}^{−\mathrm{1}} \:\mathrm{and}\:\mathrm{O}_{\mathrm{2}} \left(\mathrm{g}\right)\::\:\mathrm{29}.\mathrm{37}\:\mathrm{J} \\ $$$$\mathrm{mol}^{−\mathrm{1}} \:\mathrm{K}^{−\mathrm{1}} .\:\mathrm{Assume}\:\mathrm{heat}\:\mathrm{capacity}\:\mathrm{to}\:\mathrm{be} \\ $$$$\mathrm{independent}\:\mathrm{of}\:\mathrm{temperature}. \\ $$

Question Number 23130 Answers: 0 Comments: 1

A baloon filled with helium rises against gravity increasing its potential energy. The speed of the baloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that density of air is constant.

$$\mathrm{A}\:\mathrm{baloon}\:\mathrm{filled}\:\mathrm{with}\:\mathrm{helium}\:\mathrm{rises}\:\mathrm{against} \\ $$$$\mathrm{gravity}\:\mathrm{increasing}\:\mathrm{its}\:\mathrm{potential}\:\mathrm{energy}. \\ $$$$\mathrm{The}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{baloon}\:\mathrm{also}\:\mathrm{increases} \\ $$$$\mathrm{as}\:\mathrm{it}\:\mathrm{rises}.\:\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{reconcile}\:\mathrm{this} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{law}\:\mathrm{of}\:\mathrm{conservation}\:\mathrm{of} \\ $$$$\mathrm{mechanical}\:\mathrm{energy}?\:\mathrm{You}\:\mathrm{can}\:\mathrm{neglect} \\ $$$$\mathrm{viscous}\:\mathrm{drag}\:\mathrm{of}\:\mathrm{air}\:\mathrm{and}\:\mathrm{assume}\:\mathrm{that} \\ $$$$\mathrm{density}\:\mathrm{of}\:\mathrm{air}\:\mathrm{is}\:\mathrm{constant}. \\ $$

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